文件名pyre.cpython-36.pyc，说明是在python3.6环境下编译的，要把pyc反编译成py

但是显示失败了，结合题的名字文件的应该是文件头部被破坏

把第一行改为33 0D 0D 0A 0C 63 4A 63 61 02 00 00 E3 00 00 00

之后就能反编译了，得到源码如下

# uncompyle6 version 3.8.0
# Python bytecode 3.6 (3379)
# Decompiled from: Python 3.6.0 (v3.6.0:41df79263a11, Dec 23 2016, 08:06:12) [MSC v.1900 64 bit (AMD64)]
# Embedded file name: pyre.py
# Compiled at: 2022-10-15 15:36:44
# Size of source mod 2**32: 609 bytes
from ctypes import *
from Crypto.Util.number import bytes_to_long
from Crypto.Util.number import long_to_bytes
 
def encrypt(v, k):
    v0 = c_uint32(v[0])
    v1 = c_uint32(v[1])
    sum1 = c_uint32(0)
    delta = 195935983
    for i in range(32):
        v0.value += (v1.value << 4 ^ v1.value >> 7) + v1.value ^ sum1.value + k[(sum1.value & 3)]
        sum1.value += delta
        v1.value += (v0.value << 4 ^ v0.value >> 7) + v0.value ^ sum1.value + k[(sum1.value >> 9 & 3)]
 
    return (
     v0.value, v1.value)
 
 
if __name__ == '__main__':
    flag = input('please input your flag:')
    k = [255, 187, 51, 68]
    if len(flag) != 32:
        print('wrong!')
        exit(-1)
    a = []
    for i in range(0, 32, 8):
        v1 = bytes_to_long(bytes(flag[i:i + 4], 'ascii'))
        v2 = bytes_to_long(bytes(flag[i + 4:i + 8], 'ascii'))
        a += encrypt([v1, v2], k)
 
    enc = [
     4006073346, 2582197823, 2235293281, 558171287, 2425328816, 1715140098, 986348143, 1948615354]
    for i in range(8):
        if enc[i] != a[i]:
            print('wrong!')
            exit(-1)
 
    print('flag is flag{%s}' % flag)
# okay decompiling d:\pyre.cpython-36.pyc

由此得到解密代码：

#include<stdio.h>
#define  u_int unsigned int
int main()
{
	u_int enc[] = { 4006073346, 2582197823, 2235293281, 558171287,
				2425328816, 1715140098, 986348143, 1948615354 };
	u_int k[] = { 255, 187, 51, 68 };
	for (int i = 0; i < 8; i += 2)
	{
		u_int delta = 195935983;
		u_int s1 = delta * 32;
		for (int j = 0; j < 32; j++)
		{
			enc[i + 1] -= (enc[i] << 4 ^ enc[i] >> 7) + enc[i] ^ s1 + k[s1 >> 9 & 3];
			s1 -= delta;
			enc[i] -= (enc[i + 1] << 4 ^ enc[i + 1] >> 7) + enc[i + 1] ^ s1 + k[s1 & 3];
		}
	}
	char flag[32] = { 0 };
	for (int i = 0; i < 8; i++)
	{
		flag[4 * i] = enc[i] >> 24;
		flag[4 * i + 1] = (enc[i] >> 16) ^ 0xFF000000;
		flag[4 * i + 2] = (enc[i] >> 8) ^ 0xFF000000;
		flag[4 * i + 3] = enc[i] ^ 0xFF000000;
	}
	for (int i = 0; i < 32; i++)
	{
		printf("%c", flag[i]);
	}
}

得到：flag{Th1s_1s_A_Easy_Pyth0n__R3veRse_0}