用于记录
通过分析代码得出主要信息
s=斐波那契数列的第 1000个数
out文本中数值是最终结果c
m=(flag+随机字符串)通过libnum.s2n()函数转为数值
c=m^s
所以最终结果为
贴上代码给自己留个备份
import os
import libnum
from secret import flag
from Crypto.Util.number import *
import math
def fibonacci(n):
if n <= 0:
return "Input should be a positive integer."
fib_seq = [0, 1]
while len(fib_seq) < n:
fib_seq.append(fib_seq[-1] + fib_seq[-2])
return fib_seq[n - 1]
result = fibonacci(1000)
s = fibonacci(1001)
c=43104378128345818181217961835377190975779804452524643191544804229536124095677294719566215359919831933542699064892141754715180028183150724886016542388159082125737677224886528142312511700711365919689756090950704
m = c^s
b=libnum.n2s(m)
print()
print(b)