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题目描述

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.

输入

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.

输出

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.

样例输入

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

样例输出

13.333
31.500

提示

杭电1009

代码如下

#include"stdio.h"
double a[1001][3]={0};
int found(int n)
{
       double max=0;
       int j=0,i;
       for(i=0;i<n;i++)
      {
             if(a[i][2]>max)
              {
                   max=a[i][2];
                      j=i;
             }
      }
     if(max==0)
           return -1;
    else
          return j;

}
int main()
{
       double m,sum=0;
       int n,t=0,i;
       int f, j;
       scanf("%lf %d",&m,&n);
         while(m!=-1&&n!=-1)
        {
	      for(i=0;i<n;i++)
            {
                 scanf("%lf %lf",&a[i][0],&a[i][1]);
                  a[i][2]=a[i][0]/a[i][1];
            }
            while(m!=0)
            {
                    t=found(n);
                   if(t==-1)
                   {
                          break;
                   }
                   if(m>=a[t][1])
                  {
                          m-=a[t][1];
                         sum+=a[t][0];
                        a[t][2]=0;
                   }
                  else
                  {
                       sum+=((double)m)/((double)a[t][1])*a[t][0];
                          break;
                  }
           }
          printf("%.3lf\n",sum);
	  scanf("%lf %d",&m,&n);
	  sum=0;
  
        }

    return 0;
}

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