hdu1009

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.333
31.500

数据类型用double

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    double j,f;
    double x;
}cl[1010];
bool cmp(node x,node y)
{
    return x.x>y.x;
}
int main()
{
    int n;
    double m;
    double sum;
    while(~scanf("%lf%d",&m,&n)&&m!=-1&&n!=-1)
    {
        sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&cl[i].j,&cl[i].f);
            cl[i].x=(double)cl[i].j/cl[i].f;
        }
        sort(cl,cl+n,cmp);
        for(int i=0;i<n;i++)
        {
            if(cl[i].f<m)
            {
                sum+=cl[i].j;
                m-=cl[i].f;
                //cout<<"f"<<cl[i].f<<endl;
                //cout<<"sum"<<sum<<' '<<"m"<<m<<endl;
            }
            else
            {
                sum+=cl[i].x*m;
                break;
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}