FatMouse's Trade(猫鼠交易）

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1247: FatMouse's Trade

时间限制: 1 Sec 内存限制: 32 MB
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题目描述

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.

输入

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.

输出

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.

样例输入

样例输出

2.286
2.500

#include<cstdio>
#include<algorithm>
using namespace std;
struct off
{
   double a;
   double b;
   double c;
}team[1001];
int cmp(off x,off y)
{
   return x.c>y.c;
}
int main()
{
   int m,n;
   while(scanf("%d %d",&m,&n))
   {
       if(m==-1&&n==-1)
       {
           break;
       }
       int i,j;
       for(i=0;i<n;i++)
       {
           scanf("%lf %lf",&team[i].a,&team[i].b);
           team[i].c =team[i].a /team[i].b ;
       }
       sort(team,team+n,cmp);
       double sum=0;
       for(i=0;i<n;i++)
       {
           if(m>team[i].b)
           {
               m-=team[i].b;
               sum+=team[i].a;
           }
           else
           {
               sum+=m*team[i].c ;
               break;
           }
       }
       printf("%.3lf\n",sum);
   }
   return 0;
}