Codeforces Round #266 (Div. 2) A

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概述

题目：

Ann has recently started commuting by subway. We know that a one ride subway ticket costs a rubles. Besides, Ann found out that she can buy a special ticket for m rides (she can buy it several times). It costs b rubles. Ann did the math; she will need to use subway n times. Help Ann, tell her what is the minimum sum of money she will have to spend to make n 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 the m ride ticket, the price of a one ride ticket and the price of an m ride ticket.

Output

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

Sample test(s)

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 three m ride tickets.

题意分析：

很简单的贪心，但是边缘数据和取整注意细节。

代码：

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <numeric>
using namespace std;
int main()
{
int n, m, a, b;
cin >> n >> m >> a >> b;
double f = (double)b/(double)m;
if (f < (double)a)
{
int k1 = n/m;
int k2 = n-k1*m;
int k3 = (k1+1)*b;
cout << min(k1*b+k2*a, k3);
}
else
{
cout << n*a << endl;
}
return 0;
}