HDU 1009

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

 

大致题意：

  胖老鼠有 M 的猫粮，告诉你好多房间，里面可以用猫粮换吃的，以及换的方式。问你最多换多少。

解题思路：

  贪心，按比率排个序取最划算就好。

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
struct P{
    int f,j;
    double r;
}s[1005];
bool cmp(P x,P y){
    return x.r>y.r;
}
int main(){
    int m,n;
    while(cin>>m>>n,n+m!=-2){
        for(int i=0;i<n;i++){
            cin>>s[i].f>>s[i].j;
            s[i].r=1.0*s[i].f/s[i].j;
        }
        sort(s,s+n,cmp);
           int j=0;
        double ans=0;
        while(m>0&&j<n){
            if(m>=s[j].j){
               m-=s[j].j;
               ans+=s[j].f;
            }
            else{
                ans+=1.0*m/s[j].j*s[j].f;
                m=0;
            }
            j++;
        }
        printf("%.3lf
",ans);
    } return 0;
}