杭电 1009 FatMouse' Trade （贪心）

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

 

 

 

题目大意：

老鼠有M磅猫食。有N个房间，每个房间前有一只猫，房间里有老鼠最喜欢的食品JavaBean,J[i]。若要引开猫，必须付出相应的猫食F[i]。当然这只老鼠没必要每次都付出所有的F[i]。若它付出F[i]的a%，则得到J[i]的a%。求老鼠能吃到的做多的JavaBean。

#include<cstdio>
#include<algorithm>
using namespace std;
struct stu
{
    double a,b;  //不用float 
}a[50000];
bool cmp(stu a,stu b)
{
    return (a.a/a.b)>(b.a/b.b);
}
int main()
{
    int m,n,i,j;
    double sum;        
    while(scanf("%d %d",&m,&n)&&(m!=-1)&&(n!=-1))
    {
        sum=0;
        for(i=0; i < n; i++)
        {
            scanf("%lf %lf",&a[i].a,&a[i].b);
        }
        sort(a,a+n,cmp);
        for(i=0;i<n;i++)
        {
            if(m > a[i].b)
            {
                sum+=a[i].a;
                m-=a[i].b;
            }
            else
            {
                sum+=((m/a[i].b)*a[i].a);
                break;
            }
        }
        printf("%.3lf
",sum);    
    }
}