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.

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  .

For each test case, output the minimum cost needed.

2
1 2 2 1
1 2 3 4

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.
 

BestCoder 1st Anniversary ($)

hujie   |   We have carefully selected several similar problems for you:  5315 5314 5313 5312 5311 

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;
}

中文题目在以下：

今天是BestCoder一周年纪念日. 比赛管理员Soda想要给每一个參赛者准备一个纪念品. 商店里纪念品的单位价格是元, 同一时候也能够花元购买纪念品套装, 一个套装里有个纪念品.

今天总共同拥有个參赛者, Soda想要知道最少须要花多少钱才干够给每一个人都准备一个纪念品.

输入有多组数据. 第一行有一个整数 , 表示測试数据组数. 然后对于每组数据:

一行包括4个整数  .

对于每组数据输出最小花费.

2
1 2 2 1
1 2 3 4

1
3

对于第一组数据, Soda能够1元购买一个套装. 对于第二组数据, Soda能够直接花3元购买一个纪念品.