HDOJ 1009 FatMouse' Trade

FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 60482    Accepted Submission(s): 20335
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
 
この問題は簡単な欲張り問題です.ここでは一度ではなく比例で交換できるので、一度で取り尽くさなければならないのではないでしょうか.計算する

#include<stdio.h>
int main()
{
    int a,b;
    int m[1000],n[1000];
    double k[1000],s,t;
    int i,j;
    while(scanf("%d%d",&a,&b)&&(a+1)||(b+1))
    {
    	s=0;
    	for(i=0;i<b;i++)
    	{
        	scanf("%d%d",&m[i],&n[i]);
            k[i]=(double)m[i]/n[i];
    	}
    	for(i=0;i<b-1;i++)
        	for(j=0;j<b-i-1;j++)
            {
                if(k[j]<k[j+1])
                {
                    t=k[j]; k[j]=k[j+1]; k[j+1]=t;
                    t=m[j]; m[j]=m[j+1]; m[j+1]=t;
                    t=n[j]; n[j]=n[j+1]; n[j+1]=t;
                }
            }
        i=0;
        while(a>=n[i]&&i<b)
        {
            s+=m[i];
            a-=n[i];
            i++;         
        }
        if(a>0&&i<b)
        s+=(double)a/n[i]*m[i];
        printf("%.3llf
",s);
    }
    return  0;
}

それぞれの性価比を並べ替えて高から低にすればよい.