[MoeCTF 2022]ezTea

原题

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

void encrypt (uint32_t* v, uint32_t* k) {                       // 主要加密函数，试着搞定它
    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++) {                           // 这一段主要是把 v 按单字节输出，另外可以了解一下 “大小端序” 在这题是如何体现的
            for (int k = 0; k < 4; k++) {
                printf("%#x, ", v[j] & 0xff);
                v[j] >>= 8;
            }
        }
    }
    return 0;
}

输出
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

先把原本输入替换成

int8_t enc[] = {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};

把下面代码input换成enc

for(int i = 0; i<32;i+=8){
uint32_t v[2] = {*(uint32_t *)&enc[i], *(uint32_t *)&enc[i+4]};
    encrypt(v, k);

把加密部分修改一下

void encrypt (uint32_t* v, uint32_t* k) {                       // 主要加密函数，试着搞定它
    uint32_t v0 = v[0], v1 = v[1];
    uint32_t delta = 0xd33b470; 
    uint32_t sum =delta *32; //delta 乘32
    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;
        
        把+改成-
    }
    v[0] = v0;
    v[1] = v1;
}