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[推荐]看雪·安恒 2020 KCTF 春季赛 | 第十题设计思路及解析
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发表于: 2020-5-11 18:05
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这道题难度较大，前三天选手们都在静静寻找突破口，但并没有收获。直到第四天，金箍棒配色的泰迪熊战队的Riatre终于找到解题的方法，并在上午8点成功提交Flag。
值得一提的是，Riatre是该战队的创建者也是唯一一名成员，虽独自参赛但有智有勇，截至目前已成功破解7道题，并登上防守方排行榜的第一名。
目前攻击方排名如下：
出题团队简介
设计思路
设计思路
解题
算法
//
// Created by ZT on 2020/4/10.
//
#include <cstdio>
#include <cstdint>
static unsigned const char sbox[256] = {
0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67, 0x2b, 0xfe, 0xd7, 0xab, 0x76,
0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, 0x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0,
0xb7, 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1, 0x71, 0xd8, 0x31, 0x15,
0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05, 0x9a, 0x07, 0x12, 0x80, 0xe2, 0xeb, 0x27, 0xb2, 0x75,
0x09, 0x83, 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3, 0x29, 0xe3, 0x2f, 0x84,
0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b, 0x6a, 0xcb, 0xbe, 0x39, 0x4a, 0x4c, 0x58, 0xcf,
0xd0, 0xef, 0xaa, 0xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x02, 0x7f, 0x50, 0x3c, 0x9f, 0xa8,
0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc, 0xb6, 0xda, 0x21, 0x10, 0xff, 0xf3, 0xd2,
0xcd, 0x0c, 0x13, 0xec, 0x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d, 0x64, 0x5d, 0x19, 0x73,
0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee, 0xb8, 0x14, 0xde, 0x5e, 0x0b, 0xdb,
0xe0, 0x32, 0x3a, 0x0a, 0x49, 0x06, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62, 0x91, 0x95, 0xe4, 0x79,
0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4, 0xea, 0x65, 0x7a, 0xae, 0x08,
0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, 0xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f, 0x4b, 0xbd, 0x8b, 0x8a,
0x70, 0x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e, 0x61, 0x35, 0x57, 0xb9, 0x86, 0xc1, 0x1d, 0x9e,
0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, 0x94, 0x9b, 0x1e, 0x87, 0xe9, 0xce, 0x55, 0x28, 0xdf,
0x8c, 0xa1, 0x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f, 0xb0, 0x54, 0xbb, 0x16
};
void reverse(const unsigned char dest[256], int index, unsigned char expKey[16]) {
bool found = false;
int key = 0;
for (size_t lastKey = 0; lastKey < 256; ++lastKey) {
for (size_t curKey = 0; curKey < 256; ++curKey) {
for (size_t k = 0; k < 256; ++k) {
unsigned char tmp = sbox[k ^ lastKey];
size_t st = tmp ^ curKey;
if (st != dest[k]) {
found = false;
break;
}
found = true;
key = k;
}
if (found) {
expKey[index] = curKey;
expKey[index - 16] = lastKey;
printf("found[%d] lastKey=0x%02x, x=%d, curKey=0x%02x !\n", index, lastKey, key, curKey);
return;
}
}
}
}
// uint8_t y[4] -> uint32_t x
#define LOAD32H(x, y) \
do { (x) = ((uint32_t)((y)[0] & 0xff)<<24) | ((uint32_t)((y)[1] & 0xff)<<16) | \
((uint32_t)((y)[2] & 0xff)<<8) | ((uint32_t)((y)[3] & 0xff));} while(0)
// uint32_t x -> uint8_t y[4]
#define STORE32H(x, y) \
do { (y)[0] = (uint8_t)(((x)>>24) & 0xff); (y)[1] = (uint8_t)(((x)>>16) & 0xff); \
(y)[2] = (uint8_t)(((x)>>8) & 0xff); (y)[3] = (uint8_t)((x) & 0xff); } while(0)
// 从uint32_t x中提取从低位开始的第n个字节
#define BYTE(x, n) (((x) >> (8 * (n))) & 0xff)
// uint32_t x循环左移n位
#define ROF32(x, n) (((x) << (n)) | ((x) >> (32-(n))))
// uint32_t x循环右移n位
#define ROR32(x, n) (((x) >> (n)) | ((x) << (32-(n))))
#define MIX(x) (((sbox[BYTE(x, 2)] << 24) & 0xff000000) ^ ((sbox[BYTE(x, 1)] << 16) & 0xff0000) ^ \
((sbox[BYTE(x, 0)] << 8) & 0xff00) ^ (sbox[BYTE(x, 3)] & 0xff))
//逆行移位, 第一行不变，第二行循环右移8位/一个字节，第三行循环右移16位/2个字节，第四行循环右移24位/3个字节。
int invShiftRows(unsigned char (*state)[4]) {
uint32_t block[4] = {0};
// printf("逆行移位[%d]:\n", ++shiftRowsCount);
/* i: row */
for (int i = 0; i < 4; ++i) {
LOAD32H(block[i], state[i]);
block[i] = ROR32(block[i], 8*i);
STORE32H(block[i], state[i]);
printf("0x%02x, 0x%02x, 0x%02x, 0x%02x, \n", state[i][0], state[i][1], state[i][2], state[i][3]);
}
return 0;
}
void rightShift(unsigned char expKey[16]) {
unsigned char tmp[4][4];
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 4; ++j) {
tmp[j][i] = expKey[i*4 + j];
}
}
invShiftRows(tmp);
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 4; ++j) {
expKey[i*4 + j] = tmp[j][i];
}
}
}
constexpr uint32_t rcon[] = {
0x01000000, 0x02000000, 0x04000000,
0x08000000, 0x10000000, 0x20000000,
0x40000000, 0x80000000, 0x1b000000,
0x36000000, 0x6c000000, 0xd8000000,
0xab000000, 0x4d000000, 0x9a000000
};
void reverse(unsigned char expKey[16], int round) {
uint32_t w[4];
uint32_t top[4];
for (int i = 0; i < 4; ++i) {
LOAD32H(w[i], expKey + 4*i);
}
top[3] = w[3] ^w[2];
top[2] = w[2] ^w[1];
top[1] = w[1] ^w[0];
printf("0x%02x ^ 0x%02x ^ 0x%02x\n", w[0], rcon[round -1], MIX(top[3]));
top[0] = w[0] ^rcon[round -1] ^ MIX(top[3]);
unsigned char *tmp = expKey - 16;
for (int j = 0; j < 4; ++j) {
tmp[j*4] = top[j] >> 24;
tmp[j*4 + 1] = (top[j] >> 16) ;
tmp[j*4 + 2] = top[j] >> 8;
tmp[j*4 + 3] = top[j];
}
}
void check(const unsigned char expKey[16]) {
uint32_t top[8];
uint32_t *w = top;
for (int i = 0; i < 8; ++i) {
LOAD32H(top[i], expKey + 4*i);
}
for (int i = 0; i < 1; ++i) {
if (w[4] != (w[0] ^ MIX(w[3]) ^ rcon[9])) {
printf("error expkey!\n");
break;
}
if (w[5] != (w[1] ^ w[4])){
printf("error expkey!\n");
break;
}
if (w[6] != (w[2] ^ w[5])) {
printf("error expkey!\n");
break;
}
if(w[7] != (w[3] ^ w[6])){
printf("error expkey!\n");
break;
}
printf("right expkey!\n");
w += 4;
}
}
int main() {
unsigned char expKey[11][16] = {0};
//把最后的一组int数组提取出来转成字节数组，放入
const unsigned char Tboxes[16][256] = {...};
//暴破最后两组16字节的扩展key
for (int i = 0; i < 16; ++i) {
reverse(Tboxes[i], i, expKey[10]);
}
rightShift(expKey[9]);
check(expKey[9]);
//倒着推出上一组key
for (int j = 9; j > 0; --j) {
reverse(expKey[j], j);
}
printf("expKey:\n");
for (int k = 0*16; k < 11*16; k++ ) {
if (k && k % 4 == 0) {
printf("\n");
if (k % 16 == 0) {
printf("\n");
}
}
printf("0x%02x, ", expKey[0][k]);
}
printf("\nend expKey\n");
return 0;
}得到初始key，"you get the key!"，之后解密0b6f04a3427089858041945e70082508得到flag。
补充
flag
[招生]科锐逆向工程师培训(2024年11月15日实地，远程教学同时开班, 第51期)
