FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 381   Accepted Submission(s) : 112

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

13.33331.500

 
这是贪心中最入门的问题，还是挂上来了
这种问题一眼就能看出来，求性价比最高的
首先按性价比排序，然后计算
#include<stdio.h>
#include<algorithm>
using namespace std;
struct E{
    double weight;
    double price;
    double cost;
}a[1001];
bool cmp(E a,E b){
    return a.cost>b.cost;
}
int main(){
    freopen("in.txt","r",stdin);
    double all;
    int n;
    while(scanf("%lf%d",&all,&n)!=EOF&&all!=-1&&n!=-1){
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&a[i].weight,&a[i].price);
            a[i].cost=a[i].weight/a[i].price;
        }
        sort(a,a+n,cmp);
        double ans=0;
        for(int i=0;i<n;i++){
            if(all>=a[i].price){
                ans+=a[i].weight;
                all-=a[i].price;
            }
            else{
                ans+=all/(a[i].price)*(a[i].weight);
                all=0;
            }
            if(all==0){
                break;
            }
        }
        printf("%.3lf\n",ans);
        
    
    }
    return 0;
}