HDU-1009（简单贪心）

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

 

 

分析：按J[i]/F[i]的价值从大到小贪。

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1010;
struct javabean{
    int j,f;
    double v;
}value[maxn];
bool cmp(javabean a,javabean b){
    return a.v>b.v;
}
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)){
        if(m==-1&&n==-1) break;
        for(int i=0;i<n;i++){
            scanf("%d%d",&value[i].j,&value[i].f);
            value[i].v=(double)value[i].j/(value[i].f*1.0);
        }
        sort(value,value+n,cmp);
        double sum=0.0;
        for(int i=0;i<n;i++){
            if(m>=value[i].f){
                sum+=value[i].j;
                m-=value[i].f;
            }
            else{
                sum+=value[i].v*m;
                break;
            }
        }
        printf("%.3f
",sum);
    }
    return 0;
}