hdu5310 Souvenir（模拟）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5310

Souvenir

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 214 Accepted Submission(s): 140



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

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 ($)



AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#define LL long long
#define MAXN 1000010
using namespace std;
int num[MAXN];
LL ans=0;
int main()
{
	int T,n,m,p,q,s;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d%d",&n,&m,&p,&q);
		ans=0;
		if(m>=n)
		{
			if(q<=p*n)
			{
				ans=q;
			}
			else
			{
				ans=n*p;
			}
		}
		else
		{
			if(q<=p*m)
			{
				s=n/m;
				ans=s*q;
				LL ans1=ans+(n-s*m)*p;
				LL ans2=ans+q;
				ans=min(ans1,ans2);
			}
			else
			{
				ans=n*p;
			}
		}
		printf("%I64d\n",ans);
	}
	return 0;
 }