【转】CTF各种加密算法解密集成（java，支持Python）

#coding=utf-8 #author:MrLyn、我擦咧什么鬼 #team:米斯特安全 def peig(data): code = dict(A='aaaaa',B='aaaab',C='aaaba',D='aaabb',E='aabaa',F='aabab', G='aabba',H='aabbb',I='abaaa',J='abaab',K='ababa',L='ababb', M='abbaa',N='abbab',O='abbba',P='abbbb',Q='baaaa',R='baaab', S='baaba',T='baabb',U='babaa',V='babab',W='babba',X='babbb', Y='bbaaa',Z='bbaab',a='AAAAA',g='AABBA',n='ABBAA',t='BAABA', b='AAAAB',h='AABBB',o='ABBAB',u='BAABB',c='AAABA',i='ABAAA', j='ABAAA',p='ABBBA',d='AAABB',k='ABAAB',q='ABBBB',x='BABAB', e='AABAA',l='ABABA',r='BAAAA',y='BABBA',f='AABAB',m='ABABB', s='BAAAB',z='BABBB',v='BAABB',w='BABAA') num1 = 0 #aaaab aaaaa aaaba abbab abbaa abaaa baaab abbaa abbab baaba aabab abbab abbab aaabb s = '' l=[] answer = '' for i in range(len(data)/5): l.append(data[num1:num1+5]) num1+=5 answer = ''.join(x[0] for i in l for x in code.items() if i == x[1] and x[0] != 'j') answer_list = [''] num = 2 for i in answer: if i != 'i': for key,value in enumerate(answer_list): answer_list[key] = answer_list[key] + i elif i == 'i': for key,value in enumerate(answer_list): answer_list.append(value + 'j') answer_list[key] = answer_list[key] + 'i' if len(answer_list) == num: num = num * 2 break return answer_list def run(data): ans = '\n'.join(peig(data)) return ans