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();
     vector<room> v;
     while(cin >> m >> n){
         if(m == - && n == -)
             break;
         v.clear();
         for(int i = ; 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 = ;
         for(int i = ; 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 ;
 }