Hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43381    Accepted Submission(s): 14499

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

 

 

Author

CHEN, Yue

 

贪心题：题目的意思是FatMouse有m磅的cat food，它想用这些cat food去换JavaBean，然后有n个房间，每个房间有相应的JavaBean和需要多少cat food。问m磅cat food最多可以换得多少JavaBean？

首先要将这些数据按照性价比（即JavaBean除以cat food 注意性价比有可能是小数）从大到小排序，然后问题就迎刃而解了。

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<iomanip>
using namespace std;
#define N 1005

struct Trade{
    int J, F;       //J代表JavaBeans，F代表cat food
    double price;   //性价比
}trade[N];

int cmp(Trade a,Trade b)
{
    return a.price>b.price;
}
int main()
{
    double maximum;
    int m, n, i;
    while(cin>>m>>n)
    {
        maximum = 0;
        if(m==-1 && n==-1)
            break;
        maximum = 0;
        for(i=0; i<n; i++)
        {
            cin>>trade[i].J>>trade[i].F;
            trade[i].price = trade[i].J*1.0/trade[i].F;
        }
        sort(trade,trade+n,cmp);
        for(i=0; i<n; i++)
        {
            if(m == 0)
                break;
            else if(trade[i].F<=m)
            {
                maximum += trade[i].J;
                m -= trade[i].F;
            }
            else
            {
                maximum += m*trade[i].price;
                m = 0;
            }
        }
        printf("%.3lf\n",maximum);
    }
    return 0;
}