FatMouse' Trade

FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 21   Accepted Submission(s) : 13

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Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

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 J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

Output

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

Sample Input

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

Sample Output

13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    int M,N,i,J[10000],F[10000],k,j,tmp,sign;
    double SUM;
    while(scanf("%d%d",&M,&N)!=EOF)
    {
        if(M==-1&&N==-1)
            break;
        k=0;SUM=0;
        while(N--)
        {
            scanf("%d%d",&J[k],&F[k]);
            k++;
        }
        for(i=k-1;i>=0;i--)
        {
             sign=i;
             for(j=i-1;j>=0;j--)
                if(J[sign]*F[j]>J[j]*F[sign])sign=j;
             if(sign!=i)
             {
                 tmp=J[i];
                 J[i]=J[sign];
                 J[sign]=tmp;

                tmp=F[i];
                 F[i]=F[sign];
                 F[sign]=tmp;
             }
        }

        for(i=0;i<k;i++)
        {
            M-=F[i];
            if(M<0)
                {SUM+=((J[i]*1.0)/F[i])*(M+F[i]);break;}
            SUM+=J[i];
        }
        printf("%.3lf
",SUM);

    }
    return 0;
}

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