Codeforces 466A Cheap Travel【水题】暴力

A. Cheap Travel

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ann has recently started commuting by subway. We know that a one ride subway ticket costsa rubles. Besides, Ann found out that she can buy a special ticket form rides (she can buy it several times). It costsb rubles. Ann did the math; she will need to use subwayn times. Help Ann, tell her what is the minimum sum of money she will have to spend to maken rides?

Input

The single line contains four space-separated integers n, m, a, b (1 ≤ n, m, a, b ≤ 1000) — the number of rides Ann has planned, the number of rides covered by them ride ticket, the price of a one ride ticket and the price of anm ride ticket.

Output

Print a single integer — the minimum sum in rubles that Ann will need to spend.

Examples

Input

6 2 1 2

Output

6

Input

5 2 2 3

Output

8

Note

In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy threem ride tickets.

题目大意：

主人公一共需要坐n次列车，其中有两种票，单票是a元一个，能坐一次列车，m票是b元一个，能坐m次列车。

问怎样购买能够坐n次列车并且花费最小？求这个最小花费。

思路：

观察到n并不大，我们可以通过暴力枚举单票和m票购买的个数来枚举出这个最小花费。

枚举过程中对可行解的最小花费进行统计即可。

Ac代码：

#include<stdio.h>#include<iostream>#include<string.h>using namespace std;int main(){    int n,m,a,b;    while(~scanf("%d%d%d%d",&n,&m,&a,&b))    {        int output=0x3f3f3f3f;        for(int i=0;i<=1000;i++)        {            for(int j=0;j<=1000;j++)            {                if(i+j*m>=n)                {                    output=min(output,a*i+b*j);                }            }        }        printf("%d\n",output);    }}