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[原创]第五题分析
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发表于: 2016-11-10 15:11 2172
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401000 关键函数的逻辑
下面是伪代码的
int n = 0;
for (int i = 0; i < 0x100; i++)
{
n += input_buf[i%6] + init_buf[i];
char ch = init_buf[i];
init_buf[i] = init_buf[n & 0xFF];
init_buf[n & 0xFF] = ch;
}
for(int i=0;i<0x80;i++)
{
char ch = init_buf[i]+init_buf[0xFF-i];
byte_406030[i] ^= ch;
}
第一个循环根据输入对初始化好的缓冲区数据做交换,然后再用第二个循环解码 byte_406030 的数据
4010C0 函数
把byte_406030的0x80个字节解码后的值加起来判断是否等于2979h
整体思路除了穷举没想到其他的更好的解法
穷举的代码
#include "stdafx.h"
#include <iostream>
#include <string>
#include <memory>
#include <vector>
#include <Windows.h>
using namespace std;
unsigned char byte_406030[128] = {
0xF4, 0x12, 0x9D, 0x60, 0x45, 0xF8, 0x20, 0x6A, 0x6F, 0x67, 0x04, 0x71, 0xC0, 0x9B, 0x0C, 0x5A,
0x1D, 0x18, 0x6C, 0x96, 0x69, 0x01, 0x1C, 0xF4, 0x7F, 0x28, 0x5A, 0xFB, 0x29, 0x07, 0x40, 0x8B,
0xD3, 0xE1, 0xB1, 0x12, 0xFB, 0xCA, 0x7C, 0x89, 0xB9, 0x5A, 0x30, 0x70, 0x9D, 0x95, 0x2B, 0x95,
0x3C, 0x8D, 0x2E, 0x45, 0xEF, 0x70, 0xC6, 0xA3, 0xB9, 0xB2, 0x5A, 0x63, 0x5F, 0x03, 0x33, 0xB8,
0x64, 0x4A, 0x8F, 0xBC, 0xF7, 0x91, 0x69, 0x6A, 0x56, 0x2E, 0xD4, 0x6E, 0x82, 0x93, 0xE9, 0x76,
0xDC, 0xA3, 0x6C, 0x5E, 0x6B, 0x72, 0x64, 0x37, 0xE7, 0x15, 0x17, 0xAC, 0x64, 0x78, 0xD5, 0x4A,
0x60, 0x2D, 0xF0, 0x54, 0xA6, 0xF3, 0xE8, 0xE0, 0xE0, 0xB9, 0x8F, 0x85, 0x90, 0xE4, 0xEA, 0xD6,
0xBB, 0xB7, 0x15, 0x9E, 0x2A, 0x44, 0xE7, 0x31, 0x63, 0xAC, 0x80, 0x6C, 0x34, 0x82, 0xE9, 0xCF
};
int _tmain(int argc, _TCHAR* argv[])
{
char szbuf[32];
char init_buf[0x100];
for (int i = 0; i < 0x100; i++)
{
init_buf[i] = i;
}
char init_buf_tmp[0x100];
for (std::uint64_t i = 0; i < 1000000; i++)
{
sprintf_s(szbuf, sizeof(szbuf), "%06d", i);
memcpy_s(init_buf_tmp, sizeof(init_buf_tmp), init_buf, sizeof(init_buf));
int n = 0;
for (int i = 0; i < 0x100; i++)
{
n += szbuf[i%6] + init_buf_tmp[i];
char ch = init_buf_tmp[i];
init_buf_tmp[i] = init_buf_tmp[n & 0xFF];
init_buf_tmp[n & 0xFF] = ch;
}
unsigned char byte_buf_tmp[128];
memcpy_s(byte_buf_tmp, sizeof(byte_buf_tmp), byte_406030, sizeof(byte_406030));
for (int i = 0; i < 0x80; i++)
{
char ch = init_buf_tmp[i] + init_buf_tmp[0xFF - i];
byte_buf_tmp[i] ^= ch;
}
int n1 = 0;
for (int i = 0; i < 0x80; i++)
{
n1 += byte_buf_tmp[i];
}
if (n1 == 0x2979)
{
cout << szbuf << endl;
}
}
return 0;
}
下面是伪代码的
int n = 0;
for (int i = 0; i < 0x100; i++)
{
n += input_buf[i%6] + init_buf[i];
char ch = init_buf[i];
init_buf[i] = init_buf[n & 0xFF];
init_buf[n & 0xFF] = ch;
}
for(int i=0;i<0x80;i++)
{
char ch = init_buf[i]+init_buf[0xFF-i];
byte_406030[i] ^= ch;
}
第一个循环根据输入对初始化好的缓冲区数据做交换,然后再用第二个循环解码 byte_406030 的数据
4010C0 函数
把byte_406030的0x80个字节解码后的值加起来判断是否等于2979h
整体思路除了穷举没想到其他的更好的解法
穷举的代码
#include "stdafx.h"
#include <iostream>
#include <string>
#include <memory>
#include <vector>
#include <Windows.h>
using namespace std;
unsigned char byte_406030[128] = {
0xF4, 0x12, 0x9D, 0x60, 0x45, 0xF8, 0x20, 0x6A, 0x6F, 0x67, 0x04, 0x71, 0xC0, 0x9B, 0x0C, 0x5A,
0x1D, 0x18, 0x6C, 0x96, 0x69, 0x01, 0x1C, 0xF4, 0x7F, 0x28, 0x5A, 0xFB, 0x29, 0x07, 0x40, 0x8B,
0xD3, 0xE1, 0xB1, 0x12, 0xFB, 0xCA, 0x7C, 0x89, 0xB9, 0x5A, 0x30, 0x70, 0x9D, 0x95, 0x2B, 0x95,
0x3C, 0x8D, 0x2E, 0x45, 0xEF, 0x70, 0xC6, 0xA3, 0xB9, 0xB2, 0x5A, 0x63, 0x5F, 0x03, 0x33, 0xB8,
0x64, 0x4A, 0x8F, 0xBC, 0xF7, 0x91, 0x69, 0x6A, 0x56, 0x2E, 0xD4, 0x6E, 0x82, 0x93, 0xE9, 0x76,
0xDC, 0xA3, 0x6C, 0x5E, 0x6B, 0x72, 0x64, 0x37, 0xE7, 0x15, 0x17, 0xAC, 0x64, 0x78, 0xD5, 0x4A,
0x60, 0x2D, 0xF0, 0x54, 0xA6, 0xF3, 0xE8, 0xE0, 0xE0, 0xB9, 0x8F, 0x85, 0x90, 0xE4, 0xEA, 0xD6,
0xBB, 0xB7, 0x15, 0x9E, 0x2A, 0x44, 0xE7, 0x31, 0x63, 0xAC, 0x80, 0x6C, 0x34, 0x82, 0xE9, 0xCF
};
int _tmain(int argc, _TCHAR* argv[])
{
char szbuf[32];
char init_buf[0x100];
for (int i = 0; i < 0x100; i++)
{
init_buf[i] = i;
}
char init_buf_tmp[0x100];
for (std::uint64_t i = 0; i < 1000000; i++)
{
sprintf_s(szbuf, sizeof(szbuf), "%06d", i);
memcpy_s(init_buf_tmp, sizeof(init_buf_tmp), init_buf, sizeof(init_buf));
int n = 0;
for (int i = 0; i < 0x100; i++)
{
n += szbuf[i%6] + init_buf_tmp[i];
char ch = init_buf_tmp[i];
init_buf_tmp[i] = init_buf_tmp[n & 0xFF];
init_buf_tmp[n & 0xFF] = ch;
}
unsigned char byte_buf_tmp[128];
memcpy_s(byte_buf_tmp, sizeof(byte_buf_tmp), byte_406030, sizeof(byte_406030));
for (int i = 0; i < 0x80; i++)
{
char ch = init_buf_tmp[i] + init_buf_tmp[0xFF - i];
byte_buf_tmp[i] ^= ch;
}
int n1 = 0;
for (int i = 0; i < 0x80; i++)
{
n1 += byte_buf_tmp[i];
}
if (n1 == 0x2979)
{
cout << szbuf << endl;
}
}
return 0;
}
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