刷题笔记:[2019UNCTF]666

前言

挺简单的，考的逆向思维，我也做的来了。好耶

题解

enflag = [0x69, 0x7A, 0x77, 0x68, 0x72, 0x6F, 0x7A, 0x22, 0x22,
          0x77, 0x22, 0x76, 0x2E, 0x4B, 0x22, 0x2E, 0x4E, 0x69, ]
flag = [""] * 18
key = 18

v3 = [0] * 32
v4 = [0] * 32
v5 = [0] * 40
for i in range(0, key, 3):
    # v5[i] = enflag[i]
    # v4[i + 1] = enflag[i + 1]
    # v3[i + 2] = enflag[i + 2]
    flag[i] = (enflag[i] ^ key) - 6
    flag[i + 1] = (enflag[i + 1] ^ key) + 6
    flag[i + 2] = (enflag[i + 2] ^ key) ^ 6
    # 加减优先级比异或要高

for i in flag:
    print(chr(i), end="")