hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67291    Accepted Submission(s): 22911

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

 

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef struct {
    int j, f;
    double aver;
}food;

bool cmp(food a, food b){
    return a.aver >= b.aver;
}

food javabean[1005];
int main(){
    int m, n;
    while(scanf("%d%d", &m, &n) != EOF){
        if((m == -1) && (n == -1))
            break;
        
        for(int i = 0; i < n; i++){
            scanf("%d %d", &javabean[i].j, &javabean[i].f);
            javabean[i].aver = (double)javabean[i].j / javabean[i].f;
        }
        sort(javabean, javabean+n, cmp);
        double sum = 0;
        for(int i = 0; i < n; i++){
            if(m >= javabean[i].f){
                sum += javabean[i].j;
                m -= javabean[i].f;
            }
            else {
                sum += (double)javabean[i].j / javabean[i].f * m;
                break;
            }
        }
        printf("%.3lf
", sum);
    }
    return 0;
}