FatMouse' Trade


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

 

题解：我建立了一个结构体，然后将食物价值用r表示，然后把结构体排序，依次从最大的开始递减，可以保证价值最大。

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
struct food
{
    double j;
    double f;
    double r;
}fo[1500];
bool compare(food a,food b)
{
    return a.r>b.r;
}
int main()
{
    int m,n;
    while(scanf("%d %d",&m,&n),m!=-1,n!=-1)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lf %lf",&fo[i].j,&fo[i].f);
            fo[i].r=fo[i].j/fo[i].f;
        }
        sort(fo,fo+n,compare);
        double ans=0;
        for(int i=0;i<n;i++)
        {
            if(m>=fo[i].f)
            {
                ans+=fo[i].j;
                m-=fo[i].f;
            }
            else
            {
                ans+=m*fo[i].r;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}