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[原创]TEA、XTEA、XXTEA加解密过程及案例
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发表于: 2021-4-9 02:30
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更新原因: 那就回过头提升下文章质量吧,之前确实写的不好
简介：一种分组加密算法，TEA算法使用64位(也就是8字节一组)的明文分组和128位(16字节)的密钥，使用一个神秘常数作为倍数(也可更改)
关键加解密函数:
(可以看见加密和解密函数传入的v都是大小为2的四字节数组,uint32_t v0和v1,两个四字节加密迭代32轮直到最后)
下面这个脚本是2021mrctf逆向的Dynamic Debug题目的脚本
题目下载链接:
链接：https://pan.baidu.com/s/1mU70Z2-7tDToTBRbEU0NSA
提取码：0syj
评论有人在问这道题的解题wp,这题没啥好写wp的,调试到对应加密函数对拍就行
下面是某题flag2WP:
魔改tea的一些方法:
这些都简单对拍下就行了
但是实际上还有一些加密模式的TEA，比如CBC模式的TEA
CBC模式(Cipher Block Chaining)：其实主要就是将明文分组与前一个密文分组进行异或运算，然后再进行加密，对于第一组的话就设置一个初始值来和第一组明文异或
比如：(CBC模式循环加密64字节，每次循环加密8字节(v0和v1各4字节))
每一轮是get_data取我们v0和v1，data1, data2和v0，v1异或，异或之后的data1和data2传入指针进行tea加密，之后再将加密之后的赋值回我们的v0和v1
在理解了CBC模式的TEA之后，我们该如何去逆向它呢，首先我们是有每轮加密后的v0和v1的，就是加密数据，TEA我们也是能够逆向的，那么就剩下逆向每轮的data1 ^= v0和data2 ^= v1了，而后面的轮数的data1和data2是会被我们的加密结果更新的，我们只有第一轮的data1和data2，那么我们就从这个地方下手，先解密第一轮，得到第一轮的v0和v1明文，再重新去加密更新data1和data2用于下一轮的解密
解题脚本:
XTEA是TEA的升级版，增加了更多的密钥表，移位和异或操作等等，设计者是Roger Needham, David Wheeler
加密过程：
说白了,主要原因就是因为之前取key的时候是固定下标取的,现在通过计算来取
算法实现：
XTEA实例：
2021虎符Re-GoEncrypt(WriteUp by syj)
https://bbs.pediy.com/thread-266935.htm
XXTEA，又称Corrected Block TEA，是XTEA的升级版 ，设计者是Roger Needham, David Wheeler
特点归纳：
①：原字符串长度可以不是4的倍数了
加密过程：
解题模板：
XXTEA例题: 2021hgame-alpacha
链接：https://pan.baidu.com/s/1H9jq_VoynN5_2rS3Kh205A
提取码：0syj
下面这个解题脚本没按照上面的模板写,你们可以自己改写下,懒得去改了
#include <stdio.h>#include <stdint.h>void encrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], sum=0, i;
uint32_t delta=0x9e3779b9;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i < 32; i++) {
sum += delta;
v0 += ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
v1 += ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
}
v[0]=v0; v[1]=v1;
}void decrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], i;
uint32_t delta=0x9e3779b9;
uint32_t sum = delta*32;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i<32; i++) {
v1 -= ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
v0 -= ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
sum -= delta;
}
v[0]=v0; v[1]=v1;
}#include <stdio.h>#include <stdint.h>void encrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], sum=0, i;
uint32_t delta=0x9e3779b9;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i < 32; i++) {
sum += delta;
v0 += ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
v1 += ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
}
v[0]=v0; v[1]=v1;
}void decrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], i;
uint32_t delta=0x9e3779b9;
uint32_t sum = delta*32;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i<32; i++) {
v1 -= ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
v0 -= ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
sum -= delta;
}
v[0]=v0; v[1]=v1;
}#include <stdio.h>#include <stdint.h>void encrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], sum=0, i;
uint32_t delta=0x9e3779b9;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i < 32; i++) {
sum += delta;
v0 += ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
v1 += ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
}
v[0]=v0; v[1]=v1;
}void decrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], i;
uint32_t delta=0x9e3779b9;
uint32_t sum = delta*32;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i<32; i++) {
v1 -= ((v0<<4) + k1) ^ (v0 + sum) ^ ((v0>>5) + k0);
v0 -= ((v1<<4) + k3) ^ (v1 + sum) ^ ((v1>>5) + k2);
sum -= delta;
}
v[0]=v0; v[1]=v1;
}int main()
{ uint32_t enflag[] = {0x5585A199, 0x7E825D68, 0x944D0039, 0x71726943, 0x6A514306, 0x4B14AD00, 0x64D20D3F, 0x9F37DB15};
uint32_t key[4] = {0x67626463, 0x696D616E, 0x79645F65, 0x6B696C69};
for(int i=0;i<8;i+=2)
{
uint32_t temp[2]; //定义来解密
temp[0] = enflag[i];
temp[1] = enflag[i+1];
decrypt(temp, key);
//printf("%X%X",temp[0],temp[1]);
printf("%c%c%c%c%c%c%c%c",*((char*)&temp[0]+0),*((char*)&temp[0]+1),*((char*)&temp[0]+2),*((char*)&temp[0]+3),*((char*)&temp[1]+0),*((char*)&temp[1]+1),*((char*)&temp[1]+2),*((char*)&temp[1]+3));
}
return 0;
}#include <stdio.h>#include <stdint.h>void encrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], sum=0, i;
uint32_t delta=0x9e3779b9;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i < 32; i++) {
sum += delta;
v0 += ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
v1 += ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
}
v[0]=v0; v[1]=v1;
}void decrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], i;
uint32_t delta=0x9e3779b9;
uint32_t sum = delta*32;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i<32; i++) {
v1 -= ((v0<<4) + k1) ^ (v0 + sum) ^ ((v0>>5) + k0);
v0 -= ((v1<<4) + k3) ^ (v1 + sum) ^ ((v1>>5) + k2);
sum -= delta;
}
v[0]=v0; v[1]=v1;
}int main()
{ uint32_t enflag[] = {0x5585A199, 0x7E825D68, 0x944D0039, 0x71726943, 0x6A514306, 0x4B14AD00, 0x64D20D3F, 0x9F37DB15};
uint32_t key[4] = {0x67626463, 0x696D616E, 0x79645F65, 0x6B696C69};
for(int i=0;i<8;i+=2)
{
uint32_t temp[2]; //定义来解密
temp[0] = enflag[i];
temp[1] = enflag[i+1];
decrypt(temp, key);
//printf("%X%X",temp[0],temp[1]);
printf("%c%c%c%c%c%c%c%c",*((char*)&temp[0]+0),*((char*)&temp[0]+1),*((char*)&temp[0]+2),*((char*)&temp[0]+3),*((char*)&temp[1]+0),*((char*)&temp[1]+1),*((char*)&temp[1]+2),*((char*)&temp[1]+3));
}
return 0;
}该题文件以及详细解题过程：链接：https://pan.baidu.com/s/1TbGl2XOgqkDp60U8CliThQ
提取码：0syj
```c#include <stdio.h>#include <stdlib.h>#include <stdint.h>void decrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], i;
uint32_t delta=0x6C7A6E62;
uint32_t sum = delta*32;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i<32; i++) {
v1 ^= (31-i);
v0 ^= (31-i);
v1 -= ((v0<<2) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
v0 -= ((v1<<2) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
sum -= delta;
}
v[0]=v0; v[1]=v1;
}int main()
{ //异或后我们在内存中得到的数据dafbde108b5962f40e786cf1c994830d29bc0a29fea23e0bdc87ea0513c7350d
uint32_t enflag[] = {0x10defbda,0xf462598b,0xf16c780e,0x0d8394c9,0x290abc29,0x0b3ea2fe,0x05ea87dc,0x0d35c713};
uint32_t key[4] = {0x6C637953,0x76656C6F,0xD754C061,0x023BA78B};
for(int i=0;i<8;i+=2)
{
uint32_t temp[2]; //定义来解密
temp[0] = enflag[i];
temp[1] = enflag[i+1];
decrypt(temp, key);
//printf("%X%X",temp[0],temp[1]);
printf("%c%c%c%c%c%c%c%c",*((char*)&temp[0]+0),*((char*)&temp[0]+1),*((char*)&temp[0]+2),*((char*)&temp[0]+3),*((char*)&temp[1]+0),*((char*)&temp[1]+1),*((char*)&temp[1]+2),*((char*)&temp[1]+3));
}
return 0;
}该题文件以及详细解题过程：链接：https://pan.baidu.com/s/1TbGl2XOgqkDp60U8CliThQ
提取码：0syj
```c#include <stdio.h>#include <stdlib.h>#include <stdint.h>void decrypt (uint32_t* v, uint32_t* k) {
uint32_t v0=v[0], v1=v[1], i;
uint32_t delta=0x6C7A6E62;
uint32_t sum = delta*32;
uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];
for (i=0; i<32; i++) {
v1 ^= (31-i);
v0 ^= (31-i);
v1 -= ((v0<<2) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
v0 -= ((v1<<2) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
sum -= delta;
}
v[0]=v0; v[1]=v1;
}int main()
{ //异或后我们在内存中得到的数据dafbde108b5962f40e786cf1c994830d29bc0a29fea23e0bdc87ea0513c7350d
uint32_t enflag[] = {0x10defbda,0xf462598b,0xf16c780e,0x0d8394c9,0x290abc29,0x0b3ea2fe,0x05ea87dc,0x0d35c713};
uint32_t key[4] = {0x6C637953,0x76656C6F,0xD754C061,0x023BA78B};
for(int i=0;i<8;i+=2)
{
uint32_t temp[2]; //定义来解密
temp[0] = enflag[i];
temp[1] = enflag[i+1];
decrypt(temp, key);
//printf("%X%X",temp[0],temp[1]);
printf("%c%c%c%c%c%c%c%c",*((char*)&temp[0]+0),*((char*)&temp[0]+1),*((char*)&temp[0]+2),*((char*)&temp[0]+3),*((char*)&temp[1]+0),*((char*)&temp[1]+1),*((char*)&temp[1]+2),*((char*)&temp[1]+3));
}
return 0;
}#include <stdio.h>#include <stdint.h>#include <stdlib.h>#include <string.h>uint32_t data1 = 0x5F797274;
uint32_t data2 = 0x64726168; //初始的data 值
void encrypt(uint32_t* v, uint32_t* k) {
uint32_t v0 = v[0], v1 = v[1], sum = 0, i;
data1 ^= v0;
data2 ^= v1;
v0 = data1;
v1 = data2;
uint32_t delta = 0x6e75316c;
uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
for (i = 0; i < 32; i++)
{
sum += delta;
v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1) ^ (sum + i);
v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3) ^ (sum + i);
}
data1 = v0;
data2 = v1;
}void decrypt(uint32_t* v, uint32_t* k)
{ uint32_t delta = 0x6e75316c;
uint32_t v0 = v[0], v1 = v[1], sum = (delta * 32) & 0xffffffff, i;
uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
for (i = 0; i < 32; i++)
{
v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3) ^ (sum + (31 - i));
v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1) ^ (sum + (31 - i));
sum -= delta;
}
v0 = v0 ^ data1;
v1 = v1 ^ data2;
v[0] = v0; v[1] = v1;
}int main()
{ uint32_t array[] = {0x9b28ed45, 0x145ec6e9, 0x5b27a6c3, 0xe59e75d5, 0xe82c2500, 0xa4211d92, 0xcd8a4b62, 0xa668f440};
uint32_t key[4] = {0x65766967, 0x756F795F, 0x7075635F, 0x6165745F};
for (int i = 0; i < 8; i += 2)
{
uint32_t temp[2];
temp[0] = array[i];
temp[1] = array[i + 1];
decrypt(temp, key);
printf("%c%c%c%c%c%c%c%c", *((char*)&temp[0] + 0), *((char*)&temp[0] + 1), *((char*)&temp[0] + 2), *((char*)&temp[0] + 3), *((char*)&temp[1] + 0), *((char*)&temp[1] + 1), *((char*)&temp[1] + 2), *((char*)&temp[1] + 3));
//更新data
encrypt(temp, key);
}
return 0;
}#include <stdio.h>#include <stdint.h>#include <stdlib.h>#include <string.h>uint32_t data1 = 0x5F797274;
uint32_t data2 = 0x64726168; //初始的data 值
void encrypt(uint32_t* v, uint32_t* k) {
uint32_t v0 = v[0], v1 = v[1], sum = 0, i;
data1 ^= v0;
data2 ^= v1;
v0 = data1;
v1 = data2;
uint32_t delta = 0x6e75316c;
uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
for (i = 0; i < 32; i++)
{
sum += delta;
v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1) ^ (sum + i);
v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3) ^ (sum + i);
}
data1 = v0;
data2 = v1;
}void decrypt(uint32_t* v, uint32_t* k)
[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)
最后于 2022-10-4 19:39
被SYJ-Re编辑
，原因： 内容更新
