zoj 2109 FatMouse' Trade

FatMouse' Trade

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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:贪心法，找性价比最高的房间。

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

bool cmp(const room &a, const room &b){
    return a.aver > b.aver;
}

int main(){
    int m, n;
    cout.precision(3);
    vector<room> v;
    while(cin >> m >> n){
        if(m == -1 && n == -1)
            break;
        v.clear();
        for(int i = 0; i < n; i++){
            room rm;
            v.push_back(rm);
            cin >> v[i].j >> v[i].f;
            v[i].aver = double(v[i].j) / v[i].f;
        }
        sort(v.begin(), v.end(), cmp);
        double sum = 0;
        for(int i = 0; i < n; i++){
            if(m > v[i].f){
                sum += v[i].j;
                m -= v[i].f;
            } else {
                sum += (double)m * v[i].j / v[i].f;
                break;
            }
        }
        cout << fixed << sum << endl;
    }
    return 0;
}