WA#2

 double sum=0;
        for(int i=0;i<n;i++)
        {
                cin >> c[i];
                sum+=c[i];
        }
sum=(an1+1.0*n*a0-2*sum)/(1.0*n+1);
        printf("%.2f\n",sum);


Edited by author 02.11.2006 18:40