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  yuan and the price for a set of souvenirs if  yuan. There‘s  souvenirs in one set.

There‘s  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  , indicating the number of test cases. For each test case:

There‘s a line containing 4 integers  .

 

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.
 

 

Source

BestCoder 1st Anniversary ($)

 

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BestCoder官方解析：

1001 Souvenir

本题是一个简单的数学题. 假设套装优惠的话就尽量买套装, 否则单件买. 注意一下假设一直用套装的话可能在最后的零头不如单买好, 即.

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int t,n,m,p,q;//单位价格p元,套装q元，一个套装有m个纪念品，总共n个參赛者
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d",&n,&m,&p,&q);
        int price = 0;
        if(q/m<p)//假设套装优惠的话尽量买套装
        {
            if((n%m)*p<q)//假设在买套装最后零头的处理不如单位价格买廉价
            {
                price = (n/m)*q+(n%m)*p;//就在最后零头买单位价格
            }
            else
            {
                price = (n/m+1)*q;//否则多买一个套装
            }
        }
        else//否则直接单位价格买
        {
            price = n*p;
        }

        printf("%d\n",price);
    }
    return 0;
}

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