[HDOJ#1009]FatMouse' Trade

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

 

 

My solution:

 

#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;
typedef struct Pecent
{
    double a;
    double b;
    double p;
} node;

int main(void)
{
    int m,n;
    Pecent pe[1001];
    double ob;
    while (cin>>m>>n)
    {
        ob=0.0;
        if (m<0) break;
        for (int i=0;i<n;i++)
        {
            cin>>pe[i].a>>pe[i].b;
            pe[i].p=pe[i].a/pe[i].b;
        }
    bool cmp(const node a,const node b);
    sort(pe,pe+n,cmp);
    for (int i=0;i<n&&m>=0;i++)
    {        
        if (m<pe[i].b&&m>0)
        {
            ob+=1.0*m/pe[i].b*pe[i].a; 
            m=0;
        }
        if (m>=pe[i].b) 
        {
            m-=pe[i].b;
            ob+=pe[i].a;
        }

    }
    cout<<setiosflags(ios::fixed)<<setprecision(3)<<ob<<endl;
    }
    return 0;
}
bool cmp(const node a,const node b)
{
    return a.p>b.p;
}