找水王2

1.設計思路：這次要找3個水王，是以我這次在上次的基礎上設定了3個水王的權值。

2.代碼

#include <iostream>
#include<string>
using namespace std;
#define N 100
int main() {
    int n, i, X[N];
    string ID[N];
    cout << "請輸入文章總數：" << endl;
    cin >> n;
    cout << "請輸入文章序号以及其相對應的ID:" << endl;
    for (i = 0; i < n; i++)
    {
        cin >> X[i];
        cin >> ID[i];
    }
    cout << "文章表為：" << endl;
    cout << "序号" << '\t' << "ID" << endl;
    for (i = 0; i < n; i++)
    {
        cout << "\t" << X[i] << '\t' << ID[i] << endl;
    }
    int quanzhi[3] = { 0 }, water[3] = { 0,1,2 };
    for (int i = 0; i<n; i++)
    {
        if (ID[water[0]] == ID[i])
        {
            quanzhi[0]++;
        }
        else if (ID[water[1]] == ID[i])
        {
            quanzhi[1]++;
        }
        else if (ID[water[2]] == ID[i])
        {
            quanzhi[2]++;
        }
        else if (quanzhi[0] == 0)
        {
            ID[water[0]] = ID[i];
        }
        else if (quanzhi[1] == 0)
        {
            ID[water[1]] = ID[i];
        }
        else if (quanzhi[2] == 0)
        {
            ID[water[2]] = ID[i];
        }
        else
        {
            quanzhi[0]--;
            quanzhi[1]--;
            quanzhi[2]--;
        }
    }
    for (int i = 0; i<3; i++)
    {
        cout << "第 " << i + 1 << " 個水王是：" << ID[water[i]] << endl;
    }
    
}      

3.程式截圖