解析

对于输入数据，每行输入的分别是当天获得的金牌、银牌和铜牌，所以输入应该这样处理：

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int Au, Ag, Cu;
        cin >> Au >> Ag >> Cu;
        
    }
    return 0;
}

处理好了每一行的输入后，建立三个sum，分别存储对应的奖牌即可。

题解

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int sumAu = 0;
    int sumAg = 0;
    int sumCu = 0;
    for (int i = 1; i <= n; i++)
    {
        int Au, Ag, Cu;
        cin >> Au >> Ag >> Cu;
        sumAu += Au;
        sumAg += Ag;
        sumCu += Cu;
    }
    cout << sumAu << " " << sumAg << " " << sumCu << " " << sumAu + sumAg + sumCu;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n, gold = 0, silver = 0, copper = 0, sum = 0;
    cin >> n;
    int g, s, c;
    for (int i = 1; i <= n; i++)
    {
        cin >> g >> s >> c;
        gold += g;
        silver += s;
        copper += c;
        sum += g + s + c;
    }
    cout << gold << " " << silver << " " << copper << " " << sum;
    return 0;
}