FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 24   Accepted Submission(s) : 10
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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>
struct BEEn
{
    int x,y;
    double rate;


}F[1000];
void sort1(struct BEEn *F,int n)
{int i,j;
struct BEEn t;
    for(i=0;i<n-1;i++)
        for(j=0;j<=n-i-2;j++)
    {
        if(F[j].rate<F[j+1].rate)
         {

         t=F[j];
            F[j]=F[j+1];
            F[j+1]=t;

         }


    }
}
double sum(struct BEEn *F,int m, int n)
{ double i=0;
int j=0;
    while(m>0&&n-->0)
    { if(F[j].y>m){   i+=F[j].x*(m*1.0/F[j].y);  m=0;   }
    else {   m-=F[j].y;  i+=F[j].x;        }
        j++;


    }
    return i;
}
int main()
{
    int i,n,m;
    while(scanf("%d%d",&n,&m)&&n>0&&m>0)
    {
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&F[i].x,&F[i].y);
            F[i].rate=F[i].x*1.0/F[i].y;




        }
 sort1(F,m);

  printf("%.3lf\n",sum(F,n,m));


    }

    return 0;
}

习惯不好。。。以后改

n 是那个豆仓库的数量  ，就是可能猫粮还没换完，  那个豆没有了  每次从一个仓库拿豆去换 。 猫粮足够就全换。
每次换一个仓库的n--；

明白，谢谢！！