Greedy Mouse

描述

A fat mouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his

favorite food:peanut. The warehouse has N rooms.The ith room containsW[i] pounds of peanut and requires 

F[i] pounds of cat food. Fatmouse does not have to trade for all the peanut in the room,instead,he may get 

 W[i]*a% pounds of peanut if he pays F[i]*a% pounds of cat food.The mouse is a stupid mouse,so can you tell 

him the maximum amount of peanut he can obtain.

輸入

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers W[i] and F[i] respectively. The test case is terminated by two -1. All integers are not greater than 1000.

輸出

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of penaut that FatMouse can obtain.

樣例輸入

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

樣例輸出

13.333
31.500

上傳者

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int cmp(double a,double b)
{
    return a>b;
}
int main()
{
    int m,n,w[1111],f[1111];
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(m==-1&&n==-1)
            break;
        int i,j,t1,t2;
        double ans=0,sum=0,a[1111];
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&w[i],&f[i]);
            a[i]=(double)w[i]/f[i];
        }
        for(i=0; i<n-1; i++)
            for(j=0; j<n-1-i; j++)
                if(a[j]<a[j+1])
                {
                    ans=a[j],t1=w[j],t2=f[j];
                    a[j]=a[j+1],w[j]=w[j+1],f[j]=f[j+1];
                    a[j+1]=ans,w[j+1]=t1,f[j+1]=t2;
                }
        for(i=0; i<n; i++)
        {
            if(m>f[i])
            {
                sum+=w[i];
                m=m-f[i];
            }
            else
            {
                sum+=(double)m*w[i]/f[i];
                break;
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}