C++ 闯关游戏：代码示例与实现步骤

下面是一个简单的用 C++ 实现的闯关游戏的示例：

#include <iostream>
#include <cstdlib>
#include <ctime>

using namespace std;

void playGameLevel(int levelDifficulty) {
    // 生成随机数
    const int codeA = rand() % levelDifficulty + levelDifficulty;
    const int codeB = rand() % levelDifficulty + levelDifficulty;
    const int codeC = rand() % levelDifficulty + levelDifficulty;

    const int codeSum = codeA + codeB + codeC;
    const int codeProduct = codeA * codeB * codeC;

    // 输出提示信息
    cout << "【等级 ' << levelDifficulty << "】" << endl;
    cout << "+ 有 3 个数字" << endl;
    cout << "+ 数字之和为: ' << codeSum << endl;
    cout << "+ 数字之积为: ' << codeProduct << endl;

    int guessA, guessB, guessC;
    cin >> guessA >> guessB >> guessC;

    int guessSum = guessA + guessB + guessC;
    int guessProduct = guessA * guessB * guessC;

    // 判断玩家输入是否正确
    if (guessSum == codeSum && guessProduct == codeProduct) {
        cout << "恭喜你过关了！" << endl;
    } else {
        cout << "很遗憾，你输入的数字有误，再试一次吧！" << endl;
    }    
}

int main() {
    srand(time(NULL)); // 初始化随机种子

    int levelDifficulty = 1;
    int maxDifficulty = 5;

    while (levelDifficulty <= maxDifficulty) {
        playGameLevel(levelDifficulty);
        cin.clear(); // 清除输入缓冲区
        cin.ignore(); // 忽略之前的输入

        levelDifficulty++;
    }

    cout << "恭喜你通关了所有关卡！" << endl;

    return 0;
}

在这个游戏中，玩家需要猜测 3 个数字的和与积。每个关卡的数字范围和难度会逐渐增加。如果玩家输入的数字与关卡要求一致，则过关，否则需要再次尝试。玩家需要通过 5 个关卡才能通关。