hdu 1009

FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49098 Accepted Submission(s): 16485

 

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.

 

 

简单贪心....

需要注意的是数据是非负，所以有0的情况要考虑周全，基本都要特殊处理。

多WA了三次，不知道为什么交C++可以过，交G++就不行。

#include <iostream>

#include <cstring>

#include <algorithm>

#include <cstdio>

#include <iomanip>



using namespace std;



int main()

{

    int  m,n,f[1500],j[1500];

    double scale[1500];

    double ans,sum;

    while(scanf("%d %d",&m,&n)!=EOF&&(m!=-1))

    {  ans=0;

      //  if (m==0)

         memset(scale,0,sizeof(scale));

         memset(f,0,sizeof(f));

         memset(j,0,sizeof(j));

       //  bool flag=false;

       for (int i=1;i<=n;i++)

        {

            scanf("%d %d",&j[i],&f[i]);

            if (f[i]!=0)

            scale[i]=(double)j[i]*1.0/f[i];

            else if (j[i]!=0)

            {

                  ans=ans+j[i];



            }

        }

       // if (flag) {cout<<fixed<<setprecision(3)<<ans<<endl;continue;}

        for (int i=1;i<=n-1;i++)

          for (int k=i+1;k<=n;k++)

          if (scale[i]<scale[k])

        {

            swap(scale[i],scale[k]);

            swap(j[i],j[k]);

            swap(f[i],f[k]);

        }

        sum=0;

       int i=1;

        while (m>=sum&&i<=n)

        {

          ans=ans+j[i];

          sum=sum+f[i];

          i++;

        }

        i--;

        ans=ans-j[i];

        sum=sum-f[i];

        ans=ans+(m-sum)*scale[i];

        cout<<fixed<<setprecision(3)<<ans<<endl;





    }

    return 0;

}