FatMouse' Trade（杭电ACM---1009）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103515    Accepted Submission(s): 36159

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

 

题意：

这一道题意思就是老鼠用猫食物换取自己最喜爱的食物javaBean的过程，当然换取的最终结果是保证最后的JavaBean是最多的，

而且是当自己手中的猫食物小于每个仓库所需交换的猫食物时候，可以手中有多少就交换多少。

所以在解这道题时候要想到按照每个仓库javaBean最大的比率排序才能保证最后的交换的javaBean是最大的。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1e3+10;
int m,n;
struct node{
    int j,f;
    double r;
} food[maxn];
class cmp{
public:
    bool operator()(node a,node b)const{
        return a.r>b.r;
    }
};
int main(){
    while(~scanf("%d%d",&m,&n)){
        if(m==-1) break;
        for(int i=0; i<n; i++ ){
            cin>>food[i].j>>food[i].f;
            food[i].r=1.0*food[i].j/food[i].f;
        }
        sort(food,food+n,cmp());
        double ans=0;
        for( int i=0; i<n; i++ ){
            if(m>=food[i].f){
                ans+=food[i].j;
                m-=food[i].f;
            }
            else{
                if(m==0) break;
                else{
                    ans+=m*food[i].r;
                    break;
                }
            }
        }
        printf("%.3f\n",ans);
    }
    return 0;
}