hdu_1009_FatMouse' Trade_201310280910

FatMouse' Trade

http://acm.hdu.edu.cn/showproblem.php?pid=1009

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 35250    Accepted Submission(s): 11553

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>

typedef struct ST
{
    int j;
    int f;
    double t;
}ST;
ST s[1010];

int cmp(const void *a,const void *b)
{
    return (*(ST *)a).t > (*(ST *)b).t ? 1 : -1;
}

int main()
{
    int m,n;
    while(scanf("%d %d",&m,&n),(m!=-1&&n!=-1))
    {
        int i,j;
        int num;
        double sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&s[i].j,&s[i].f);
            s[i].t = s[i].j*1.0/s[i].f;
        }
        qsort(s,n,sizeof(s[0]),cmp);
        num=m;
        for(i=n-1;i>=0;i--)
        {
            if(num>s[i].f)
            {
                sum+=s[i].j;
                num-=s[i].f;
            }
            else
            {
                sum+=s[i].t * num;
                num=0;
            }
            if(num==0)
            break;
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}