bnu 45519

Om Nom and Candies

Time Limit: 1000ms

Memory Limit: 262144KB

This problem will be judged on  CodeForces. Original ID:  526C
64-bit integer IO format:  %I64d      Java class name:  (Any)

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                   Tag it!

A sweet little monster Om Nom loves candies very much. One day he found himself in a rather tricky situation that required him to think a bit in order to enjoy candies the most. Would you succeed with the same task if you were on his place?

One day, when he came to his friend Evan, Om Nom didn't find him at home but he found two bags with candies. The first was full of blue candies and the second bag was full of red candies. Om Nom knows that each red candy weighs Wr grams and each blue candy weighs Wb grams. Eating a single red candy gives Om Nom Hr joy units and eating a single blue candy gives Om Nom Hb joy units.

Candies are the most important thing in the world, but on the other hand overeating is not good. Om Nom knows if he eats more than C grams of candies, he will get sick. Om Nom thinks that it isn't proper to leave candy leftovers, so he can only eat a whole candy. Om Nom is a great mathematician and he quickly determined how many candies of what type he should eat in order to get the maximum number of joy units. Can you repeat his achievement? You can assume that each bag contains more candies that Om Nom can eat.

Input 

The single line contains five integers C, Hr, Hb, Wr, Wb (1 ≤ C, Hr, Hb, Wr, Wb ≤ 109).

Output 

Print a single integer — the maximum number of joy units that Om Nom can get.

Sample Input 

Input

10 3 5 2 3

Output

16

以为用完全背包解决 自己还是太天真  暴力两次 j<100000纯属投机取巧 刚开始时间超限 就在j上做了点文章

#include <stdio.h>

#include "stdio.h"

long long int max(long long int a,long long int b){return a>b?a:b;}

int main()

{

    long long int vmax,w1,w2,v1,v2,k,maxn=0;

    scanf("%lld%lld%lld%lld%lld",&vmax,&v1,&v2,&w1,&w2);

    k=vmax/w1;

    for(int j=0;j<=k&&j<=100000;j++)

        maxn=max(maxn,j*v1+(vmax-j*w1)/w2*v2);

    k=vmax/w2;

    for(int j=0;j<=k&&j<=100000;j++)

        maxn=max(maxn,j*v2+(vmax-j*w2)/w1*v1);

    printf("%lld\n",maxn);

    return 0;

}