hdu 1009:FatMouse' Trade（贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36632    Accepted Submission(s): 12064

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

 

Source

ZJCPC2004

 

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水题，读不懂题意是个坑，英语是个大问题啊……

 

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
    int N,M;
    while(cin>>M>>N,M!=-1 || N!=-1){
        int J[1001],F[1001];
        for(int i=1;i<=N;i++){
            cin>>J[i]>>F[i];
        }
        //ÅÅÐò
        for(int i=1;i<=N-1;i++)
            for(int j=1;j<=N-i;j++){
                if((double)J[j]/F[j]<(double)J[j+1]/F[j+1]){
                    int t;
                    t=F[j];F[j]=F[j+1];F[j+1]=t;
                    t=J[j];J[j]=J[j+1];J[j+1]=t;
                }
            }
        int fs=1,fe=1;

        //¼ÆËã
        double res=0;
        for(int i=1;i<=N;i++){
            if(M-F[i]>=0){
                res+=J[i];
                M-=F[i];
            }
            else{
                res+=M/(double)F[i]*(double)J[i];
                break;
            }
        }
        cout<<setiosflags(ios::fixed);
        cout<<setprecision(3);
        cout<<res<<endl;
    }
    return 0;
}

 Freecode : www.cnblogs.com/yym2013