首先拿到题目 分析可得第一行可得是斐波那契数列得乱序(加强数感),于是可以猜测神秘字符串可能也是flag得乱序,且两者得乱序规则是一样得,接着通过编程可以得到flag字符串
li1=[1 ,233 ,3 ,2584 ,1346269, 144 ,5 ,196418, 21 ,1597, 610 ,377 ,10946, 89 ,514229, 987, 8 ,55, 6765, 2178309, 121393, 317811, 46368, 4181, 1, 832040 ,2 ,28657, 75025, 34, 13, 17711 ] li2=[3,6,9,6,8,8,5,3,8,8,2,1,1,6,7,2,5,5,4,7,3,4,2,1,7,6,9,5,2,2,8,6] li4=[] lians=[] for i in range(32): li4.append(0) lians.append(0) li3=[1,1] for i in range(2,32): li3.append(li3[i-1]+li3[i-2]) print(li3) for i in range(32): li4[i]=li3.index(li1[i]) li4[24]=1 for i in range(32): lians[li4[i]]=li2[i] ans="flag{" for i in range(32): ans=ans+str(lians[i]) ans=ans+'}' print(ans)得到flag flag{37995588256861228614165223347687}
