杭电1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 36211    Accepted Submission(s): 11928

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

 

Recommend

JGShining

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贪心。   求能得到最多的食物

单位价格比最高的先选出来的。

#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
    int n;
    double a[1010],b[1010];
    int j,k,l,i,m;
    double sum,c,d;
    while(cin>>m>>n)
    {
        if(m==-1&&n==-1) break;
        i=n;
        sum=0.0;
        if(n==0)
        {
            cout<<setiosflags(ios::fixed)<<setprecision(3)<<sum<<endl;
            continue;
        }  //没有食物的情况
        while(n)
        {
            cin>>a[i-n]>>b[i-n];
            n-=1;
        }
        if(m!=0)
        {
            for(j=0; j<i-1; j++)
                for(k=j+1; k<i; k++)
                {
                    c=((double)(a[j])/(double)(b[j]));
                    d=((double)(a[k])/(double)(b[k]));
                    if(c<d)
                    {
                        l=a[j];
                        a[j]=a[k];
                        a[k]=l;
                        l=b[j];
                        b[j]=b[k];
                        b[k]=l;
                    }    //把单位价格从高到底排序
                }
            for(j=0; j<i; j++)
            {
                m-=b[j];
                if(m<=0)
                {
                    break;
                }
                sum+=(double)(a[j]);
            }
            m+=b[j];
            sum+=((double)m/(double)b[j])*(double)a[j];
        }
        else                 / /没钱的时候就只能找免费的咯~
        {
            for(j=0; j<i; j++)
            {
                if(b[j]==0)
                    sum+=a[j];
            }
        }
        cout<<setiosflags(ios::fixed)<<setprecision(3)<<sum<<endl;
    }
}