A. Optimal Currency Exchange
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.

Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.

Input
The first line of the input contains one integer n (1≤n≤108) — the initial sum in rubles Andrew has.

The second line of the input contains one integer d (30≤d≤100) — the price of one dollar in rubles.

The third line of the input contains integer e (30≤e≤100) — the price of one euro in rubles.

Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.

Input

100
60
70

Output

40

Input

410
55
70

Output

5

题意：有n个卢布，要换成美元和欧元，使手上剩余的卢克最少。一美元价值d卢布，一欧元价值e卢克。

思路：欧元的面值最小为5，则只需找5的倍数，美元最小面值为1。

写法：枚举。

AC代码：

//欧元的面值最小为5，其实的都少5的倍数，美元最小面值为1。
/*
1 dollar---> d rubles
1 euro ----> e rubles
 
dollar  1 2 5 10 20 50 100
euro     5 10 20 50 100 200
 
*/
#include<bits/stdc++.h>
 
using namespace std;
 
#define int long long
 
signed main(){
    int n,d,e;
    cin>>n>>d>>e;
    int ans=n;
    int x1=n/e;
    if(x1<5){
        printf("%lld\n",n-(n/d)*d);return 0;
    }else{
        int t1=n/(e*5);
        for(int i=0;i<=t1;i++){
            int temp=n;
            temp=temp-e*i*5-((n-e*i*5)/d)*d;
            //int sum2=temp-(temp/d)*d;
            ans=min(ans,temp);
        }
        printf("%lld\n",ans);
        
    }
    return 0;
}