corCTF2021 复盘

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babyrev

  1. memfrob函数：将字符串的每个字符与42进行异或实现加密。 解密脚本:
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def is_prime(n: int) -> bool:
    if n <= 1:
        return False
    i = 2
    while i <= n // 2:
        if n % i == 0:
            return False
        i += 1
    return True


def rot_n(c, n):
    ASCII_UPPER = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
    ASCII_LOWER = 'abcdefghijklmnopqrstuvwxyz'

    if c in ASCII_UPPER:
        return ASCII_UPPER[(ASCII_UPPER.index(c) + n) % 26]
    elif c in ASCII_LOWER:
        return ASCII_LOWER[(ASCII_LOWER.index(c) + n) % 26]
    else:
        return c


check = [0x5f, 0x40, 0x5a, 0x15, 0x75, 0x45, 0x62, 0x53, 0x75, 0x46, 0x52, 0x43, 0x5f, 0x75, 0x50, 0x52, 0x75, 0x5f,
         0x5c, 0x4f]
# memfrob
check = [c ^ 42 for c in check]
flag = []

for i in range(len(check)):
    p = 4 * i
    while not is_prime(p):
        p += 1
    print(rot_n(chr(check[i]), -p), end='')