[網鼎杯2020青龍組] you_raise_me_up

解題過程

1. 題目給的代碼如下：

#!/usr/bin/env python
# -*- coding: utf-8 -*-
from Crypto.Util.number import *
import random

n = 2 ** 512
m = random.randint(2, n-1) | 1
c = pow(m, bytes_to_long(flag), n)
print 'm = ' + str(m)
print 'c = ' + str(c)

# m = 391190709124527428959489662565274039318305952172936859403855079581402770986890308469084735451207885386318986881041563704825943945069343345307381099559075
# c = 6665851394203214245856789450723658632520816791621796775909766895233000234023642878786025644953797995373211308485605397024123180085924117610802485972584499

讀完代碼可以得到一個表達式：

　　c = m^{bytes_to_long(flag)} mod n

2. 那么現在已知c, m, n求離散對數：

　　bytes_to_long(flag) = log_{(m mod n)} (c mod n)

sage中的discrete_log可以幫我們計算出結果^[1]，使用方法為:

　　bytes_to_long(flag) = discrete_log((c mod n), (m mod n))  或

　　bytes_to_long(flag) = discrete_log(c, (m mod n))

3. 將得到的數值long_to_bytes(flag)即可得到：flag{5f95ca93-1594-762d-ed0b-a9139692cb4a}

from Crypto.Util.number import *
flag = 56006392793405651552924479293096841126763872290794186417054288110043102953612574215902230811593957757
print long_to_bytes(flag)

Reference

[1] https://blog.csdn.net/qq_39642801/article/details/104158699?utm_medium=distribute.pc_relevant.none-task-blog-baidujs-4