var i,j,k,n:longint;
begin
 read(n);

  for i:=1 to 250 do
   if sqr(i)=n then begin write(1); exit; end;

  for i:=1 to 250 do
   for j:=1 to 250 do
    if sqr(i)+sqr(j)=n then begin write(2); exit; end;

  for i:=1 to 250 do
   for j:=1 to 250 do
    for k:=1 to 250 do
     if sqr(i)+sqr(j)+sqr(k)=n then begin write(3); exit; end;

write(4);
end.

If N was bigger, you would have time limit.

I don't think so. My code got AC.
#include<stdio.h>

int i,j,k;
long int n;

int main()
{
           scanf("%d",&n);

           for(i=1;(i*i)<=n;i++)
                               if((i*i)==n)
                               {
                                            printf("%d",1);
                                            return 0;
                               }
           for(i=1;i*i<=n;i++)
           for(j=1;j*j<=n;j++) if((i*i+j*j)==n)
                               {
                                printf("%d",2);
                                return 0;
                               }

           for(i=1;i*i<=n;i++)
           for(j=1;j*j<=n;j++)
           for(k=1;k*k<=n;k++) if((i*i+j*j+k*k)==n)
                               {
                                printf("%d",3);

                                return 0;
                               }

           printf("%d",4);
           return 0;


}
But this solution is bad.

I think it's DP in this problem O(n*sqrt(n));

this is the most epic answer I've read....hhhh :) u are awesome
thanks dud and good luck

for(i=1;(i*i)<=n;i++)
                               if((i*i)==n)
                               {
                                            printf("%d",1);
                                            return 0;
                               }
you can just write if (sqrt(n) * sqrt(n) == n)