资源修改器

1.简介

《命令与征服：红色警戒(报)3》作为经典游戏，是不少人儿时的记忆，同时也是博主童年最爱的一款游戏，为此最近我重新下载了该款游戏，红警3的下载地址如下：

---

链接：https://pan.baidu.com/s/1yCDmmmYpiDfW4TnwSGme9g?pwd=ajsi 
提取码：ajsi

---

本文针对金币和电量无线修改的源码和相关教程

2.破解方法

1.获取红色警戒3游戏程序的句柄

const char* text = "命令与征服：红色警报3";
    HWND hwnd = FindWindow(NULL, text);
    if (hwnd == NULL)
    {
        MessageBox(NULL, TEXT("RA3 is not found! ACanary must find it again!"), TEXT("crack_ra3"), MB_ICONSTOP);
        return 1;
    }

    DWORD pid;
    GetWindowThreadProcessId(hwnd, &pid);
    HANDLE handle = OpenProcess(PROCESS_ALL_ACCESS, FALSE, pid);

2.通过游戏金币初始值（默认为10000，可在开始游戏界面修改）查找存储金币数值的地址

由于查找到的地址较多，因此需要在此基础上继续搜索

    int value[1024] = { -1 };
    int value_input = 1;
    cout << "请输入需要寻找的值：";
    cin >> value_input;
    int ans[1024] = { 0 };
    int num = 0;
    for (int addr = 0x00000000; addr < 0x08000000; )
    {
        ReadProcessMemory(handle, (LPVOID)addr, &value, 4096, NULL);
        for (int i = 0; i < 1024; i++) {
            if (value[i] == value_input)
            {
                ans[num++] = addr + 4 * i;
                cout << num << " . 0x";
                printf("%x\n", addr + 4 * i);
            }
        }
        addr += 4096;
    }
    cout << "find addr number is: " << num << endl;

3.使用金币（可以建造建筑或是训练士兵？这个无所谓，主要目的是使得金币数改变）

4.遍历步骤2中找到的地址，筛选出地址存储值为当前金币数的地址，直到地址条数为较少数值时停止筛选

    int v = -1;
    int n = 0;
    int res[512];
    n = 10;
    while (n > 4) {
        n = 0;
        cout << "请输入需要寻找的值(设置的值不得为-1，即-1为退出查找)：";
        cin >> value_input;
        for (int i = 0; i < num; i++)
        {
            ReadProcessMemory(handle, (LPVOID)ans[i], &v, 4, NULL);
            if (v == value_input)
            {
                res[n++]= ans[i];
                cout << n << " . 0x";
                printf("%x\n", ans[i]);
            }
        }
    }

5.修改金币存储地址处的数值

    int money = 1000000000;
    cout << "ACanary,what money do you want:";
    cin >> money;
    switch(n){
    case 1:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        break;
    case 2:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[1], &money, 4, NULL);
        break;
    case 3:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[1], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[2], &money, 4, NULL);
        break;
    case 4:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[1], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[2], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[3], &money, 4, NULL);
        break;
    }

关闭程序句柄，破解成功

3.破解测试

1.打开游戏遭遇战或局域网（初始资源为10000）

 

 2.初次搜索（当前资源金币为10000）

 3.使用资源（建造发电站）

 3.第二次搜索（当前资源金币为9200）

4.输入修改值99999999，游戏资源修改为99999999

 【注意：】更改游戏资源金币后并不会立刻刷新资源金币，需要再次使用金币资源，此时游戏资源才会刷新

 4.完整代码（如果无法运行那就是需要以管理员身份运行exe文件，权限不够）

#include<iostream>
#include<Windows.h>
using namespace std;

int main()
{
    const char* text = "命令与征服：红色警报3";
    HWND hwnd = FindWindow(NULL, text);
    if (hwnd == NULL)
    {
        MessageBox(NULL, TEXT("RA3 is not found! ywh must find it again!"), TEXT("crack_ra3"), MB_ICONSTOP);
        return 1;
    }

    DWORD pid;
    GetWindowThreadProcessId(hwnd, &pid);
    HANDLE handle = OpenProcess(PROCESS_ALL_ACCESS, FALSE, pid);
    if (!handle)
    {
        printf("ErrorCode=%x\n", GetLastError());
    }
    int value[1024] = { -1 };
    int value_input = 1;
    cout << "请输入需要寻找的值：";
    cin >> value_input;
    int ans[1024] = { 0 };
    int num = 0;
    for (int addr = 0x00000000; addr < 0x08000000; )
    {
        ReadProcessMemory(handle, (LPVOID)addr, &value, 4096, NULL);
        for (int i = 0; i < 1024; i++) {
            if (value[i] == value_input)
            {
                ans[num++] = addr + 4 * i;
                cout << num << " . 0x";
                printf("%x\n", addr + 4 * i);
            }
        }
        addr += 4096;
    }
    cout << "find addr number is: " << num << endl;
    int v = -1;
    int n = 0;
    int res[512];
    n = 10;
    while (n > 4) {
        n = 0;
        cout << "请输入需要寻找的值(设置的值不得为-1，即-1为退出查找)：";
        cin >> value_input;
        for (int i = 0; i < num; i++)
        {
            ReadProcessMemory(handle, (LPVOID)ans[i], &v, 4, NULL);
            if (v == value_input)
            {
                res[n++]= ans[i];
                cout << n << " . 0x";
                printf("%x\n", ans[i]);
            }
        }
    }

    int money = 1000000000;
    cout << "ACanary,what money do you want:";
    cin >> money;
    switch(n){
    case 1:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        break;
    case 2:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[1], &money, 4, NULL);
        break;
    case 3:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[1], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[2], &money, 4, NULL);
        break;
    case 4:
        WriteProcessMemory(handle, (LPVOID)res[0], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[1], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[2], &money, 4, NULL);
        WriteProcessMemory(handle, (LPVOID)res[3], &money, 4, NULL);
        break;
    }

    CloseHandle(handle);
    return 0;
}

---

 文章原创，禁止私自转载，转载前请联系博客获得同意，本文仅供学习交流