Transactions between mice and cats

A mouse liked cheese, but it was put in N rooms, and there was a cat guard in these rooms. Now it is ready to trade with the cats. It has m-pound cat food and wants to use this m-pound cat food for cheese. In every room guarded by a cat, there is a cheese J [I] pound, and the cat needs f [I] pound food. If the teacher gives the cat f [I] * a % of cat food, then it can get J [I] * A % cheese. Now we know the demand for cat food and the number of cheese in each room for each cat. How can a mouse get the most cheese?

Input

Enter M and N in the first line, followed by N rows (the number of cheeses and the demand for cat food in each room ). The program is terminated when m and n are-1 and-1.

Output

The branch outputs the maximum number of cheese obtained by the mouse and retains three decimal places.

Sample Input

5 37 24 35 220 325 1824 1515 10-1 -1

Sample output

13.33331.500

#include <stdlib.h>#include <stdio.h>int main(){    int m;    int n;    int k;    int i;    int a[1001],b[1001];    double sum=0;    double p[1001];  //    double max = 0;        while(scanf("%d %d",&m,&n)&&m!=-1&&n!=-1)    {                sum=0;  //        for(i=0;i<n;i++)        {            scanf("%d %d",&a[i],&b[i]);            p[i]=(double)a[i]/b[i];        }        while(m!=0&&n!=0)        {                        max=0;            for(i=0;i<n;i++)            {                if(p[i]>max)                {                    max=p[i];                    k=i;                }            }            if(m-b[k]>=0)            {                m=m-b[k];                sum+=a[k];                p[k]=0;            }            else            {                sum+=(double)p[k]*m;                m=0;            }        }        printf("%.3lf\n",sum);    }        return 0;}

 

Transactions between mice and cats