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): 36309    Accepted Submission(s): 11957

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

贴上代码：

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

const int maxn = 1001;
struct Node{
    double j, f , value;
    bool operator < ( const Node& v )  //从大到小
    {
        return value > v.value;
    }
}room[maxn];

int main(){
    int N;
    double jb , food ;
    while ( cin>>food>>N )
    {
        if ( food==-1&& N==-1 )
            break;
        jb=0;
        for ( int i=0;i<N;i++ ){
            cin>>room[i].j>>room[i].f;
            room[i].value = room[i].j / room[i].f;
        }
        sort ( room , room+N );
        for ( int i=0; i<N; i++ ){
            if ( room[i].f <= food )
            {
                food-=room[i].f;
                jb+=room[i].j;
            }
            else{
                jb += food*room[i].j / room[i].f;
                break;
            }
        }
        cout<<fixed<<setprecision(3)<<jb<<endl;
    }
    return 0;
}