HDU1009FatMouse' 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

Author

CHEN, Yue

Source

ZJCPC2004 

#include<stdio.h>
#include<algorithm>//sort的头文件
using namespace std; 
const int maxn=1000+10;
struct node{
    double j,f;
    double r;
}a[maxn];
bool cmp(node a,node b)//从大到小排序 
{
    return a.r>b.r;
}
int main()
{
    int m,n,i;
    double sum;
    while(scanf("%d%d",&m,&n)==2&&(m!=-1&&n!=-1))
    {
        for(i=0;i<n;i++){
            scanf("%lf%lf",&a[i].j,&a[i].f);
            a[i].r=a[i].j/a[i].f;
        }
        sort(a,a+n,cmp);
        sum=0.0;
        for(i=0;i<n;i++)
        {
            if(m>=a[i].f)
            {
                sum+=a[i].j;
                m-=a[i].f;
            }
            else
            {
                sum+=(a[i].r)*m;
                break;
            }
        }
        printf("%.3lf
",sum);
    }
    return 0;
}