HDU 1009 FatMouse' Trade（肥老鼠…

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17063    Accepted Submission(s): 5207

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

 

Recommend

JGShining

根据效益排序，然后依次减去，知道不能减为止，

代码：

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

struct room{

    int j;

    int f;

    double k;

}room[1002];

int cmp(const void *a,const void *b)

{

    struct room *c,*d;

    c=(struct room*)a;

    d=(struct room*)b;

    return c->k < d->k?1:-1;

}

int main()

{

    int m,n,i,num;

    double sum;

    while(scanf("%d%d",&m,&n),m!=-1||n!=-1)

    {

        for(i=0;i<n;i++)

        {

            scanf("%d%d",&room[i].j,&room[i].f);

            room[i].k=room[i].j*1.0/room[i].f;

        }

        qsort(room,n,sizeof(room[0]),cmp);

        sum=0;num=m;

        for(i=0;i<n;i++)

        {

            if(num<=room[i].f)

            {

                sum=sum+1.0*num*room[i].k;

                break;

            }

            else 

            {

                sum=sum+room[i].j;

                num=num-room[i].f;

            }

        }

        printf("%.3lf\n",sum);

    }

    return 0;

}