BestCoder 1st Anniversary ($) 1001 Souvenir

Souvenir

 

 Accepts: 901

 

 Submissions: 2743

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 262144/262144 K (Java/Others)

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 pp yuan and the price for a set of souvenirs if qq yuan. There's mm souvenirs in one set.

There's nn 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 TT (1 \le T \le 10^5)(1≤T≤10​5​​), indicating the number of test cases. For each test case:

There's a line containing 4 integers n, m, p, qn,m,p,q (1 \le n, m, p, q \le 10^4)(1≤n,m,p,q≤10​4​​).

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

Hint

For the first case, Soda can use 1 yuan to buy a set of 2 souvenirs. For the second case, Soda can use 3 yuan to buy a souvenir.

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
    int T,n,m,p,q,i,j,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d %d",&n,&m,&p,&q);
        if((double) p<= (double)q/m)
        {
            printf("%d\n",n*p);
        }
        else
        {
            int x=0;
            for(i=1;;i++)
            {
                x=x+m;
                if(x>=n)
                    break;
            }
            printf("%d\n",min(i*q,(i-1)*q+(n-(x-m))*p));
        }
    }
    return 0;
}