HDU1009

一 题意描述：

                                     FatMouse' Trade

                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                       Total Submission(s): 42443    Accepted Submission(s): 14123

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

 二 思路分析：

本题目主要考察贪心算法，即可以考虑用f/j的比值大小作为依据，显然f/j的值越大，老鼠和猫交易越占优势。那么我们便看可以采用f/j的数值进行排序，从前往后进行交易，直到m为0停止。

三 代码展示：

#include <iostream>
#include <algorithm>
#include<iomanip>
using namespace std;
#define maxn 1005
struct Room
{
    int f,j;
    double value;
}room[maxn];
int cmp(Room a,Room b)
{
    return a.value>b.value;
}
int main()
{
    int m,n;
    while(cin>>m>>n)
    {
        if(m==-1&&n==-1) break;
        for(int i=0;i<n;i++)
        {
         cin>>room[i].j>>room[i].f;
         room[i].value=(1.0*room[i].j) /room[i].f;
        }
        sort(room,room+n,cmp);//排序
        double total=0;
          for(int i=0;i<n;i++)
          {
              if(m<=room[i].f){total+=(room[i].j*1.0*m)/room[i].f;break;}
              else {total+=room[i].j;m-=room[i].f;}
          }
        cout<<setiosflags(ios::fixed)<<setprecision(3)<<total<<endl;

    }
    return 0;
}