SameMod

题目下载

n相同，很明显是共模攻击。

# -*- coding: utf-8 -*-

from libnum import n2s,s2n
from gmpy2 import invert
# 欧几里得算法
def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)

def main():
    n = 6266565720726907265997241358331585417095726146341989755538017122981360742813498401533594757088796536341941659691259323065631249
    c1 = 3453520592723443935451151545245025864232388871721682326408915024349804062041976702364728660682912396903968193981131553111537349
    c2 = 5672818026816293344070119332536629619457163570036305296869053532293105379690793386019065754465292867769521736414170803238309535
    e1 = 773
    e2 = 839
    s = egcd(e1, e2)
    s1 = s[1]
    s2 = s[2]
    # 求模反元素
    if s1<0:
        s1 = - s1
        c1 = invert(c1, n)
    elif s2<0:
        s2 = - s2
        c2 = invert(c2, n)

    m = pow(c1,s1,n)*pow(c2,s2,n) % n
    print m
    print hex(m)
    print n2s(m)

if __name__ == '__main__':
    main()

输出

1021089710312311910410111011910111610410511010710511610511511211111511510598108101125

看着非常像ascii码，解码即可得到flag。