题解 | #奥运排序问题#

测试用例有个人口数是0的，可能有点问题

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
struct score
{
    int id;                  // 国家id
    int gold;                // 金牌总数
    int medals;              // 奖牌总数
    double gold_to_people;   // 金牌人口比例
    double medals_to_people; // 奖牌人口比例
    int rank_1;
    int rank_2;
    int rank_3;
    int rank_4;
};

int main()
{
    int i, j, a[10], n, m;
    cin >> n >> m;
    int gold[n], medals[n], population[n];
    int yaopaide[m];
    for (i = 0; i < n; i++)
    {
        cin >> gold[i] >> medals[i] >> population[i];
        // cout << gold[i] << medals[i] << population[i] << endl;
    }
    for (i = 0; i < m; i++)
    {
        cin >> yaopaide[i];
    }
    struct score chengji[m]; // 计算各种方式下，各个要排的国家的数据，注意不是最终排名，而是存的数值
    for (i = 0; i < m; i++)
    {
        chengji[i].id = yaopaide[i];
        chengji[i].gold = gold[yaopaide[i]];
        chengji[i].medals = medals[yaopaide[i]];
        chengji[i].gold_to_people = gold[yaopaide[i]] * 1.0 / population[yaopaide[i]];
        chengji[i].medals_to_people = medals[yaopaide[i]] * 1.0 / population[yaopaide[i]];
    }
    int rank[4][m]; // 计算各种方式下，各个要排的国家的排名
    int zancun[m];
    // 第一种排序方式
    for (i = 0; i < m; i++)
    {
        zancun[i] = chengji[i].gold;
        // cout << zancun[i];
    }
    sort(zancun, zancun + m, greater<int>()); // 降序排列金牌数
    for (i = m - 1; i >= 2; i--)
    {
        if (zancun[i] == zancun[i - 1]) // 剔除重名次,变成-1
            zancun[i] = -1;
    }
    for (i = 0; i < m; i++)
    {
        for (j = 0; j < m; j++)
        {
            if (chengji[i].gold == zancun[j])
            {
                chengji[i].rank_1 = j + 1;
                break;
            }
        }
        // cout << chengji[i].rank_1 << endl;
    }
    // 第二种排序方式
    for (i = 0; i < m; i++)
    {
        zancun[i] = chengji[i].medals;
        // cout << zancun[i];
    }
    sort(zancun, zancun + m, greater<int>()); // 降序排列金牌数
    for (i = m - 1; i >= 2; i--)
    {
        if (zancun[i] == zancun[i - 1]) // 剔除重名次,变成-1
            zancun[i] = -1;
    }
    for (i = 0; i < m; i++)
    {
        for (j = 0; j < m; j++)
        {
            if (chengji[i].medals == zancun[j])
            {
                chengji[i].rank_2 = j + 1;
                break;
            }
        }
        // cout << chengji[i].rank_2 << endl;
    }
    double temp[m];
    // 第三张排序方式
    for (i = 0; i < m; i++)
    {
        temp[i] = chengji[i].gold_to_people;
        // cout << zancun[i];
    }
    sort(temp, temp + m, greater<int>()); // 降序排列金牌数
    for (i = m - 1; i >= 2; i--)
    {
        if (temp[i] == temp[i - 1]) // 剔除重名次,变成-1
            temp[i] = -1.0;
    }
    for (i = 0; i < m; i++)
    {
        for (j = 0; j < m; j++)
        {
            if (chengji[i].gold_to_people == temp[j])
            {
                chengji[i].rank_3 = j + 1;
                break;
            }
        }
        // cout << chengji[i].rank_3 << endl;
    }
    // 第四张排序方式
    for (i = 0; i < m; i++)
    {
        temp[i] = chengji[i].medals_to_people;
        // cout << zancun[i];
    }
    sort(temp, temp + m, greater<int>()); // 降序排列金牌数
    for (i = m - 1; i >= 2; i--)
    {
        if (temp[i] == temp[i - 1]) // 剔除重名次,变成-1
            temp[i] = -1.0;
    }
    for (i = 0; i < m; i++)
    {
        for (j = 0; j < m; j++)
        {
            if (chengji[i].medals_to_people == temp[j])
            {
                chengji[i].rank_4 = j + 1;
                break;
            }
        }
        // cout << chengji[i].rank_4 << endl;
    }
    int min, minsosrt;
    for (i = 0; i < m; i++)
    {
        min = chengji[i].rank_1;
        minsosrt = 1;
        if (min > chengji[i].rank_2)
        {
            min = chengji[i].rank_2;
            minsosrt = 2;
        }
        if (min > chengji[i].rank_3)
        {
            min = chengji[i].rank_3;
            minsosrt = 3;
        }
        if (min > chengji[i].rank_4)
        {
            min = chengji[i].rank_4;
            minsosrt = 4;
        }
        cout << min << ':' << minsosrt << endl;
    }

    return 0;
}