[网鼎杯2020青龙组] you_raise_me

[网鼎杯2020青龙组] you_raise_me_up

解题过程

1. 题目给的代码如下：

 1 #!/usr/bin/env python
 2 # -*- coding: utf-8 -*-
 3 from Crypto.Util.number import *
 4 import random
 5 
 6 n = 2 ** 512
 7 m = random.randint(2, n-1) | 1
 8 c = pow(m, bytes_to_long(flag), n)
 9 print 'm = ' + str(m)
10 print 'c = ' + str(c)
11 
12 # m = 391190709124527428959489662565274039318305952172936859403855079581402770986890308469084735451207885386318986881041563704825943945069343345307381099559075
13 # c = 6665851394203214245856789450723658632520816791621796775909766895233000234023642878786025644953797995373211308485605397024123180085924117610802485972584499

读完代码可以得到一个表达式：

　　c = m^{bytes_to_long(flag)} mod n

2. 那么现在已知c, m, n求离散对数：

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

sage中的discrete_log可以帮我们计算出结果^[1]，使用方法为:

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

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

3. 将得到的数值long_to_bytes(flag)即可得到：flag{5f95ca93-1594-762d-ed0b-a9139692cb4a}

1 from Crypto.Util.number import *
2 flag = 56006392793405651552924479293096841126763872290794186417054288110043102953612574215902230811593957757
3 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