FatMouse' Trade（hdoj1009）

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

#include<stdio.h>/**此题有很多边缘数据没有想到，或者压根就没有去想/ 
#include<algorithm>
using namespace std;
typedef struct
{
    double x,y;/*数据类型错点*/
    double z;
}node;
int cmp(node p,node q)
{
    return p.z>=q.z;
}
node a[1002];
int main()
{
    double M;
    int N;
    while(~scanf("%lf%d",&M,&N))
    {
        int i,j;
        double max=0,shen=M;
        if(M==-1&&N==-1)
            break;
        if(N==0)/*room为0情况*/ 
            {
                printf("0.000
");
                continue;
            }
        else
        {
            for(i=0;i<N;i++)
                {
                    scanf("%lf%lf",&a[i].x,&a[i].y);
                    a[i].z=a[i].x/a[i].y;
                }
            sort(a,a+N,cmp);
            /*for(i=0;i<N;i++)
                printf("%d %d  %lf
",a[i].x,a[i].y,a[i].z);*/
            for(i=0;;i++)
            {
                if(shen>a[i].y)
                {
                    max+=a[i].x;
                    shen=shen-a[i].y;
                }    
                else if(shen==a[i].y)/*可能漏掉，因为存在y=0情况,此情况不能和else并在一起，看了讨论组才知道的*/ 
                {
                    max+=a[i].x;
                    shen=shen-a[i].y;
                }    
                else
                {
                    max+=(shen/a[i].y)*a[i].x;
                    break;
                }
            }
        }
        printf("%.3lf
",max);
    }
}