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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}