• 结对项目—黄金点游戏


    程序步骤:

           开始游戏—>选择轮次—>选择玩家人数—>每个玩家输入其选择的0到100之间的整数—>得出黄金点G值—>得出比分

    代码来源:两人合作编写(莫军林 沈义杰)

    运行环境:windows

    编译语言:C++

    #include <iostream>
    #include<math.h>
    using namespace std;
    void main()
    {
    int i, j, num, n, m, S[100], S2[100], S3[100] = { 0};
    double S1[100], ave,sum,G;
    double max, min;
    cout << "==============欢迎进入黄金点游戏!=============" << endl;
    cout << "============== 游戏规则 ==============" << endl;
    cout << "=========n个玩家进行游戏=========" << endl;
    cout << "==每人写一个0~100之间的整数 (不包括0或100)==" << endl;
    cout << "===========G=n位玩家平均值avg*0.618==============" << endl;
    cout << "======最高分(1): 提交数字最接近G值的玩家======" << endl;
    cout << "=====最低分(-1):提交数字距离G值最远的玩家=====" << endl;
    cout << "===========:其他所有玩家为0分============" << endl;
    cout << "请输入游戏轮次!" << endl;
    cin >> n;
    cout << "请输入玩家人数:" << endl;
    cin >> m;
    for (i = 1; i <= n; i++)
    {
    sum = 0;
    G = 0;
    cout << "请输入0 到100之间的整数." << endl;
    for (j = 1; j <= m; j++)
    {
    cout <<" 玩家" << j << " :请输入你的整数!" << endl;
    cin >> num;
    S[j] = num;
    sum = sum + S[j];
    }
    ave = sum / num;
    G = ave*0.618;
    cout << "黄金点是G=" << G << endl;
    for (j = 1; j <= m; j++)
    {
    S1[j] = abs(S[j] - G);
    }
    max = min = S1[1];
    for (j = 1; j <= m; j++)
    {
    if (S1[j] >= max)
    max = S1[j];
    else if (S1[j] < min)
    min = S1[j];
    }
    for (j = 1; j <= m; j++)
    {
    if (S1[j] == max)
    S2[j] = -1;
    else if (S1[j] == min)
    S2[j] = 1;
    else S2[j] = 0;

    }
    for (j = 1; j <= m; j++)
    {
    cout << "本轮分数" << "S2[" << j << "]=" << S2[j] << endl;
    S3[j] = S3[j] + S2[j];
    cout << "总分数 " << "S3[" << j << "]=" << S3[j] << endl;
    }
    }
    system("pause"); //按任意键返回
    }

  • 相关阅读:
    Android的依赖注入框架:Dagger
    Android第三方开发组件
    Android 应用市场链接上传地址
    Android 开源框架以及开源项目以及连接
    Android 左右侧滑
    Android Textview 跑马灯
    Android XMPP推送
    Android多分辨率设配方案
    Android特效UI
    Android 时间的转换
  • 原文地址:https://www.cnblogs.com/mjl4396/p/7648394.html
Copyright © 2020-2023  润新知