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

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

Freecode : www.cnblogs.com/yym2013