py文件,随便在网上搜个反编译器,看看源码

看下就是很简单的俩轮加密,反着写过去就行了.

贴上别人的脚本…

print(‘Welcome to Re World!’)
print(‘Your input1 is your flag~’)

code = [0x1f,0x12,0x1d, 0x28,0x30,0x34, 0x01,0x06,0x14, 0x34,
0x2c, 0x1b, 0x55,0x3f,0x6f,0x36,0x2a,0x3a, 0x01, 0x44,
0x3b,0x25, 0x13]

code.reverse()
l = len(code)

for i in range(l-1):
code[i + 1] = code[i] ^ code[i + 1]

code.reverse()

for i in range(len(code)):
if code[i] - i < 0:
code[i] = code[i] + 128 -i
else:
code[i] -= i

flag = ‘’.join(chr(i) for i in code)
print(flag)

GWHT{Just_Re_1s_Ha66y!}
运行了一下,答案就是这个啊,给我不对,日常抽风吗

GWHT{Just_Re_1s_Ha66y!}

标签：code,flag,0x01,Re,GWCTF,2019,print,pyre,Ha66y
来源： https://blog.csdn.net/weixin_52034946/article/details/112723263