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WriteUp-CTFHub-you raise me up

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2020-网鼎杯-青龙组-Crypto-you raise me up

题目考点

  • 离散对数

解题思路

  • 下载题目附件： 
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    #!/usr/bin/env python
    # -*- coding: utf-8 -*-
    from Crypto.Util.number import *
    import random

    n = 2 ** 512    # 2^512
    m = random.randint(2, n-1) | 1  # 随机生成一个奇数（使用random.randint()函数生成一个大于等于2且小于n-1的随机整数，并通过按位或操作符"| 1"将其转换为奇数）
    c = pow(m, bytes_to_long(flag), n)  # c = m^flag mod n
    print 'm = ' + str(m)
    print 'c = ' + str(c)

    # m = 391190709124527428959489662565274039318305952172936859403855079581402770986890308469084735451207885386318986881041563704825943945069343345307381099559075
    # c = 6665851394203214245856789450723658632520816791621796775909766895233000234023642878786025644953797995373211308485605397024123180085924117610802485972584499
  • 对于c = m^flag mod n，已知c，m，n，求flag，这是典型的离散对数问题。
  • 使用sympy库进行求解： 
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    from sympy.ntheory import discrete_log  # conda update sympy
    from Crypto.Util.number import long_to_bytes    # pip install pycryptodome
    m = 391190709124527428959489662565274039318305952172936859403855079581402770986890308469084735451207885386318986881041563704825943945069343345307381099559075
    c = 6665851394203214245856789450723658632520816791621796775909766895233000234023642878786025644953797995373211308485605397024123180085924117610802485972584499
    n = 2 ** 512

    flag=discrete_log(n, c, m)  # 求解离散对数

    print long_to_bytes(flag)

FLAG

1
flag{5f95ca93-1594-762d-ed0b-a9139692cb4a}

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