密码学百题计划启动！进行勇往直前！

image-20210628200701835

本次的16道题目，也是非常有意思的呢！各种各样的呢……

image-20210628200448759

0x0 浪里淘沙

看样子，应该是个脑洞题没错了

来康康这题的题目：

我有密集恐惧症，所以大家自求多福吧，把获得的单词连在一起提交即可。（我这里有一串数字：4，8，11，15，16） 注意：得到的 flag 请包上 flag{} 提交

这题目好像有点意思，可以从题目中提炼一些信息：

  1. 密集恐惧症
  2. 获得的单词连在一起提交
  3. 我这里有一串数字：4，8，11，15，16

根据提炼的数据进行分享，应该是词频统计类的题目，这题有点意思

看看附件是啥吧：

tonightsuccessnoticenoticewewesuccesstonightweexamplecryptoshouldwebackspacetonightbackspaceexamplelearnwesublimlearnbackspacetheshouldwelearnfoundsublimsystemexamplesublimfoundlearnshouldmorningsublimsystemuserlearnthecryptomorningexamplenoticetonightlearntonightlearntonightsublimenterusermorningfoundtonightweenterfoundnoticethecryptomorningthebackspacelearntonightlearnsublimtonightlearnfoundenterfoundsuccesstonightsuccessuserfoundmorningtonighttheshouldsublimentertonightenterbackspacelearnexamplenoticeexamplefoundsystemsuccesssublimsuccessshouldtonightcryptowelearncryptofoundshouldsublimsublimweentertonightsuccessshouldentertheentercryptouserbackspaceshouldentersystemsuccesssystementerfoundenterlearnexampletonightnoticemorningusertonightlearnmorningtonightfoundfoundsuccessnoticesystementerlearnexamplebackspaceshouldcryptocryptosublimweexampletonighttheshouldthemorningbackspacelearntonightsystemsuccesssuccessbackspacemorningnoticeuserfoundfoundtonightmorningenterenterthefoundbackspacelearnenterentershouldthesystemfounduserlearnlearnsystemnoticetonighttheshouldlearnuserbackspaceweusernoticeshouldthewefoundsystemwecryptocryptowethebackspacesystementershouldtonightsystemnoticemorningsystemweentermorningfoundsuccessusertonightsuccesstonightbackspaceshouldweenterthewesystemusernoticesystemthelearnexamplelearnfoundlearnnoticeexamplesystemthecryptocryptolearnsystemthecryptoenterlearnexamplemorningmorningweenterentersuccessexampleuserthebackspacenoticesublimenterbackspacesuccessbackspacethesublimexamplesystemtheexamplecryptolearnuserexamplelearnsystemusersuccessenterentersuccesstheuserbackspacelearnsuccessbackspacethesublimshouldwebackspaceexamplesuccesssuccesstonightweusershouldsuccessmorningcryptomorningfoundbackspacesublimshouldentershouldnoticesuccessmorningsuccessexamplelearnshouldsublimlearntonightshoulduserbackspacesublimlearncryptosuccessenternoticetonightmorningtonightwesuccessweuserbackspaceexamplewesystemnoticemorningsystemmorningcryptolearnsystemthethefoundcryptouserlearnusersystemwemorningenterexampleshouldlearncryptofoundenterbackspacelearnenterenterbackspaceshouldbackspacetheshouldthesystemshouldshouldsuccessmorningthefoundsystementersystemtonightcryptowelearnexampleexamplesystementerbackspaceshouldtheentersublimtonightfoundfoundsuccesssuccesssystemsublimcryptoshouldentersublimmorninglearnfoundtonightcryptobackspacesuccesscryptowebackspacefoundshouldnoticeshouldmorningnoticesystemcryptosystemlearnsystemnoticemorningsystementerwemorninglearnsuccessfoundwesuccesswetheusercryptousernoticebackspacesuccessshouldtonightmorningentermorninguserenternoticefoundmorningwetonightsystemthecryptotonightcryptosystemuserthefoundexampletonightusersystemcryptosublimmorninguserthefoundbackspaceshouldsuccesscryptotonightsystemnoticebackspaceusershouldenterthecryptomorningwesublimnoticesuccessnoticeusersuccesstonightlearnweuserenterfounduserexampleshouldshouldtonightwelearnthenoticethewefoundmorningexampleshouldexamplethesuccessnoticeenterfoundthecryptonoticeuserlearnuserweenterfoundmorningsystemweexamplenoticethebackspaceexamplesublimtheusermorningtonightthesuccesscryptosuccessusersuccesstonighttonightwelearnenterenterthemorningentersystemcryptobackspacemorningsystemexamplecryptouserexamplelearnsublimsuccessusersystemfoundmorningshouldcryptotonightsublimtheexamplemorningsystemuserexampleweexamplenoticesuccesssublimnoticecryptoshouldbackspaceshouldthetonightfoundsublimbackspacebackspacetonightshouldbackspacesuccesstonightbackspacesuccessmorningsystemcryptobackspaceentertonighttonightnoticelearnshoulduserfoundexamplesystemthesuccessweusertonightcryptousernoticeenternoticebackspaceusersystemfoundusernoticeshouldlearnuserfoundexampleusermorningshouldsuccessmorningmorningexampleexamplefoundsublimfoundenterbackspacenoticelearnfoundmorningcryptonoticecryptoshouldweshouldtonightcryptobackspacesublimcryptosublimenterentersublimentercryptonoticethethesublimexampleenterentershouldlearncryptoentershouldmorninglearnnoticeuserexamplesublimtonightshouldfoundtonightsuccessshouldmorningfoundtheweuserlearnsublimsystembackspacecryptotheusertonightcryptosublimmorningmorningexamplenoticetheenterlearnshouldmorningsublimfoundtonightsublimsublimexamplefounduserexamplethefoundwemorningnoticefoundcryptosuccesssublimsublimexamplethesuccessexamplenoticesuccessbackspacesublimlearnuserexamplesuccesssuccesssystemsuccessmorningmorninglearnexamplemorningtonightfoundbackspaceenternoticemorningentersuccessmorningusermorningbackspacelearncryptoenteruserenteruserthetonighttonightsuccesslearnenterfoundsuccesssystemfoundbackspaceenterlearnsystemsublimcryptoentermorningwetonightshouldlearnenterfoundcryptonoticelearnlearnshouldfoundsuccessexampletonightthesuccessfoundusertonightenterfoundsuccessshouldmorningusernoticemorningsystemsystemsuccessshouldwelearnenterfoundexamplewethefoundweshouldsystemsystemmorningmorningbackspaceshouldentersublimentertonightsuccesssystemsystemcryptousershouldsublimfoundwetonightnoticeexamplewewesuccessfoundusertonightfoundsystemexamplecryptofoundshouldshouldsuccessenterbackspaceexampletonightthelearnnoticeuserlearnsystemsublimfoundlearnsuccesssystemshouldsublimnoticelearnsystemnoticetonightexamplefoundusernoticeenterlearnnoticecryptousersystemmorningthewesystemfoundfoundshouldsystementerenterbackspacesystemsublimcryptousermorninglearnlearntonightsublimlearnenterenterbackspacesystemuserusercryptoentershouldtheusersublimnoticeexamplemorningexamplesublimsublimbackspacesystemexampleshouldsublimlearnfoundenterbackspacelearnmorningmorningfoundthetonightmorningnoticeenterlearnusersystemtonightbackspaceexamplelearntonightbackspaceweshouldcryptosuccessbackspaceexamplesuccesstheshouldmorninguserbackspacelearnthetheshouldcryptocryptotonightbackspacecryptocryptobackspacebackspacenoticeusertonightentermorningfoundweenterexampleenterfoundusersublimsystemtheexampleexamplesystemsuccessusersublimentermorningbackspacesystemfoundlearnsystemshouldsublimsublimentershouldtheusershouldexampleexampleshouldsuccesswelearnfoundsublimshoulduserweentertonightwenoticesublimsystemlearnshouldfoundsuccessuserentersuccessmorningcryptoenteruserfoundexampletonightlearnexampleexamplefoundlearnsuccesssystembackspacecryptonoticethefoundbackspacelearncryptothelearnlearnexamplesuccessnoticenoticesystemmorningcryptotonightnoticenoticeentersuccesscryptoenterbackspacesublimexampleenterfoundtonightcryptotonightsublimnoticesuccesssublimtheentertonighttheshouldthefoundsystemtonightuserbackspacesuccessshouldwebackspacenoticebackspacebackspacenoticewecryptobackspacebackspaceusertonightlearnsuccessmorningusertonightsuccessshouldbackspacecryptoenterentershouldsublimsystemexamplemorningcryptonoticethesuccessthebackspacenoticelearnsublimlearnsuccesscryptothesuccessenternoticecryptosystemsublimsuccesswebackspaceuserenterlearnuserwewemorningsuccesslearncryptobackspacewecryptosystemlearnenterenteruserexamplefoundsystemcryptousernoticefoundusersublimbackspacewesublimnoticemorningshouldexamplenoticecryptoshouldtonightmorningthefoundsystementerentersystemthecryptobackspacesublimlearnsuccessmorningsublimsystemcryptousersublimwesuccessmorningsublimbackspacecryptobackspacesublimthelearnsuccesssublimlearncryptoweweexamplecryptowenoticelearnfoundbackspacesystemsystemexampleshouldlearnsuccesssublimcryptobackspacetonightbackspacemorningmorningnoticeshouldnoticefoundthetheshouldtheshouldfoundfoundcryptosuccessbackspacesuccessshouldweenternoticeweweshouldmorningfoundusersuccessbackspacewenoticeusersuccessenterenterexamplelearnfoundwetonightusercryptothesublimsublimtonightsuccesslearnbackspacetonightentertonightthesublimnoticewefoundcryptobackspaceenterenterlearnlearntonightexamplesystementersublimnoticecryptoshoulduseruserbackspaceuserwesublimmorningwesystemshouldtonighttheusershouldnoticefoundusernoticeentersublimwethewefoundfoundlearnfoundwecryptosystemexamplemorningcryptocryptosublimtheexamplenoticefoundlearnwelearnmorningtheenterthesystemsublimtonightsuccesssystemlearnshouldenterbackspaceentersuccesssuccessbackspaceexamplenoticeentershouldsublimlearnbackspacetheshouldexamplelearnsystemusersublimbackspacebackspacesuccesswelearntonightexamplewecryptoenterwesystemsystemsublimexamplecryptolearnmorningsublimfoundsublimfoundbackspacefoundtonighttonightnoticesuccesssuccessexampleusersuccesstonightsublimcryptosystemweenterexamplesystemthethenoticesublimtonightbackspacenoticesystemexamplethesuccesstonightmorningsuccesstonightwenoticesublimtonightwelearntonightmorningsublimbackspaceenterthetonightenterwecryptofoundtheenternoticebackspacesuccesswesystemuserexamplebackspaceentersuccesstonightsublimwemorningsuccesssuccesswesublimsuccessnoticesublimfoundlearnlearnweexamplecryptonoticelearnweusershoulduserfoundcryptolearnfoundmorningtonightmorningmorningnoticewecryptowewesuccessfoundsublimweuserentershouldshouldshouldsublimbackspacetonightenterwesublimsuccessshouldfoundthethetonightwecryptoweenterfoundcryptoshouldcryptouseruserfoundentersublimsublimthelearntheshouldnoticebackspacefoundsuccessshouldtonightentermorningsystemmorningtonightwenoticelearnbackspaceexampleusershouldnoticesublimsublimexamplethesuccessnoticesystemmorningnoticecryptosystemsublimcryptosystemsuccessshouldmorningbackspaceshouldmorninglearnnoticenoticeshouldthewewesublimsublimnoticeusersuccessentersystemfoundshouldshouldcryptobackspaceusermorningsystemshouldshouldtonightwesublimuserfoundlearnbackspacethetonightmorningexampleuserthefoundbackspaceshouldtonightcryptocryptofounduserexamplenoticecryptousernoticethenoticeshouldweshouldfoundwemorningcryptosuccesslearnfoundtonightsublimnoticenoticewefoundwewesuccesssublimsublimcryptoweexampletonightsuccessfoundshouldsuccesstonightbackspacesystemshouldwesystemnoticebackspaceusersystembackspacewenoticelearnnoticenoticesuccesslearntonightuserlearnsuccessbackspacesuccesswesystemusercryptonoticethesystemusernoticewethesuccessweshouldfoundshouldcryptomorningtonightwethewesuccesslearntheshouldweexampletonightsuccessnoticenoticemorningfoundmorningfoundusersublimsystemsuccessbackspacesuccessmorninguserthefoundweexamplemorningsublimlearnfoundfoundnoticemorningshouldweuserwemorningexamplesuccesssuccessfoundthetheshouldweusershouldtheshouldexamplenoticefoundsuccesssystemfoundshouldsublimbackspacetonightshouldsystemtonightsuccesslearntonightsystemsublimsuccesscryptobackspacesystemsublimmorningmorningshouldmorninglearnsuccesslearnmorningusermorninglearnexamplecryptoshouldbackspacesublimshouldfoundbackspacesystemsystemweexamplesystemtonightsublimmorningmorninguserfoundcryptolearnbackspaceshouldbackspacenoticesublimfoundthecryptousershouldsuccesssystemsuccessshouldsystembackspacesublimshouldsublimsystembackspaceexampleshouldbackspacesublimnoticelearnsublimuserbackspaceusersublimsuccesssublimuserusernoticeshouldsuccessnoticenoticelearnexamplesystemweexamplesublimbackspacebackspacecryptoshouldusercryptosublimbackspacesublimshouldsystemnoticenoticethesuccesssuccesslearnsystemsublimwenoticelearnusersublimsystemusernoticeuserthesuccesslearnwelearnwenoticecryptolearncryptonoticenoticebackspacecryptothecryptousercryptobackspacesuccesslearnthesystemsuccessthesystemsystemcryptosuccessbackspacesublimlearnsublimcryptobackspacelearnsublimusersublimexamplecryptosublimsystemnoticecryptocryptousertheusernoticebackspacenoticenoticethecryptocryptosystembackspacesublimbackspacecryptocryptobackspacesystemuserthenoticesystemsystemsystemusernoticethecryptouserusersystemtheusercryptoexamplenoticecryptoexamplenoticetheexampleexamplethecryptotheusernoticetheexampleexamplecryptotheexampleexamplethenoticethecryptocryptoexampletheexamplecryptocryptothenoticeexamplecryptonoticetheexampleexampleexamplecryptocryptoexampleexamplethenoticethecryptothethethethethetheexampleexamplethetheexampletheexampletheexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexample

好多连在一起的词汇，应该是一个简单NTP题目，也就是NLP最基础的一个步骤，分词。

这道题目的大致思路应该是：

  1. 分词
  2. 词频统计
  3. 统计排序
  4. 根据排序位置找到指定的单词
  5. 拼接单词

根据这个思路，写个脚本处理一下哈：

from collections import Counter // 统计库
import wordninja // 英语单词词库
data="tonightsuccessnoticenoticewewesuccesstonightweexamplecryptoshouldwebackspacetonightbackspaceexamplelearnwesublimlearnbackspacetheshouldwelearnfoundsublimsystemexamplesublimfoundlearnshouldmorningsublimsystemuserlearnthecryptomorningexamplenoticetonightlearntonightlearntonightsublimenterusermorningfoundtonightweenterfoundnoticethecryptomorningthebackspacelearntonightlearnsublimtonightlearnfoundenterfoundsuccesstonightsuccessuserfoundmorningtonighttheshouldsublimentertonightenterbackspacelearnexamplenoticeexamplefoundsystemsuccesssublimsuccessshouldtonightcryptowelearncryptofoundshouldsublimsublimweentertonightsuccessshouldentertheentercryptouserbackspaceshouldentersystemsuccesssystementerfoundenterlearnexampletonightnoticemorningusertonightlearnmorningtonightfoundfoundsuccessnoticesystementerlearnexamplebackspaceshouldcryptocryptosublimweexampletonighttheshouldthemorningbackspacelearntonightsystemsuccesssuccessbackspacemorningnoticeuserfoundfoundtonightmorningenterenterthefoundbackspacelearnenterentershouldthesystemfounduserlearnlearnsystemnoticetonighttheshouldlearnuserbackspaceweusernoticeshouldthewefoundsystemwecryptocryptowethebackspacesystementershouldtonightsystemnoticemorningsystemweentermorningfoundsuccessusertonightsuccesstonightbackspaceshouldweenterthewesystemusernoticesystemthelearnexamplelearnfoundlearnnoticeexamplesystemthecryptocryptolearnsystemthecryptoenterlearnexamplemorningmorningweenterentersuccessexampleuserthebackspacenoticesublimenterbackspacesuccessbackspacethesublimexamplesystemtheexamplecryptolearnuserexamplelearnsystemusersuccessenterentersuccesstheuserbackspacelearnsuccessbackspacethesublimshouldwebackspaceexamplesuccesssuccesstonightweusershouldsuccessmorningcryptomorningfoundbackspacesublimshouldentershouldnoticesuccessmorningsuccessexamplelearnshouldsublimlearntonightshoulduserbackspacesublimlearncryptosuccessenternoticetonightmorningtonightwesuccessweuserbackspaceexamplewesystemnoticemorningsystemmorningcryptolearnsystemthethefoundcryptouserlearnusersystemwemorningenterexampleshouldlearncryptofoundenterbackspacelearnenterenterbackspaceshouldbackspacetheshouldthesystemshouldshouldsuccessmorningthefoundsystementersystemtonightcryptowelearnexampleexamplesystementerbackspaceshouldtheentersublimtonightfoundfoundsuccesssuccesssystemsublimcryptoshouldentersublimmorninglearnfoundtonightcryptobackspacesuccesscryptowebackspacefoundshouldnoticeshouldmorningnoticesystemcryptosystemlearnsystemnoticemorningsystementerwemorninglearnsuccessfoundwesuccesswetheusercryptousernoticebackspacesuccessshouldtonightmorningentermorninguserenternoticefoundmorningwetonightsystemthecryptotonightcryptosystemuserthefoundexampletonightusersystemcryptosublimmorninguserthefoundbackspaceshouldsuccesscryptotonightsystemnoticebackspaceusershouldenterthecryptomorningwesublimnoticesuccessnoticeusersuccesstonightlearnweuserenterfounduserexampleshouldshouldtonightwelearnthenoticethewefoundmorningexampleshouldexamplethesuccessnoticeenterfoundthecryptonoticeuserlearnuserweenterfoundmorningsystemweexamplenoticethebackspaceexamplesublimtheusermorningtonightthesuccesscryptosuccessusersuccesstonighttonightwelearnenterenterthemorningentersystemcryptobackspacemorningsystemexamplecryptouserexamplelearnsublimsuccessusersystemfoundmorningshouldcryptotonightsublimtheexamplemorningsystemuserexampleweexamplenoticesuccesssublimnoticecryptoshouldbackspaceshouldthetonightfoundsublimbackspacebackspacetonightshouldbackspacesuccesstonightbackspacesuccessmorningsystemcryptobackspaceentertonighttonightnoticelearnshoulduserfoundexamplesystemthesuccessweusertonightcryptousernoticeenternoticebackspaceusersystemfoundusernoticeshouldlearnuserfoundexampleusermorningshouldsuccessmorningmorningexampleexamplefoundsublimfoundenterbackspacenoticelearnfoundmorningcryptonoticecryptoshouldweshouldtonightcryptobackspacesublimcryptosublimenterentersublimentercryptonoticethethesublimexampleenterentershouldlearncryptoentershouldmorninglearnnoticeuserexamplesublimtonightshouldfoundtonightsuccessshouldmorningfoundtheweuserlearnsublimsystembackspacecryptotheusertonightcryptosublimmorningmorningexamplenoticetheenterlearnshouldmorningsublimfoundtonightsublimsublimexamplefounduserexamplethefoundwemorningnoticefoundcryptosuccesssublimsublimexamplethesuccessexamplenoticesuccessbackspacesublimlearnuserexamplesuccesssuccesssystemsuccessmorningmorninglearnexamplemorningtonightfoundbackspaceenternoticemorningentersuccessmorningusermorningbackspacelearncryptoenteruserenteruserthetonighttonightsuccesslearnenterfoundsuccesssystemfoundbackspaceenterlearnsystemsublimcryptoentermorningwetonightshouldlearnenterfoundcryptonoticelearnlearnshouldfoundsuccessexampletonightthesuccessfoundusertonightenterfoundsuccessshouldmorningusernoticemorningsystemsystemsuccessshouldwelearnenterfoundexamplewethefoundweshouldsystemsystemmorningmorningbackspaceshouldentersublimentertonightsuccesssystemsystemcryptousershouldsublimfoundwetonightnoticeexamplewewesuccessfoundusertonightfoundsystemexamplecryptofoundshouldshouldsuccessenterbackspaceexampletonightthelearnnoticeuserlearnsystemsublimfoundlearnsuccesssystemshouldsublimnoticelearnsystemnoticetonightexamplefoundusernoticeenterlearnnoticecryptousersystemmorningthewesystemfoundfoundshouldsystementerenterbackspacesystemsublimcryptousermorninglearnlearntonightsublimlearnenterenterbackspacesystemuserusercryptoentershouldtheusersublimnoticeexamplemorningexamplesublimsublimbackspacesystemexampleshouldsublimlearnfoundenterbackspacelearnmorningmorningfoundthetonightmorningnoticeenterlearnusersystemtonightbackspaceexamplelearntonightbackspaceweshouldcryptosuccessbackspaceexamplesuccesstheshouldmorninguserbackspacelearnthetheshouldcryptocryptotonightbackspacecryptocryptobackspacebackspacenoticeusertonightentermorningfoundweenterexampleenterfoundusersublimsystemtheexampleexamplesystemsuccessusersublimentermorningbackspacesystemfoundlearnsystemshouldsublimsublimentershouldtheusershouldexampleexampleshouldsuccesswelearnfoundsublimshoulduserweentertonightwenoticesublimsystemlearnshouldfoundsuccessuserentersuccessmorningcryptoenteruserfoundexampletonightlearnexampleexamplefoundlearnsuccesssystembackspacecryptonoticethefoundbackspacelearncryptothelearnlearnexamplesuccessnoticenoticesystemmorningcryptotonightnoticenoticeentersuccesscryptoenterbackspacesublimexampleenterfoundtonightcryptotonightsublimnoticesuccesssublimtheentertonighttheshouldthefoundsystemtonightuserbackspacesuccessshouldwebackspacenoticebackspacebackspacenoticewecryptobackspacebackspaceusertonightlearnsuccessmorningusertonightsuccessshouldbackspacecryptoenterentershouldsublimsystemexamplemorningcryptonoticethesuccessthebackspacenoticelearnsublimlearnsuccesscryptothesuccessenternoticecryptosystemsublimsuccesswebackspaceuserenterlearnuserwewemorningsuccesslearncryptobackspacewecryptosystemlearnenterenteruserexamplefoundsystemcryptousernoticefoundusersublimbackspacewesublimnoticemorningshouldexamplenoticecryptoshouldtonightmorningthefoundsystementerentersystemthecryptobackspacesublimlearnsuccessmorningsublimsystemcryptousersublimwesuccessmorningsublimbackspacecryptobackspacesublimthelearnsuccesssublimlearncryptoweweexamplecryptowenoticelearnfoundbackspacesystemsystemexampleshouldlearnsuccesssublimcryptobackspacetonightbackspacemorningmorningnoticeshouldnoticefoundthetheshouldtheshouldfoundfoundcryptosuccessbackspacesuccessshouldweenternoticeweweshouldmorningfoundusersuccessbackspacewenoticeusersuccessenterenterexamplelearnfoundwetonightusercryptothesublimsublimtonightsuccesslearnbackspacetonightentertonightthesublimnoticewefoundcryptobackspaceenterenterlearnlearntonightexamplesystementersublimnoticecryptoshoulduseruserbackspaceuserwesublimmorningwesystemshouldtonighttheusershouldnoticefoundusernoticeentersublimwethewefoundfoundlearnfoundwecryptosystemexamplemorningcryptocryptosublimtheexamplenoticefoundlearnwelearnmorningtheenterthesystemsublimtonightsuccesssystemlearnshouldenterbackspaceentersuccesssuccessbackspaceexamplenoticeentershouldsublimlearnbackspacetheshouldexamplelearnsystemusersublimbackspacebackspacesuccesswelearntonightexamplewecryptoenterwesystemsystemsublimexamplecryptolearnmorningsublimfoundsublimfoundbackspacefoundtonighttonightnoticesuccesssuccessexampleusersuccesstonightsublimcryptosystemweenterexamplesystemthethenoticesublimtonightbackspacenoticesystemexamplethesuccesstonightmorningsuccesstonightwenoticesublimtonightwelearntonightmorningsublimbackspaceenterthetonightenterwecryptofoundtheenternoticebackspacesuccesswesystemuserexamplebackspaceentersuccesstonightsublimwemorningsuccesssuccesswesublimsuccessnoticesublimfoundlearnlearnweexamplecryptonoticelearnweusershoulduserfoundcryptolearnfoundmorningtonightmorningmorningnoticewecryptowewesuccessfoundsublimweuserentershouldshouldshouldsublimbackspacetonightenterwesublimsuccessshouldfoundthethetonightwecryptoweenterfoundcryptoshouldcryptouseruserfoundentersublimsublimthelearntheshouldnoticebackspacefoundsuccessshouldtonightentermorningsystemmorningtonightwenoticelearnbackspaceexampleusershouldnoticesublimsublimexamplethesuccessnoticesystemmorningnoticecryptosystemsublimcryptosystemsuccessshouldmorningbackspaceshouldmorninglearnnoticenoticeshouldthewewesublimsublimnoticeusersuccessentersystemfoundshouldshouldcryptobackspaceusermorningsystemshouldshouldtonightwesublimuserfoundlearnbackspacethetonightmorningexampleuserthefoundbackspaceshouldtonightcryptocryptofounduserexamplenoticecryptousernoticethenoticeshouldweshouldfoundwemorningcryptosuccesslearnfoundtonightsublimnoticenoticewefoundwewesuccesssublimsublimcryptoweexampletonightsuccessfoundshouldsuccesstonightbackspacesystemshouldwesystemnoticebackspaceusersystembackspacewenoticelearnnoticenoticesuccesslearntonightuserlearnsuccessbackspacesuccesswesystemusercryptonoticethesystemusernoticewethesuccessweshouldfoundshouldcryptomorningtonightwethewesuccesslearntheshouldweexampletonightsuccessnoticenoticemorningfoundmorningfoundusersublimsystemsuccessbackspacesuccessmorninguserthefoundweexamplemorningsublimlearnfoundfoundnoticemorningshouldweuserwemorningexamplesuccesssuccessfoundthetheshouldweusershouldtheshouldexamplenoticefoundsuccesssystemfoundshouldsublimbackspacetonightshouldsystemtonightsuccesslearntonightsystemsublimsuccesscryptobackspacesystemsublimmorningmorningshouldmorninglearnsuccesslearnmorningusermorninglearnexamplecryptoshouldbackspacesublimshouldfoundbackspacesystemsystemweexamplesystemtonightsublimmorningmorninguserfoundcryptolearnbackspaceshouldbackspacenoticesublimfoundthecryptousershouldsuccesssystemsuccessshouldsystembackspacesublimshouldsublimsystembackspaceexampleshouldbackspacesublimnoticelearnsublimuserbackspaceusersublimsuccesssublimuserusernoticeshouldsuccessnoticenoticelearnexamplesystemweexamplesublimbackspacebackspacecryptoshouldusercryptosublimbackspacesublimshouldsystemnoticenoticethesuccesssuccesslearnsystemsublimwenoticelearnusersublimsystemusernoticeuserthesuccesslearnwelearnwenoticecryptolearncryptonoticenoticebackspacecryptothecryptousercryptobackspacesuccesslearnthesystemsuccessthesystemsystemcryptosuccessbackspacesublimlearnsublimcryptobackspacelearnsublimusersublimexamplecryptosublimsystemnoticecryptocryptousertheusernoticebackspacenoticenoticethecryptocryptosystembackspacesublimbackspacecryptocryptobackspacesystemuserthenoticesystemsystemsystemusernoticethecryptouserusersystemtheusercryptoexamplenoticecryptoexamplenoticetheexampleexamplethecryptotheusernoticetheexampleexamplecryptotheexampleexamplethenoticethecryptocryptoexampletheexamplecryptocryptothenoticeexamplecryptonoticetheexampleexampleexamplecryptocryptoexampleexamplethenoticethecryptothethethethethetheexampleexamplethetheexampletheexampletheexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexample"
lm = wordninja.LanguageModel('my_dict.txt.gz') // 通过调试写出的简单字典
data_list = lm.split(data)
data_count = Counter(data_list)
index = 1
hint_list = [4,8,11,15,16]
m = ""
for i in data_count.most_common(20)[::-1]:
    if index in hint_list:
        m += i[0]
    index += 1
print(m)

运行一下脚本：

weshouldlearnthecrypto

游戏结束！得到flag：

flag{weshouldlearnthecrypto}

0x1 [AFCTF2018]Vigenere

维吉尼亚密码，多表替换密码。来瞧瞧题目的附件：

两个附件文件，一个加密的flag_encode.txt和加密算法的Encode.c

Encode.c

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
	freopen("flag.txt","r",stdin);
	freopen("flag_encode.txt","w",stdout);
	char key[] = /*SADLY SAYING! Key is eaten by Monster!*/;
	int len = strlen(key);
	char ch;
	int index = 0;
	while((ch = getchar()) != EOF){
		if(ch>='a'&&ch<='z'){
			putchar((ch-'a'+key[index%len]-'a')%26+'a');
			++index;
		}else if(ch>='A'&&ch<='Z'){
			putchar((ch-'A'+key[index%len]-'a')%26+'A');
			++index;
		}else{
			putchar(ch);
		}
	}
	return 0;
}

flag_encode.txt

Yzyj ia zqm Cbatky kf uavin rbgfno ig hnkozku fyyefyjzy sut gha pruyte gu famooybn bhr vqdcpipgu jaaju obecu njde pupfyytrj cpez cklb wnbzqmr ntf li wsfavm azupy nde cufmrf uh lba enxcp, tuk uwjwrnzn inq ksmuh sggcqoa zq obecu zqm Lncu gz Jagaam aaj qx Hwthxn'a Gbj gfnetyk cpez, g fwwang xnapriv li phr uyqnvupk ib mnttqnq xgioerry cpag zjws ohbaul drinsla tuk liufku obecu ovxey zjwg po gnn aecgtsneoa.

Cn poyj vzyoe gxdbhf zq ty oeyl-ndiqkpl, ndag gut mrt cjy yrrgcmd rwwsf, phnz cpel gtw yjdbcnl bl zjwcn Cekjboe cklb yeezjqn htcdcannhum Rvmjlm, phnz juoam vzyoe nxn Tisk, Navarge jvd gng honshoc wf Ugrhcjefy. — Cpag zq kyyuek cpefk taadtf, Mxdeetowhps nxn qnfzklopeq gvwnt Sgf, xarvbrvg gngal fufz ywwrxu xlkm gnn koaygfn kf gnn ooiktfyz, — Tugc ehrtgnyn aae Owrz uh Yireetvmng hguiief jnateaelcre bl cpefk gfxo, ig ob bhr Xkybp os zqm Prurdy po nrcmr bx vg uxoyobp ig, gpv nk iaycqthzg fys Gbbnznzkpl, fwyvtp qtf lqmhzagoxv oa ywub lrvtlqpyku shz oemjvimopy cps cufmrf op koyh suau, af zq lbam fnjtl fkge gksg rrseye vg ybfric bhrot Kubege jvd Ugrhcjefy. Yzuqkpuy, enqknl, wvrn vcytnzn bhnz Igparasnvtf rqfa asggktifngv mdohrm vog hg ubwntkm noe rkybp aaj czaaykwhp cnabms; ntf swyoejrvgye cdf axckaqeaig zuph fnnen gncl gwnxowl aek ogla dvyywsrj vg mqfska, ehvrg wpelf gam shlhwlwbyk cpaa zq jcchg zqmmfknnyo bl gkwlvyjahc tuk owrzy vg qdipn cpel gtw uychycwmrj. Dmn shrt j toam vjuen bl jjufku shz ufaaxagoqfm, lueydqnt opnuninhug tuk usga Oopnkt rbkfwas n jnaitt vg ladhin bhrs wfxar nhbwlhzg Vyopbzram, vz kk ndevx aqguz, kl co tukrz dhza, li pheuf wfs ywub Coikavmrtv, shz tb vawvvjg fys Ghgals sut lbaie ldbuek uwwqrvzh. — Aupn jsm xert cpe cgvayjt faoneegpuy kf gnnae Pungheef; gwl shij am joj zqm nrigkmetl cqqcu iqfmprnowa tuko li wlgka bhrot xinmrx Bgsgkok ib Gbbnznzkpl. Nde uobboee qx nde cxnaeaz Mahc os Mamag Htanwia ob i hvyvglu os xnxenzgv cjjhxrms ntf mmqrcgcqoay, cdf daiowo ia jkjyyt bhsmcg zjw yotnhuqsusgfn kf nt jjsbrwly Pyegwvy bbgj ndefk Bbagku. Li lrbbn bhvy, nwn Bapzb je fadecptrj cw a pgpvcz wbxul.

Hr nck lafhynl hvy Ckmang zx Tajy, vzy iofz fpoykugga aaj wmcryuslu fbx cpe caddcy gbum.

Pe ugu xinbvjmmn uou Yireetxzs gu rsmo Lncb wf vsowxeagk jvd cxgkment ovxoezcfwa, uarnas fauhyjdrj rv tukkj ileegcqoa zkdf dif Gbaeaz uziqlq hn wbggkfyz; aaj fpea yq kooprtmmd, uk jsm qtgkaty akidyytrj cw agzgfx po gnnu.

Hr nck lafhynl tb vckm ktuka Tajy hgl phr glkozsqvupibt xn lnxiw xesgxrktf uh hykpyk, dvlryu lbksr vnwpyk ygohd ekuqndakkb phr xrohg uh Jylrrynvtnzkgh en gnn Tetoudupuek, j zitnv ahasgovibyk vg ndez gwl fbxoaxwbyk cw tlxcfno oarh.

Pe ugu uuhlrj cwgrzjwl hetobtagoxw vkdvkb it crcuyo uaabcay, apuiifbxcibyk, cfx zifzjvt sxqe nde qkywsvzqjs kf gnnqr Caddcy Rrixzdf, lqj nde fuum phxrgma os ljbitakfa phrs rvtb iqejhintlm wvzj zco mrgbcrry.

Jw bws qobaoybgv Lapekbmnggvapa Hbabms ekrwupeqrh, noe urhioiam fqtu scffu fvxvvefy jam enigbqoay qf nde eopptf uh lba pruyte.

Uk jsm nesabmd sut s fknt zrue, nlvwl oupn mqsfunmneoay, cw cnauw iphrxb bo ok gdyytrj, fpeekdq nde Ykpqsygvapa Pbcnzs, vtesjwbyk xn Aatkzchagoxv, hnbg jypuetnl tb zjw Jaocrn it ygtyy boe zqmie kzwlyifk; cpe Fzcly nezgrviam kf nde zkjv tvsg wrlofkm bo nrn lba dntpmrf uh ahrafoxv feuo ocphbac, inq iqfpqlfoxvs jovzcj.

Hr nja eajgspkuekm bo cxgnyjt gnn xocansneoa uo bhryg Knwtry; owr gncl jqrcubm ooyvjoytvtp bhr Rcom boe Tjbuegnatwtvuw wf Sutwccnrxb; zesauahc tb vjas bzjwlo tb kwkohxcyy phroa uitxclcknf nrbhrx, cfx navyrvg gng uijdvzrwnf uh fys Acvawpeoclcknf uo Taaju.

Zy daf ukateaelyz tuk Jlmvtkknnagoxv os Pwknecr hh zesauahc hvy Jasrtv li Hajy owr ryvsvhifnrvg Wafaweaee Ywwrxu.

Zy daf sjle Wafyyo drvnvdrtv gh dif Crtl nrqfy boe zqm trtwjy kf gnnqr blhawas, ntm bhr gogojt ntm xalsgfn kf gnnqr fgnsleef.

luig vy cxwpf{Jnxwobuqg_O_Cogiqi!}

Hr nck ynepznl a zanlcpuqk xn Nrc Qxzecry, jvd fkpl betuka awnxok ib Oslrkeey vg bwrnyb wue vggjhe ntm mag uwl ndevx bcbfzcfwa.

Hr nja krvv sgknt ab, qn goowm kf ckjke, Fzcfxent Gauiry yandohz cpe Pupkyjt bl xcr ykiamhagaams.

Uk jsm wfsklbeq zq jyjdrx cpe Zonanwrl owleckpvyjt bl jvd farwleoe zx bhr Iknch Pbcnz.

Hr nck wkmoowmd jovz iphrxb bo fadbyyt hy cw a watamzipzrwn sutwccn gu xcr pupknethzrwn, ntf mhwcxtxelrjiwx xy baa tajy; iapent nra Afygfn po gnnqr Nivk ib pekcmnqkf Dycifrjbibt:

Hgl munxcmrvti dungr hxliry qx unmrj czobvu sgknt ab:

Noe vtgnacgowo tuko, ts w mbit Brvgn xlkm cawqsusgfn boe gwg Mhxfwlo wuolp tuka kbkuyj lwmzov gh phr Owpaoovshps bl cpefk Ulupef:

Lxz chzvahc osl xcr Gxcvy sign jtl cgtlm kf gnn eoerf:

Xin izvxaiam Vsras bt da wvzjgop ohx Lwnfkpl:

Zkr qkyziiopy oo ia sjvy pguwm, kf gnn jeakhan kf Gxril oe Lmlu:

Fbx czaayrglpiam da breqfx Oeny cw br ztayz fbx yzegkpvyz oslnvcry:

Hgl wbbrrahvti lba fekn Ayfzge ib Eamuqsu Rcom en n tnqguhqmlent Vawvvtew, yotnhuqsuopy ndeekrv aa Gttcprnxh ooiktfgang, gwl earcjaent oca Bbapvuniry bw af zq jyjdrx rb ag upuy wn rdjupyk cfx big owateaowhp fbx rvteufmwent zqm snsg svooyacm rhrg ahpo gnnae Pungheef

Lxz tnqkfa wwne xcr Pncjnarf, gkwlvyjahc ohx vwsg bcdowbyk Uiwf gpv uhtrxrvg sapvuieazjtll zjw Zkrzy xn ohx Igparasnvtf:

Lqj mqsckwliam qml kwa Rnoifrclonef, gwl drinslent zqmmfknnyo iabnatrj yand pbcnz tb rgycolnzn noe au ah wly ijaef cjsnoorbnz.

Hr nck uxdvijbeq Mqnynnzkwb hrxg, ts zeprjziam wk iqt bl qqs Cxqlyytvuw inq ccycjg Jga ignopkn qs.

Uk qis crwfxarrj xcr fkck, lwvnmnl ohx eguotf, hdzng uwj nkway, jvd qkullkyrj cpe yoxwm kf baa xebvnw.

Ba if gc bhvy vaga tegwapbxvahc lnxpm Aeskwm kf suamitt Owlyeagaqef zq uiipykjb tuk yglgs bl mmagn, fwmklnzrwn, ntf lsnaath, ilekcvs xetaw eign ealyuzycinpku gz Yrhkuby & Cktxczy fijzcrra hunayrnteq op lba mbyc jaehcjiqs nmna, aaj vgnwlye dvwbxvzs phr Nnid bl c ucriyoimd agvaij.

Hr nja cbtullwiakm wue lgdfkw Pocqzrtu lugea Ijxtvbg gh phr nroh Fkck nk brga Irzy cyuenfz cpevx Egojtee, cw briqey phr kgmchzkgharf uo bhrot xleeajb inq Htwndrrt, xz tb lcdf phrsbmliku ts phroa Paaju.

Zy daf kgkigkf viiefzrk iaywjlacgoxvs nsqfaot hy, jvd ugu whzenbxcrrj vg vniam xv tuk kfbwbvzjvtf uh gon feuwbirxu, lba mrxlqlryu Ahzint Bivnmgk qdofk tvojt tmfa os cjzfnxg, am wn htmqsgopyoesukm lefztmwpibt xn ayr cyyo, srdna aaj eghzigoxvs.

Vt gnyny fzjoe bl vzyoe Bvyzefykgho Wr njde Ckvaneoakm noe Xgvlasf ow bhr sqkn duzhum trxok: Iqr ekymagkf Hypigoxvs ugxw vaea gwawrxgv ijll hh zeckclyz iapdzy. N Vtahye, jnxae pncjuytrx ra tuau eunkrj kg eiktq uyt jnrkh zga vybiak j Byegpl, co ualrb tb hg lba rhrnz os g hjya pruyte.

Aut zure Jk kmea ccfnent ow itgkplcknf zx wue Htanesu hamtuxgf. Qa hnbn eaetgv ndez lawm goow nk tvsn wf nzvwgltf hh bhrot dycifrjbuek vg yttrtm in htyslnaazjjlr pwjcodvicqoa uxwl qs. Jk qivr xgecjdrj cpez uh lba cvxlcmfzcfwas bl xcr rskylwtvuw inq yglnhezkwb hrxg. Oy daik jxprgnwx po gnnqr agvapa jhycqcr gpv gwgagwqmvza, shz wr njde pupboneq zqmm oe vzy piry xn ohx eggioa qrvdekf li zifgeww gngky qshxyitvupk, qdipn fwuyj kfyriggkty vtvwlnucz xcr pupfyytvuwa aaj eglnefvxvdrtew. Ndel zxw hnbg tyan qkjn tb zjw pkipk xn jhyvawa aaj xn cbtushcuvtrby. Jk ommp, tukamfbxg, swmuvkbke vt vzy jepkbaige, yzcyh qkwwuaigk iqr Fkyirnzkgh, wnq nxtd gnge, uo wr nxtd gng jyot bl vinxopv, Yjezona ia Ccj, cj Prglm Feogfxo.

Wr, zqmrrlqjy, phr Xnxrrygfnwtvbna os zjw ojigkm Atnzgk ib Azkaqcn, op Yyjeegu Koamtwmo, Afynubykf, sjlenrrvg gu vzy Oucxnue Wafyy kf gnn eoerf xin tuk amcgovmxa os udz iazgfneoay, mw, ia zjw Hwmr, gwl bl Gwlbkrvzh wf gng yikd Ckxxlr uh lbasr Ixtoaogk, mklrswty caddcoh ntm leprcjy, Phnz cpefk wfcpeq Ixtoaogk une, ntm wf Eoizn kutnc bo ok Hjya aaj Rvdrvgfxang Ycitry, vzup tukh irr Gdkihvrj ozoz gnd Uhlrmrinpk vg nde Oxrbifn Ejisn, ntm bhnz cdf loyocqcnr eghjepzrwn okvoyan gnnu aaj vzy Otnzn wf Txgsn Xrvzjqn, vy cfx kutnc bo ok vgnwlye mqsfunnyz; aaj cpag gu Xlae ntm Qnqkrwhzeaz Bbagku, lbay ugem fhrn Hisee zx teie Ysl, yoaiucdr Vgswa, cbtczapz Cdfeaaina, efzctfesu Ixumrxew, ujd gu mw ayr qlbar Nica aaj Vzcjgf cqqcu Opvyleajnvt Fzclyo mne xn rvmjl xk. — Aaj owr gng kolpbxc wf gnkk Xacygaitvup, ocph n lrzm eknaujcr uw bhr vtgnacgoxv os Jkncje Cxxdiqkpuy, se zaccayra hfadtk cw enij gndee udz Lvbgk, iqr Suabuaku, shz ohx bicekf Zijoe.

直接去维吉尼亚密码破解网站进行破解就好，进行破解得到：

When in the Course of human events it becomes necessary for one people to dissolve the political bands which have connected them with another and to assume among the powers of the earth, the separate and equal station to which the Laws of Nature and of Nature's God entitle them, a decent respect to the opinions of mankind requires that they should declare the causes which impel them to the separation.

We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness. — That to secure these rights, Governments are instituted among Men, deriving their just powers from the consent of the governed, — That whenever any Form of Government becomes destructive of these ends, it is the Right of the People to alter or to abolish it, and to institute new Government, laying its foundation on such principles and organizing its powers in such form, as to them shall seem most likely to effect their Safety and Happiness. Prudence, indeed, will dictate that Governments long established should not be changed for light and transient causes; and accordingly all experience hath shewn that mankind are more disposed to suffer, while evils are sufferable than to right themselves by abolishing the forms to which they are accustomed. But when a long train of abuses and usurpations, pursuing invariably the same Object evinces a design to reduce them under absolute Despotism, it is their right, it is their duty, to throw off such Government, and to provide new Guards for their future security. — Such has been the patient sufferance of these Colonies; and such is now the necessity which constrains them to alter their former Systems of Government. The history of the present King of Great Britain is a history of repeated injuries and usurpations, all having in direct object the establishment of an absolute Tyranny over these States. To prove this, let Facts be submitted to a candid world.

He has refused his Assent to Laws, the most wholesome and necessary for the public good.

He has forbidden his Governors to pass Laws of immediate and pressing importance, unless suspended in their operation till his Assent should be obtained; and when so suspended, he has utterly neglected to attend to them.

He has refused to pass other Laws for the accommodation of large districts of people, unless those people would relinquish the right of Representation in the Legislature, a right inestimable to them and formidable to tyrants only.

He has called together legislative bodies at places unusual, uncomfortable, and distant from the depository of their Public Records, for the sole purpose of fatiguing them into compliance with his measures.

He has dissolved Representative Houses repeatedly, for opposing with manly firmness his invasions on the rights of the people.

He has refused for a long time, after such dissolutions, to cause others to be elected, whereby the Legislative Powers, incapable of Annihilation, have returned to the People at large for their exercise; the State remaining in the mean time exposed to all the dangers of invasion from without, and convulsions within.

He has endeavoured to prevent the population of these States; for that purpose obstructing the Laws for Naturalization of Foreigners; refusing to pass others to encourage their migrations hither, and raising the conditions of new Appropriations of Lands.

He has obstructed the Administration of Justice by refusing his Assent to Laws for establishing Judiciary Powers.

He has made Judges dependent on his Will alone for the tenure of their offices, and the amount and payment of their salaries.

flag is afctf{Whooooooo_U_Gotcha!}

He has erected a multitude of New Offices, and sent hither swarms of Officers to harass our people and eat out their substance.

He has kept among us, in times of peace, Standing Armies without the Consent of our legislatures.

He has affected to render the Military independent of and superior to the Civil Power.

He has combined with others to subject us to a jurisdiction foreign to our constitution, and unacknowledged by our laws; giving his Assent to their Acts of pretended Legislation:

For quartering large bodies of armed troops among us:

For protecting them, by a mock Trial from punishment for any Murders which they should commit on the Inhabitants of these States:

For cutting off our Trade with all parts of the world:

For imposing Taxes on us without our Consent:

For depriving us in many cases, of the benefit of Trial by Jury:

For transporting us beyond Seas to be tried for pretended offences:

For abolishing the free System of English Laws in a neighbouring Province, establishing therein an Arbitrary government, and enlarging its Boundaries so as to render it at once an example and fit instrument for introducing the same absolute rule into these Colonies

For taking away our Charters, abolishing our most valuable Laws and altering fundamentally the Forms of our Governments:

For suspending our own Legislatures, and declaring themselves invested with power to legislate for us in all cases whatsoever.

He has abdicated Government here, by declaring us out of his Protection and waging War against us.

He has plundered our seas, ravaged our coasts, burnt our towns, and destroyed the lives of our people.

He is at this time transporting large Armies of foreign Mercenaries to compleat the works of death, desolation, and tyranny, already begun with circumstances of Cruelty & Perfidy scarcely paralleled in the most barbarous ages, and totally unworthy the Head of a civilized nation.

He has constrained our fellow Citizens taken Captive on the high Seas to bear Arms against their Country, to become the executioners of their friends and Brethren, or to fall themselves by their Hands.

He has excited domestic insurrections amongst us, and has endeavoured to bring on the inhabitants of our frontiers, the merciless Indian Savages whose known rule of warfare, is an undistinguished destruction of all ages, sexes and conditions.

In every stage of these Oppressions We have Petitioned for Redress in the most humble terms: Our repeated Petitions have been answered only by repeated injury. A Prince, whose character is thus marked by every act which may define a Tyrant, is unfit to be the ruler of a free people.

Nor have We been wanting in attentions to our British brethren. We have warned them from time to time of attempts by their legislature to extend an unwarrantable jurisdiction over us. We have reminded them of the circumstances of our emigration and settlement here. We have appealed to their native justice and magnanimity, and we have conjured them by the ties of our common kindred to disavow these usurpations, which would inevitably interrupt our connections and correspondence. They too have been deaf to the voice of justice and of consanguinity. We must, therefore, acquiesce in the necessity, which denounces our Separation, and hold them, as we hold the rest of mankind, Enemies in War, in Peace Friends.

We, therefore, the Representatives of the united States of America, in General Congress, Assembled, appealing to the Supreme Judge of the world for the rectitude of our intentions, do, in the Name, and by Authority of the good People of these Colonies, solemnly publish and declare, That these united Colonies are, and of Right ought to be Free and Independent States, that they are Absolved from all Allegiance to the British Crown, and that all political connection between them and the State of Great Britain, is and ought to be totally dissolved; and that as Free and Independent States, they have full Power to levy War, conclude Peace, contract Alliances, establish Commerce, and to do all other Acts and Things which Independent States may of right do. — And for the support of this Declaration, with a firm reliance on the protection of Divine Providence, we mutually pledge to each other our Lives, our Fortunes, and our sacred Honor.

在破解得到文字中找到flag：

afctf{Whooooooo_U_Gotcha!}

0x2 [NPUCTF2020]这是什么🐎

看题目应该是个脑洞题，古典密码学题目，瞧瞧附件吧：

附件是一个attachment文件，没有文件扩展名……那就先探测一下：

file attachment

得到探测输出：

attachment: Zip archive data, at least v2.0 to extract

应该是一个压缩文件，修改文件名为attachment.zip，然后进行解压，得到了一个图片：

flag

（这是杂项题目，越做越不像密码学）

这个图片应该就是密码，认真观察一下图片，发现图片底部有密文。

对密文进行分析：

  1. 日历的周是按英文来进行显示的
  2. 密文数组中的首字母好像对应着英文的周

因此可以根据以下思路进行解题：

  1. 根据密码需要对应数字
  2. 将数字映射到字母表
  3. 讲字母拼接

根据思路，写个脚本求解：

import string
c = "F1 W1 S22 S21 T12 S11 W1 S13"
def encode_cal(data):
    encode_dict = {
            'M':[],
            'T1':[],
            'W':[],
            'T2':[],
            'F':[],
            'S1':[],
            'S2':[]
            }
    dict_index=['M','T1','W','T2','F','S1','S2']
    index = 2
    for i in range(30):
        encode_dict[dict_index[(index+i)%7]].append(i+1)

    result=""

    for m in data:
        if len(m)==2:
            result += string.ascii_lowercase[encode_dict[m[0]][int(m[1])-1]-1]
        elif len(m)==3:
            result += string.ascii_lowercase[encode_dict[m[0:2]][int(m[2])-1]-1]
    return result

if __name__ == "__main__":
    c_list = c.split(" ")
    flag = encode_cal(c_list)
    print(flag)

运行脚本，游戏结束！

calendar

0x3 easyrsa

rsa题目，来，让我们暴打出题人！

附件是个python源码程序

from Crypto.Util.number import getPrime,bytes_to_long
from sympy import Derivative
from fractions import Fraction
from secret import flag

p=getPrime(1024)
q=getPrime(1024)
e=65537
n=p*q
z=Fraction(1,Derivative(arctan(p),p))-Fraction(1,Derivative(arth(q),q))
m=bytes_to_long(flag)
c=pow(m,e,n)
print(c,z,n)
'''
output:
7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035
32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482
15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441

'''

简单审计一下，发现就是道数学题，微分方程的题目。这python程序的关键代码也只有一行：

z=Fraction(1,Derivative(arctan(p),p))-Fraction(1,Derivative(arth(q),q))

应该就是求解方程的事情，因为涉及数学运算，所以写个sage脚本求解吧

from Crypto.Util.number import long_to_bytes

c = 7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035
z = 32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482
n = 15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441
e = 65537

var('p, q')

f1 = (1/diff(arctan(p),p))-(1/diff(arctanh(q),q))
f2 = p*q

result = sorted(solve([f1==z,f2==n],(p,q)))
p = result[1][0]
q = result[1][1]
p =int(str(p)[5:])
q =int(str(q)[5:])

phi = (p-1)*(q-1)

d = inverse_mod(e,phi)
m = pow(c,d,n)
flag = long_to_bytes(m)
print(flag)

由于sagemath是基于python3的开源数学软件，进行sage运行sage脚本是会转换成python格式进行运行，代码非常像python3

运行脚本，暴打出题人！

b'BJD{Advanced_mathematics_is_too_hard!!!}'

0x4 babyRSA

又一个RSA题目，希望这个不单单只是数学运算了，看看题目：

from Crypto.Util.number import *
from flag import flag

def nextPrime(n):
    n += 2 if n & 1 else 1
    while not isPrime(n):
        n += 2
    return n

p = getPrime(1024)
q = nextPrime(p)
n = p * q
e = 0x10001
d = inverse(e, (p-1) * (q-1))
c = pow(bytes_to_long(flag.encode()), e, n)

# d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
# c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804

这道题目的输出是d、c，在算法程序里面暴露出了e。但是n不知道，这就有点麻烦了。简单推导一下公式： $$ \phi = (p-1)\times(q-1) $$

$$ d \equiv e^{-1} \text{mod}\phi $$

$$ d \cdot e \equiv 1\text{mod} \phi $$

$$ d\cdot e -1 = k \cdot(p-1)\cdot(q-1) $$ 根据推导，可以肯定的是这道题目需要进行爆破。为了提高爆破效率，需要进行缩小范围来进行更快求解。缩小范围需要先进行判断：p和q都是1024位，n就是2048位。进行简单的测试发现ed-1是2064位，那么可以判断出k一定是小于16位的，根据目前的推断来写脚本进行求解：

import libnum
import sympy
e = 0x10001
d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804

ed_1 = e*d-1
n = 0
for k in range(pow(2,15),pow(2,16)):
    if ed_1 % k == 0:
        phi = ed_1//k
        base = libnum.nroot(phi,2)
        p = sympy.nextprime(base)
        q = sympy.prevprime(base)
        if (p-1)*(q-1)*k == ed_1:
            n = p*q
            break

m = pow(c,d,n)
flag = libnum.n2s(m)
print flag

求解方式比较暴力，执行脚本就拿到flag了：

NCTF{70u2_nn47h_14_v3ry_gOO0000000d}

0x5 [AFCTF2018]你能看出这是什么加密么

脑洞题，看看题目推测一下这是什么加密：

p=0x928fb6aa9d813b6c3270131818a7c54edb18e3806942b88670106c1821e0326364194a8c49392849432b37632f0abe3f3c52e909b939c91c50e41a7b8cd00c67d6743b4f

q=0xec301417ccdffa679a8dcc4027dd0d75baf9d441625ed8930472165717f4732884c33f25d4ee6a6c9ae6c44aedad039b0b72cf42cab7f80d32b74061

e=0x10001

c=0x70c9133e1647e95c3cb99bd998a9028b5bf492929725a9e8e6d2e277fa0f37205580b196e5f121a2e83bc80a8204c99f5036a07c8cf6f96c420369b4161d2654a7eccbdaf583204b645e137b3bd15c5ce865298416fd5831cba0d947113ed5be5426b708b89451934d11f9aed9085b48b729449e461ff0863552149b965e22b6

好常规的RSA题目，这是签到题目吧

写个脚本解决：

import libnum

p=0x928fb6aa9d813b6c3270131818a7c54edb18e3806942b88670106c1821e0326364194a8c49392849432b37632f0abe3f3c52e909b939c91c50e41a7b8cd00c67d6743b4f

q=0xec301417ccdffa679a8dcc4027dd0d75baf9d441625ed8930472165717f4732884c33f25d4ee6a6c9ae6c44aedad039b0b72cf42cab7f80d32b74061

e=0x10001

c=0x70c9133e1647e95c3cb99bd998a9028b5bf492929725a9e8e6d2e277fa0f37205580b196e5f121a2e83bc80a8204c99f5036a07c8cf6f96c420369b4161d2654a7eccbdaf583204b645e137b3bd15c5ce865298416fd5831cba0d947113ed5be5426b708b89451934d11f9aed9085b48b729449e461ff0863552149b965e22b6

phi = (p-1)*(q-1)

n = p*q

d = libnum.invmod(e,phi)

m = pow(c,d,n)
flag = libnum.n2s(m)

print flag

运行脚本，拿到flag：

afctf{R54_|5_$0_$imp13}

0x6 [ACTF新生赛2020]crypto-rsa3

RSA算法的题目，有点意思哦！(*^_^*)

瞅一瞅题目附件：

有两个附件，比较常规的形式，一个是output.txt的输出文件，一个rsa3.py的算法文件

rsa3.py：

from flag import FLAG
from Cryptodome.Util.number import *
import gmpy2
import random

e=65537
p = getPrime(512)
q = int(gmpy2.next_prime(p))
n = p*q
m = bytes_to_long(FLAG)
c = pow(m,e,n)
print(n)
print(c)

output.txt：

177606504836499246970959030226871608885969321778211051080524634084516973331441644993898029573612290095853069264036530459253652875586267946877831055147546910227100566496658148381834683037366134553848011903251252726474047661274223137727688689535823533046778793131902143444408735610821167838717488859902242863683
1457390378511382354771000540945361168984775052693073641682375071407490851289703070905749525830483035988737117653971428424612332020925926617395558868160380601912498299922825914229510166957910451841730028919883807634489834128830801407228447221775264711349928156290102782374379406719292116047581560530382210049

非常常规的RSA题目，给出了n，c，e。加密算法也比较常规，其中p和q的数值比较接近，可以直接对n开方来得到p和q的数值。根据这一思路，写个脚本求解：

import libnum
import sympy
n = 177606504836499246970959030226871608885969321778211051080524634084516973331441644993898029573612290095853069264036530459253652875586267946877831055147546910227100566496658148381834683037366134553848011903251252726474047661274223137727688689535823533046778793131902143444408735610821167838717488859902242863683
c = 1457390378511382354771000540945361168984775052693073641682375071407490851289703070905749525830483035988737117653971428424612332020925926617395558868160380601912498299922825914229510166957910451841730028919883807634489834128830801407228447221775264711349928156290102782374379406719292116047581560530382210049
e = 65537

base = libnum.nroot(n,2)
p = sympy.prevprime(base)
q = sympy.nextprime(base)

assert(p*q == n)
phi = (p-1)*(q-1)
d = libnum.invmod(e,phi)
m = pow(c,d,n)
flag = libnum.n2s(m)
print flag

运行脚本，得到flag：

actf{p_and_q_should_not_be_so_close_in_value}

0x7 鸡藕椒盐味

题目比较奇怪，那看看题目究竟什么样的题目吧：

公司食堂最新出了一种小吃，叫鸡藕椒盐味汉堡，售价八块钱，为了促销，上面有一个验证码，输入后可以再换取一个汉堡。但是问题是每个验证码几乎都有错误,而且打印的时候倒了一下。小明买到了一个汉堡，准备还原验证码，因为一个吃不饱啊验证码如下：1100 1010 0000 ,而且打印的时候倒了一下。把答案哈希一下就可以提交了。(答案为正确值(不包括数字之间的空格)的32位md5值的小写形式) 注意：得到的 flag 请包上 flag{} 提交

可以从题目中提炼出一些关键信息：

  1. 验证码：1100 1010 0000
  2. 打印的时候倒了一下
  3. 哈希提交。(答案为正确值(不包括数字之间的空格)的32位md5值的小写形式)
  4. 而且打印的时候倒了一下

这些信息咋看不出来什么东西呀，嘤嘤嘤~

仔细看看题目，鸡藕椒盐味是不是很像是奇偶校验位，应该是这个。可以沿着奇偶校验位的这个方向沿着方向走一下，看看能否找到突破的入口。沿着奇偶校验位进行探索，发现有个海明校验码的东西，而且还有一个海明纠错码。这道题目，应该是使用海明纠错码来进行纠错的，那么什么是海明码呢？可以看看这篇知乎文章大致原理都有了，根据这篇文章阐述的原理进行脚本编写，应该就可以找出答案了。写下脚本：

import hashlib

c = "110010100000"
c_r = c[::-1]
c_list =[]

for i in c_r:
    c_list.append(int(i))

h1 = c_list[2]^c_list[4]^c_list[6]^c_list[8]^c_list[10] ^ c_list[0]
h2 = c_list[2]^c_list[5]^c_list[6]^c_list[9]^c_list[10] ^ c_list[1]
h3 = c_list[4]^c_list[5]^c_list[6]^c_list[11] ^ c_list[3]
h4 = c_list[8]^c_list[9]^c_list[10]^c_list[11] ^ c_list[7]

mistake_bit = int(str(h4)+str(h3)+str(h2)+str(h1),2)
if c_list[mistake_bit-1]==1:
    c_list[mistake_bit-1] = 0
else:
    c_list[mistake_bit-1] = 1

m = ""
for i in c_list:
    m += str(i)
m = m[::-1]
flag = hashlib.md5(m).hexdigest()
print flag

运行一下脚本，flag就有了：

d14084c7ceca6359eaac6df3c234dd3b

0x8 [ACTF新生赛2020]crypto-classic0

看题目就知道，这个是古典密码学的题目，看看究竟是在考察些什么古典密码吧：

附件文件里面有三个文件，一个cipher文本文件，一个hint.txt提示文件和一个howtoencrypt.zip加密压缩文件。

hint.txt:

哼，压缩包的密码？这是小Z童鞋的生日吧==

应该是压缩包文件的提示内容

另一个cipher文件需要进行探测一下，使用file命令简单探测一下

cipher: International EBCDIC text, with no line terminators

是一个纯文本文件，这样可以使用记事本直接打开：

Ygvdmq[lYate[elghqvakl}

这个提示应该告诉压缩文件的密码长度，这里可以使用压缩爆破工具进行爆破，经过爆破可以得到密码是19990306。使用密码进行解压得到一个c语言源程序，应该是加密算法。

classic0.c：

#include<stdio.h>

char flag[25] = ***

int main()
{
	int i;
	for(i=0;i<25;i++)
	{
		flag[i] -= 3;
		flag[i] ^= 0x7;
		printf("%c",flag[i]);
	}
	return 0; 
}

根据这个c程序算法设计一个逆算法应该就可以进行求解，这里写个c语言逆算法：

#include <stdio.h>
int main()
{
    char flag[25];
    FILE *fp = NULL;
    fp = fopen("cipher.txt","r");
    fgets(flag,25,fp);

    for(int i=0;i<25;i++)
    {
        flag[i] ^= 0x7;
        flag[i] += 3;
        printf("%c", flag[i]);
    }
    fclose(fp);
    return 0;
}

编译并执行得到flag：

actf{my_naive_encrytion}

0x9 [RoarCTF2019]babyRSA

RSA题目，看看这次又有什么样的惊喜！

import sympy
import random

def myGetPrime():
    A= getPrime(513)
    print(A)
    B=A-random.randint(1e3,1e5)
    print(B)
    return sympy.nextPrime((B!)%A)
p=myGetPrime()
#A1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467234407
#B1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467140596

q=myGetPrime()
#A2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858418927
#B2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858351026

r=myGetPrime()

n=p*q*r
#n=85492663786275292159831603391083876175149354309327673008716627650718160585639723100793347534649628330416631255660901307533909900431413447524262332232659153047067908693481947121069070451562822417357656432171870951184673132554213690123308042697361969986360375060954702920656364144154145812838558365334172935931441424096270206140691814662318562696925767991937369782627908408239087358033165410020690152067715711112732252038588432896758405898709010342467882264362733
c=pow(flag,e,n)
#e=0x1001
#c=75700883021669577739329316795450706204502635802310731477156998834710820770245219468703245302009998932067080383977560299708060476222089630209972629755965140317526034680452483360917378812244365884527186056341888615564335560765053550155758362271622330017433403027261127561225585912484777829588501213961110690451987625502701331485141639684356427316905122995759825241133872734362716041819819948645662803292418802204430874521342108413623635150475963121220095236776428
#so,what is the flag?

代码挺长的，考察的内容是阶乘？是不是要考虑一下递归的方式进行求解？那估计慢的不得了，肯定有更好的算法进行求解，垃圾的我去翻翻了王小云院士写的数学书，发现有个叫威尔逊定理的东西可以用于算法的求解，这道题目就是考察威尔逊定理的内容。

威尔逊定理的内容很简单： $$ (p-1)! \equiv -1 (\text{mod} p) $$ 其中p为素数

根据威尔逊定理就可以对这道题目进行简单的推导求解：

首先，了解一下算法的加密过程： $$ p \equiv B_1! \text{mod}A_1 $$

$$ q\equiv B_2! \text{mod}A_2 $$

$$ r \equiv B_3! \text{mod}A_3 $$

$$ n = p \cdot q \cdot r $$

$$ c \equiv m ^e \text{mod} n $$

加密过程也是非常清晰，问题是如何更快地计算出p和q的数值，这里就需要用到威尔逊定理来进行相应的计算：

根据代码中的算法，可以看出有一句代码表明：

B=A-random.randint(1e3,1e5)

因此，这里可以根据这一信息使用威尔逊定理进行推导： $$ (A-1)! \equiv -1\text{mod}A $$

$$ (A-1)\cdot(A-2)\cdot(A-3)\dots(B+1)\cdot(B)\dots2\cdot1\equiv-1\text{mod} A $$

$$ (A-1)\cdot(A-2)\cdot(A-3)\dots(B+1)\cdot(B!)\equiv-1\text{mod} A $$

$$ (A-1)\cdot(A-2)\cdot(A-3)\dots(B+1)\cdot(B!)\equiv(A-1)\text{mod} A $$

$$ (A-2)\cdot(A-3)\dots(B+1)\cdot(B!)\equiv1\text{mod} A $$

因此要求(B!)%A可以通过求$(A-2)\cdot(A-1)\dots(B+1)$的逆元来进行优化求解，提高运算的效率，这样分析完，这道题目的大致思路也就有了。根据这个思路，写个脚本求解就好：

import libnum
import sympy

def wilson(A,B):
    tmp = 1
    for i in range(B+1,A-1):
        tmp *= i
        tmp %= A
    tmp_inv = libnum.invmod(tmp,A)
    result = sympy.nextprime(tmp_inv)
    return result

A1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467234407
B1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467140596
A2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858418927
B2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858351026
n=85492663786275292159831603391083876175149354309327673008716627650718160585639723100793347534649628330416631255660901307533909900431413447524262332232659153047067908693481947121069070451562822417357656432171870951184673132554213690123308042697361969986360375060954702920656364144154145812838558365334172935931441424096270206140691814662318562696925767991937369782627908408239087358033165410020690152067715711112732252038588432896758405898709010342467882264362733
e=0x1001
c=75700883021669577739329316795450706204502635802310731477156998834710820770245219468703245302009998932067080383977560299708060476222089630209972629755965140317526034680452483360917378812244365884527186056341888615564335560765053550155758362271622330017433403027261127561225585912484777829588501213961110690451987625502701331485141639684356427316905122995759825241133872734362716041819819948645662803292418802204430874521342108413623635150475963121220095236776428
p = wilson(A1, B1)
q = wilson(A2, B2)
r = n //(p*q)
phi = (p-1)*(q-1)*(r-1)

d = libnum.invmod(e,phi)
m = pow(c,d,n)
flag = libnum.n2s(m)
print flag

运行脚本，拿到flag

RoarCTF{wm-CongrAtu1ation4-1t4-ju4t-A-bAby-R4A}

这道题目还是挺有意思的，考的内容比较综合，是一道不错的RSA题目。

0xA [AFCTF2018]可怜的RSA

这次又有什么花样呢？可怜的RSA

瞧瞧题目：

附件里面有两个文件，一个flag.enc的密文文件和一个public.enc的公钥文件

这道题目的思路应该是读取公钥文件，然后对公钥文件中的modulus进行大数分解，最后使用实际应用的会用到的填充方法进行解密。

思路很简单，根据思路写个脚本：

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
from base64 import b64decode
from sympy.ntheory.factor_ import smoothness
import sympy
import libnum

with open("public.key","rb") as f:
    rsakey = RSA.importKey(f.read())
    n = rsakey.n
    e = rsakey.e

result = sympy.factorint(n)
p = long(result.keys()[0])
q = long(result.keys()[1])

phi = (p-1)*(q-1)
d = libnum.invmod(e,phi)

key_info = RSA.construct((n,e,d,p,q))
key = RSA.importKey(key_info.exportKey())
key = PKCS1_OAEP.new(key)

with open("flag.enc","rb") as f:
    c = b64decode(f.read())
    flag = key.decrypt(c)

print flag

运行脚本，得到flag：

afctf{R54_|5_$0_B0rin9}

这道题目比较偏向于应用方面，涉及到了RSA的加密算法填充方法，不再是裸露再外面的RSA算法。可以进一步了解一下。

0xB 救主捷径

救主捷径，应该是一个脑洞题目，看看题目：

一个名叫CPU的神秘大陆有26个国家，有些国家之间会有一条无向路，每条路径都有不同的长度和一段神秘代码，救世主尼奥要从国家1出发，赶往国家26拯救大陆，请你帮助救世主选择最短路径，而走过的路的神秘代码连接起来便是flag。 以下是数行数据，每行第一个，第二个数字代表这条路的两个端点国家，第三个数字代表路途长度，最后一个字符串便是神秘代码。路在附件中~ 帮助救世主尼奥吧，他快被吓尿了。。。 注意：得到的 flag 请包上 flag{} 提交

看样子应该是广度优先算法的应用，题目中只说到了距离，没有涉及权重，应该就是广度优先算法，看看附件是什么吧！

1 2 100 FLAG{
2 3 87 AFQWE
2 4 57 ETKLS
2 5 50 WEIVK
2 6 51 AWEIW
3 7 94 QIECJF
3 8 78 QSXKE
3 9 85 QWEIH
4 13 54 WQOJF
4 14 47 KDNVE
4 15 98 QISNV
5 10 43 AEWJV
5 11 32 QWKXF
5 12 44 ASJVL
6 16 59 ASJXJ
6 17 92 QJXNV
6 18 39 SCJJF
6 23 99 SJVHF
7 19 99 WJCNF
8 20 96 SKCNG
9 20 86 SJXHF
10 21 60 SJJCH
11 21 57 SJHGG
12 22 47 SJCHF
14 10 55 EJFHG
16 17 59 ASJVH
18 12 53 SJFHG
18 24 93 SHFVG
21 22 33 SJFHB
19 25 88 ASHHF
20 25 96 SJVHG
22 25 23 SJVHJ
25 26 75 SDEV}

根据附件文件绘制个图，也方便进行广度优先算法的使用和计算：

image-20210630202812243

根据图示来使用广度优先算法的设计，这里可以使用比较经典的Dijkstra算法来求解最优问题，使用python进行算法实现：

import networkx as nx

def Dijkstra(G, start, end):
    RG = G.reverse();
    dist = {};
    previous = {}
    for v in RG.nodes():
        dist[v] = float('inf')
        previous[v] = 'none'
    dist[end] = 0
    u = end
    while u != start:
        u = min(dist, key=dist.get)
        distu = dist[u]
        del dist[u]
        for u, v in RG.edges(u):
            if v in dist:
                alt = distu + RG[u][v]['weight']
                if alt < dist[v]:
                    dist[v] = alt
                    previous[v] = u
    path = (start,)
    last = start
    while last != end:
        nxt = previous[last]
        path += (nxt,)
        last = nxt
    return path

G = nx.DiGraph()
datalist = []
datadict = {}
with open("data.txt","rb") as f:
    lines = f.readlines()
    for line in lines:
        line = line.strip("\r\n")
        datalist.append(line)

for i in datalist:
    data = i.split(" ")
    datadict[data[0]+"-"+data[1]] = data[3]
    G.add_edge(int(data[0]),int(data[1]),weight=int(data[2]))
rs = Dijkstra(G, 1, 26)
flag = ""
for i in range(len(rs)-1):
    flag += datadict[str(rs[i])+"-"+str(rs[i+1])]

print flag
print flag.lower()

运行脚本，即可获得flag：

FLAG{WEIVKASJVLSJCHFSJVHJSDEV}
flag{weivkasjvlsjchfsjvhjsdev}

0xC [网鼎杯 2020 青龙组]boom

这道题目，boom的意思爆炸，这道题目应该会有点意思。

来看看题目呀：

看到附件内容是一个exe文件，可执行文件，有点像是逆向题目。首先执行一下exe程序，看看程序会给我们展示什么样的效果？

image-20210630213025137

按一下任意键：

image-20210630213049828

应该是一个进行答题的exe程序，这里有两条路可以走，第一条路就是通过答题来拿到flag，而第二条路就是通过逆向工具进行逆向。下面绘制一个拓扑图来看看这道题目的思路：

image-20210630214244533

无论怎么走，都需要解决题目的问题，这道题目的本意应该是考察数学技能，这道题目就使用IDA逆向工具进行分析，打开IDA看到伪代码：

int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v3; // eax
  char Str[50]; // [esp+24h] [ebp-128h] BYREF
  unsigned __int8 v6[50]; // [esp+56h] [ebp-F6h] BYREF
  _DWORD v7[16]; // [esp+88h] [ebp-C4h]
  unsigned int v8[22]; // [esp+C8h] [ebp-84h] BYREF
  __int64 v9; // [esp+120h] [ebp-2Ch] BYREF
  int v10; // [esp+12Ch] [ebp-20h] BYREF
  int v11; // [esp+130h] [ebp-1Ch] BYREF
  int v12; // [esp+134h] [ebp-18h] BYREF
  int v13; // [esp+138h] [ebp-14h]
  int i; // [esp+13Ch] [ebp-10h]

  __main();
  menu();
  system("pause");
  system("cls");
  v7[0] = 70;
  v7[1] = 229;
  v7[2] = 239;
  v7[3] = 230;
  v7[4] = 22;
  v7[5] = 90;
  v7[6] = 90;
  v7[7] = 251;
  v7[8] = 54;
  v7[9] = 18;
  v7[10] = 23;
  v7[11] = 68;
  v7[12] = 106;
  v7[13] = 45;
  v7[14] = 189;
  v7[15] = 1;
  puts("first:this string md5:46e5efe6165a5afb361217446a2dbd01");
  scanf("%s", Str);
  MD5Init(v8);
  v3 = strlen(Str);
  MD5Update((int)v8, Str, v3);
  MD5Final(v8, v6);
  v13 = 1;
  for ( i = 0; i <= 15; ++i )
  {
    if ( v6[i] != v7[i] )
    {
      v13 = 0;
      break;
    }
  }
  if ( v13 != 1 )
  {
    printf("Game over");
    system("pause");
    exit(0);
  }
  puts("Great next level");
  system("pause");
  system("cls");
  puts("This time:Here are have some formulas");
  puts("3x-y+z=185");
  puts("2x+3y-z=321");
  puts("x+y+z=173");
  printf("input: x = ");
  scanf("%d", &v12);
  printf("input: y = ");
  scanf("%d", &v11);
  printf("input : z = ");
  scanf("%d", &v10);
  if ( 3 * v12 - v11 + v10 != 185 || 2 * v12 + 3 * v11 - v10 != 321 || v11 + v12 + v10 != 173 )
  {
    printf("Game over");
    exit(0);
  }
  printf("Great last level coming...");
  printf("pause");
  system("cls");
  puts("Last time: Kill it");
  puts("x*x+x-7943722218936282=0");
  printf("input x: ");
  scanf("%lld", &v9);
  if ( v9 * (v9 + 1) != 0x1C38C5F50DD7DALL )
  {
    printf("Game over");
    exit(0);
  }
  puts("Great This is your FLAG");
  return 0;
}

可以从逆向的伪代码中找到一些比较关键的代码：

原始内存数据

  system("pause");
  system("cls");
  v7[0] = 70;
  v7[1] = 229;
  v7[2] = 239;
  v7[3] = 230;
  v7[4] = 22;
  v7[5] = 90;
  v7[6] = 90;
  v7[7] = 251;
  v7[8] = 54;
  v7[9] = 18;
  v7[10] = 23;
  v7[11] = 68;
  v7[12] = 106;
  v7[13] = 45;
  v7[14] = 189;
  v7[15] = 1;

level 1关键代码

puts("first:this string md5:46e5efe6165a5afb361217446a2dbd01");
  scanf("%s", Str);
  MD5Init(v8);
  v3 = strlen(Str);
  MD5Update((int)v8, Str, v3);
  MD5Final(v8, v6);
  v13 = 1;
  for ( i = 0; i <= 15; ++i )
  {
    if ( v6[i] != v7[i] )
    {
      v13 = 0;
      break;
    }
  }
  if ( v13 != 1 )
  {
    printf("Game over");
    system("pause");
    exit(0);
  }
  puts("Great next level");
  system("pause");
  system("cls");

level 2关键代码

  puts("This time:Here are have some formulas");
  puts("3x-y+z=185");
  puts("2x+3y-z=321");
  puts("x+y+z=173");
  printf("input: x = ");
  scanf("%d", &v12);
  printf("input: y = ");
  scanf("%d", &v11);
  printf("input : z = ");
  scanf("%d", &v10);
  if ( 3 * v12 - v11 + v10 != 185 || 2 * v12 + 3 * v11 - v10 != 321 || v11 + v12 + v10 != 173 )
  {
    printf("Game over");
    exit(0);
  }
  printf("Great last level coming...");
  printf("pause");
  system("cls");

level 3关键代码

  puts("Last time: Kill it");
  puts("x*x+x-7943722218936282=0");
  printf("input x: ");
  scanf("%lld", &v9);
  if ( v9 * (v9 + 1) != 0x1C38C5F50DD7DALL )
  {
    printf("Game over");
    exit(0);
  }
  puts("Great This is your FLAG");
  return 0;
}

根据三个关键代码进行代码审计，代码审计还是老基本功了

level 1的关键代码是判断md5值的，使用[cmd5](md5在线解密破解,md5解密加密 (cmd5.com))就可以解出，即：

Str = "en5oy";

level 2的关键代码是解方程组，使用sage求解：

var("x y z") 
solve([3*x-y+z==185,2*x+3*y-z==321,x+y+z==173],x,y,z)

运行就可以求解得出x，y和z的值：

[[x == 74, y == 68, z == 31]]

因此，可以得到：

v12 == 74;
v11 == 68;
v10 == 31;

leve 3的关键代码就是解一元二次方程或者可以简化，就是解v9 * (v9 + 1) == 7943722218936282，可以使用sage直接进行求解：

 solve(x * (x + 1) == 7943722218936282,x)

运行可得：

 [x == 89127561, x == -89127562]

由此，可以得到：

v9 == 89127561;

然后进行组合就可以得到flag：

flag{en5oy_746831_89127561}

0xD [RoarCTF2019]RSA

又来了一道RSA题目，看看有些什么东西吧！

A=(((y%x)**5)%(x%y))**2019+y**316+(y+1)/x
p=next_prime(z*x*y)
q=next_prime(z)
A =  2683349182678714524247469512793476009861014781004924905484127480308161377768192868061561886577048646432382128960881487463427414176114486885830693959404989743229103516924432512724195654425703453612710310587164417035878308390676612592848750287387318129424195208623440294647817367740878211949147526287091298307480502897462279102572556822231669438279317474828479089719046386411971105448723910594710418093977044179949800373224354729179833393219827789389078869290217569511230868967647963089430594258815146362187250855166897553056073744582946148472068334167445499314471518357535261186318756327890016183228412253724
n =  117930806043507374325982291823027285148807239117987369609583515353889814856088099671454394340816761242974462268435911765045576377767711593100416932019831889059333166946263184861287975722954992219766493089630810876984781113645362450398009234556085330943125568377741065242183073882558834603430862598066786475299918395341014877416901185392905676043795425126968745185649565106322336954427505104906770493155723995382318346714944184577894150229037758434597242564815299174950147754426950251419204917376517360505024549691723683358170823416757973059354784142601436519500811159036795034676360028928301979780528294114933347127
c =  41971850275428383625653350824107291609587853887037624239544762751558838294718672159979929266922528917912189124713273673948051464226519605803745171340724343705832198554680196798623263806617998072496026019940476324971696928551159371970207365741517064295956376809297272541800647747885170905737868568000101029143923792003486793278197051326716680212726111099439262589341050943913401067673851885114314709706016622157285023272496793595281054074260451116213815934843317894898883215362289599366101018081513215120728297131352439066930452281829446586562062242527329672575620261776042653626411730955819001674118193293313612128

这道题目涉及到了模线性方程，需要考虑模线性方程该如何进行求解。目前，我尚未找到除了爆破外的解法，因此，这次就用爆破来求解，但是要用爆破求解就必然需要考虑爆破时间，如何来缩短爆破时间。

再来看看题目：

题目给定了$A$ 、$n$ 、$c$ ，且满足

$$ A = [ (y \text{ % } x ) ^ 5 \text{ % } ( x \text{ % } y ) ] ^ {2019} + y ^ {316} + \frac{y+1}{x} $$

根据方程可以得知 $x | y+1$ 因此，可以推断出来 $ y \equiv -1 \text{mod}x$ 。而且由于 $ y+1$ 比 $x$ 大，因此，应该有$x%y=x$ 。于是：

$$ [(y\text{ % }x)^5\text{ % }(x\text{ % }y)]^{2019} = (x-1)^{2019} $$

因此，可以得到：

$$ A = (x-1)^{2019} + y^{316} + \frac{y+1}{x} $$

通过观察发现$A$的值是一个2015个bit的数。而且：

$$ A \geq (x-1)^{2019} $$

显然只有：

$$ x \leq 2 $$

因此进行讨论$x=1 $和$x =2$这两种情况时，发现存在有唯一解，即：

$$ (x,y) = (2,83) $$

现在$ x $和$ y $知道了，看一下算法的其他细节吧

p=next_prime(z*x*y)
q=next_prime(z)

由于已经知道了$x$和$y$ ，因此，可以知道： $$ n = p \cdot q $$

$$ n = z\cdot x \cdot y \cdot z $$

因此可以推出： $$ q \approx \sqrt{\frac{n}{x\cdot y}} $$

$$ p \approx x\cdot y \cdot q $$

于是可以通过爆破求出 $p$ 和 $q$ 来进行求解，但是题目没有告知 $e$ ，可以先猜测 $e = 65537$ 。根据这个思路写个完整脚本：^1^

import libnum, gmpy2, itertools

A =  2683349182678714524247469512793476009861014781004924905484127480308161377768192868061561886577048646432382128960881487463427414176114486885830693959404989743229103516924432512724195654425703453612710310587164417035878308390676612592848750287387318129424195208623440294647817367740878211949147526287091298307480502897462279102572556822231669438279317474828479089719046386411971105448723910594710418093977044179949800373224354729179833393219827789389078869290217569511230868967647963089430594258815146362187250855166897553056073744582946148472068334167445499314471518357535261186318756327890016183228412253724
n =  117930806043507374325982291823027285148807239117987369609583515353889814856088099671454394340816761242974462268435911765045576377767711593100416932019831889059333166946263184861287975722954992219766493089630810876984781113645362450398009234556085330943125568377741065242183073882558834603430862598066786475299918395341014877416901185392905676043795425126968745185649565106322336954427505104906770493155723995382318346714944184577894150229037758434597242564815299174950147754426950251419204917376517360505024549691723683358170823416757973059354784142601436519500811159036795034676360028928301979780528294114933347127
c =  41971850275428383625653350824107291609587853887037624239544762751558838294718672159979929266922528917912189124713273673948051464226519605803745171340724343705832198554680196798623263806617998072496026019940476324971696928551159371970207365741517064295956376809297272541800647747885170905737868568000101029143923792003486793278197051326716680212726111099439262589341050943913401067673851885114314709706016622157285023272496793595281054074260451116213815934843317894898883215362289599366101018081513215120728297131352439066930452281829446586562062242527329672575620261776042653626411730955819001674118193293313612128

x , y = 2, 83
A_t = (((y%x)**5)%(x%y))**2019+y**316+(y+1)/x
assert A_t == A

for q in itertools.count(gmpy2.iroot(n//166,2)[0]):
    if n%q == 0:
        break
p = n//q

assert n == p*q
e = 65537
phi = (p-1)*(q-1)
d = gmpy2.invert(e, phi)
m = gmpy2.powmod(c,d,n)
flag = libnum.n2s(m)

print flag

运行脚本，即可得到flag：

RoarCTF{wm-l1l1ll1l1l1l111ll}

这道题目的求解方法是优雅与暴力相结合的，能展现出一种特别的美感，数学推导的优雅和蛮力攻击的暴力。野蛮与优雅，展现出密码学独有美，题目还是挺不错的。

0xE [AFCTF2018]Single

单一的，看看题目给了我们什么包裹吧！

一个Cipher.txt密文文件和一个Encode.cpp加密文件。

Encode.cpp

#include <bits/stdc++.h>
using namespace std;
int main()
{
	freopen("Plain.txt","r",stdin);
	freopen("Cipher.txt","w",stdout);
	map<char, char> f;
	int arr[26];
	for(int i=0;i<26;++i){
		arr[i]=i;
	}
	random_shuffle(arr,arr+26);
	for(int i=0;i<26;++i){
		f['a'+i]='a'+arr[i];
		f['A'+i]='A'+arr[i];
	}
	char ch;
	while((ch=getchar())!=EOF){
		if(f.count(ch)){
			putchar(f[ch]);
		}else{
			putchar(ch);
		}
	}
	return 0;
}

Cipher.txt

Jmqrida rva Lfmz (JRL) eu m uqajemf seny xl enlxdomrexn uajiderc jxoqarerexnu. Rvada mda rvdaa jxooxn rcqau xl JRLu: Paxqmdyc, Mrrmjs-Yalanja mny oekay.

Paxqmdyc-urcfa JRLu vmu m jxiqfa xl giaurexnu (rmusu) en dmnza xl jmrazxdeau. Lxd akmoqfa, Wab, Lxdanuej, Jdcqrx, Benmdc xd uxoarvenz afua. Ramo jmn zmen uxoa qxenru lxd atadc uxftay rmus. Oxda qxenru lxd oxda jxoqfejmray rmusu iuimffc. Rva nakr rmus en jvmen jmn ba xqanay xnfc mlrad uxoa ramo uxfta qdatexiu rmus. Rvan rva zmoa reoa eu xtad uio xl qxenru uvxwu cxi m JRL wenad. Lmoxiu akmoqfa xl uijv JRL eu Yaljxn JRL gimfu.

Waff, mrrmjs-yalanja eu mnxrvad enradaurenz seny xl jxoqarerexnu. Vada atadc ramo vmu xwn narwxds(xd xnfc xna vxur) werv tifnmdmbfa uadtejau. Cxid ramo vmu reoa lxd qmrjvenz cxid uadtejau mny yatafxqenz akqfxeru iuimffc. Ux, rvan xdzmnehadu jxnnajru qmdrejeqmnru xl jxoqarerexn mny rva wmdzmoa urmdru! Cxi uvxify qdxrajr xwn uadtejau lxd yalanja qxenru mny vmjs xqqxnanru lxd mrrmjs qxenru. Veurxdejmffc rveu eu m ledur rcqa xl JRLu, atadcbxyc snxwu mbxir YAL JXN JRL - uxoarvenz fesa m Wxdfy Jiq xl mff xrvad jxoqarerexnu.

Oekay jxoqarerexnu omc tmdc qxuuebfa lxdomru. Er omc ba uxoarvenz fesa wmdzmoa werv uqajemf reoa lxd rmus-bmuay afaoanru (a.z. IJUB eJRL).

JRL zmoau xlran rxijv xn omnc xrvad muqajru xl enlxdomrexn uajiderc: jdcqrxzdmqvc, urazx, benmdc mnmfcueu, datadua anzanaadenz, oxbefa uajiderc mny xrvadu. Zxxy ramou zanadmffc vmta urdxnz useffu mny akqadeanja en mff rvaua euuiau.

Iuimffc, lfmz eu uxoa urdenz xl dmnyxo ymrm xd rakr en uxoa lxdomr. Akmoqfa mljrl{Xv_I_lxiny_er_neja_rDc}

看样子，应该是一个替换加密的题目。可以直接扔进词频分析网站进行词频分析来求解，得到求解结果：

	Capture the Flag (CTF) is a special kind of information security competitions. There are three common types of CTFs: Jeopardy, Attack-Defence and mixed. Jeopardy-style CTFs has a couple of questions (tasks) in range of categories. For example, Web, Forensic, Crypto, Binary or something else. Team can gain some points for every solved task. More points for more complicated tasks usually. The next task in chain can be opened only after some team solve previous task. Then the game time is over sum of points shows you a CTF winer. Famous example of such CTF is Defcon CTF quals. Well, attack-defence is another interesting kind of competitions. Here every team has own network(or only one host) with vulnarable services. Your team has time for patching your services and developing exploits usually. So, then organizers connects participants of competition and the wargame starts! You should protect own services for defence points and hack opponents for attack points. Historically this is a first type of CTFs, everybody knows about DEF CON CTF - something like a World Cup of all other competitions. Mixed competitions may vary possible formats. It may be something like wargame with special time for task-based elements (e.g. UCSB iCTF). CTF games often touch on many other aspects of information security: cryptography, stego, binary analysis, reverse engeneering, mobile security and others. Good teams generally have strong skills and experience in all these issues. Usually, flag is some string of random data or text in some format. Example afctf{Oh_U_found_it_nice_tRy}

快瞧，flag就在最后一行！这题就这么完了，有工具真是好(*^_^*)

0xF [WUSTCTF2020]B@se

看题目，应该是一道BASE的题目。看看附件会给些什么吧！

密文：MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbvD==
JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs****kxyz012789+/

oh holy shit, something is missing...

这道题目的应该是变换了编码表的base64，修改并替换一下使用base64进行解密，但是变换后的编码表中有四个字母是顺序位置，因此可能会有24种可能的情况。需要进行爆破，首先要知道是那些字符缺失了，然后对缺失字符的数组进行爆破，然后在使用base64替换来进行base64的求解。根据这个思路来写个脚本：

import base64
import string
import itertools

base64_stand = string.uppercase+string.lowercase+string.digits+"+/"
new_base64 = "JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs****kxyz012789+/"
lose_letters =""
c = "MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbvD=="
flags = []

def base64_change(c,s,n):
    result = ""
    for i in c:
        if n.find(i) != -1:
            result += s[n.find(i)]
        else:
            result += i
    result = base64.b64decode(result)
    if len(result)+2 == len(repr(result)):
        return result
    else:
        return

for i in base64_stand:
    test = new_base64.find(i)
    if test == -1:
        lose_letters += i

for i in list(itertools.permutations(lose_letters,4)):
    m = "".join(i)
    base64_new = new_base64.replace("****", m)
    flag = base64_change(c, base64_stand, base64_new)
    if flag and "2020" in flag:
        flags.append(flag)

for flag in set(flags):
    print flag

运行脚本，得到可能的flag：

wctf2020{base64_1s_v3ry_e@sy_and_fuN}
wctf2020{bare64_!r_v2ry_e@ry_and_fuN}

经过测试，flag应该是：

wctf2220{base64_1s_v3ry_e@sy_and_fuN}

本题到此结束！

参考

  1. CTF RSA题解集 (ruanx.net)