0x00 前言

CTF 加解密合集：CTF 加解密合集

0x01 题目

from secrets import randbelow
from nationalsecret import p, r, k, flag

g = 2
y = pow(g, k, p)

def gogogo():
    print("Another chance:")
    t = int(input('t = '))
    c = randbelow(p)
    print("Here is my real challenge:")
    print(f'c = {c}')
    print("Give me your answer.")
    s = int(input('s = '))
    return pow(g, s, p) == t * pow(y, c, p) % p

def dododo():
    print("Here is my gift for National's Day!")
    t = pow(g, r, p)
    print(f't = {t}')
    print("What do you want?")
    c = int(input('c = '))
    s = (r + c*k) % (p - 1)
    print("This is another gift:")
    print(f's = {s}')

print("Happy National's Day!")
print("Don't play CTF these days.")
print("Just go out and play~")
print("Don't finish this chall,")
print("although I should tell you that")
print(f"p = {p}")


dododo()

if gogogo():
    print(f"FUIYOH! You can get flag: {flag}")
    print("and then go out and play!")
else:
    print("HAIYAA! You are wrong. Just go out and play!")

0x02 Write Up

这道题实际上是考察对于数论的理解程度，首先看到结果，首先看到

pow(g, s, p) == t * pow(y, c, p) % p

并且y = pow(g, k, p)

然后进行化简
pow(g, s, p) == pow(y, c, p)
等价于

pow(g, s, p) == pow(pow(g, k, p), c, p)

等价于

pow(g, s, p) == pow(g, k * c, p)

等价于

s = k * c

那么就是说我们只要知道了k和c就知道了结果。

c是已知的，只需要算k即可。

s = (r + c*k) % (p - 1) 此时c为0的时候，可以得到s的值，并且可以计算出r的值。

得到r的值之后，再代入即可得到：

k=p-1+s-r

最终可得到答案：

以上