题解 | #记负均正#

#include <stdio.h>

int main() {
    int n,a[2010],k=0,l=0,sum=0;
    double aver;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
       {
        scanf("%d ",&a[i]);
        if(a[i]<0)
           k++;
        else if (a[i]==0) 
        {
           l++;
        }
        else
         sum+=a[i];
       }  
       if(sum==0)
       aver=0.0;
       else
      aver=(double)sum/(n-k-l);
      printf("%d %.1lf",k,aver);
    return 0;
}