[MoeCTF 2022]ezTea

#include <stdio.h>
#include <stdint.h>

void encrypt (uint32_t* v, uint32_t* k) {                       
    uint32_t v0 = v[0], v1 = v[1], sum = 0;
    uint32_t delta = 0xd33b470;
    for (int i = 0; i < 32; i++) {
        sum += delta;
        v0 += ((v1<<4) + k[0]) ^ (v1 + sum) ^ ((v1>>5) + k[1]);
        v1 += ((v0<<4) + k[2]) ^ (v0 + sum) ^ ((v0>>5) + k[3]);
    }
    v[0] = v0;
    v[1] = v1;
}

int main() {
    uint32_t k[4] = {1, 2, 3, 4};
    int8_t input[33] = {0};
    
    scanf("%32s", input);
    for (int i = 0; i < 32; i+=8) {
        uint32_t v[2] = {*(uint32_t *)&input[i], *(uint32_t *)&input[i+4]};
        encrypt(v, k);
        for (int j = 0; j < 2; j++) {                          
            for (int k = 0; k < 4; k++) {
                printf("%#x, ", v[j] & 0xff);
                v[j] >>= 8;
            }
        }
    }
    return 0;
}

exp

#include <stdio.h>
#include <stdint.h>
 
void decrypt(uint32_t* v, uint32_t* k) 
{
    uint32_t v0 = v[0], v1 = v[1], sum = 0xd33b470*32;
    uint32_t delta = 0xd33b470;
    for (int i = 0; i < 32; i++) 
    {
        v1 -= ((v0 << 4) + k[2]) ^ (v0 + sum) ^ ((v0 >> 5) + k[3]);
        v0 -= ((v1 << 4) + k[0]) ^ (v1 + sum) ^ ((v1 >> 5) + k[1]);
        sum -= delta;
    }
    v[0] = v0;
    v[1] = v1;
}
 
int main() 
{
    uint32_t k[4] = { 1, 2, 3, 4 };//uint32_t——4字节，uint8_t——1字节
    int8_t input[] = { 0x17, 0x65, 0x54, 0x89, 0xed, 0x65, 0x46, 0x32, 0x3d, 0x58, 0xa9, 0xfd, 0xe2, 0x5e,0x61, 0x97, 0xe4, 0x60, 0xf1, 0x91, 0x73, 0xe9, 0xe9, 0xa2, 0x59, 0xcb, 0x9a, 0x99,0xec, 0xb1, 0xe1, 0x7d };
    for (int i = 0; i < 32; i += 8) 
    {
        uint32_t v[2] = { *(uint32_t*)&input[i], *(uint32_t*)&input[i + 4] };//1字节强制转换为4字节
        decrypt(v, k);//加密
        for (int j = 0; j < 2; j++)//一字节一字节打印出来
        {
            for (int k = 0; k < 4; k++) 
            {
                printf("%c", v[j] & 0xff);
                v[j] >>= 8;
            }
        }
    }
    return 0;
}