一个价格的比较计算最小花费的问题，用C语言的解决

Problem Description
Today is the 1st anniversary of BestCoder. Soda, the contest manager, wants to buy a souvenir for each contestant. You can buy the souvenir one by one or set by set in the shop. The price for a souvenir is p yuan and the price for a set of souvenirs if q yuan. There's m souvenirs in one set.

There's n contestants in the contest today. Soda wants to know the minimum cost needed to buy a souvenir for each contestant.

Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤105), indicating the number of test cases. For each test case:

There's a line containing 4 integers n,m,p,q (1≤n,m,p,q≤104).

Output
For each test case, output the minimum cost needed.

Sample Input
2
1 2 2 1
1 2 3 4

Sample Output
1
3

#include<iostream>
#include"标头.h"
#include<vector>
#include"math.h"
using namespace std;
void main() {
    int cost1, cost2, m, n, p, q, T,temp;
    vector <int> min;
    while (1)
    {
        cin >> T;
        if (T>=1&&T<=105) {
            break;
        }
        else
            std::cout << "输入有误，请重新输入：（请输入1到105间的数字）" << endl;

    }
    for (int i = 1; i <=T; i++) 
    {
        cin >> n >> m >> p >> q;
        if (n>=1&&n<=104&&m>=1&&m<=104&&p>=1&&p<=104&&q>=1&&q<=104){    
            cost1 = n*p;
            if (n >= m)
            {
                int tp = ceil(n / m);
                cost2 = q*tp;
            }
            else
                cost2 = q;
            temp= compax(cost1, cost2); 
            min.push_back(temp);
        }

        else
        {
            std::cout << "输入错误，请重新输入" << endl;
            i = i - 1;
        }
    }
    for (int i = 0; i < T; i++) {
        cout << min[i]<<endl;
    }
};

这个代码你可以参考参考，写的有点复杂了，输入输出用的是c++的cin和cout，很久没有变c代码了，printf和scanf你可以换下，另外可以增加一些提示，输入字母之类的判断没做，可以补充下。