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

void decrypt (uint32_t* v, uint32_t* k) {
    uint32_t delta = 0xd33b470;               
    uint32_t v0 = v[0], v1 = v[1], sum = 32 * delta;
    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[1] = v1;
    v[0] = v0;
}
int main(){
    uint32_t k[4] = {1, 2, 3, 4};
    int8_t input[33] = {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]};
	decrypt(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;
}