c - Optimized way to handle extremely large number without using external library -

optimized way handle value of n^n (1 ≤ n ≤ 10^9)

i used long long int it's not enough value might (1000^1000)

searched , found gmp library http://gmplib.org/ , bigint class don't wanna use them. i looking numerical method handle this.

i need print first , last k (1 ≤ k ≤ 9) digits of n^n

for first k digits getting shown below (it's bit ugly way of doing it)

num = pow(n,n); while(num){     arr[i++] = num%10;     num /= 10;     digit++; } while(digit > 0){     j=digit;     j--;     if(count<k){         printf("%lld",arr[j]);         count++;     }     digit--; } 

and last k digits using num % 10^k below.

findk=pow(10,k); lastdigits = num % findk; enter code here 

maximum value of k 9. need 18 digits @ max. think of getting 18 digits without solving complete n^n expression.

any idea/suggestion??

finding least significant digits (last k digits) easy because of property of modular arithmetic, says: (n*n)%m == (n%m * n%m)%m, code shown bluepixy followed exponentiation squaring method work finding k lsds.

now, significant digits (1st k digits) of n^n can found in way:

 know,   n^n = 10^(n log n) 

so if calculate n log (n) number of format xxxx.yyyy, have use number power of 10, understandable xxxx or integer part of number add xxxx zeros after 10, not important you! means, if calculate 10^0.yyyy, significants digits looking for. solution this:

double r = n * log10 (n); r = r - (long long) r; //so taking fractional part double v = pow(10, r); int powerk = 1; (int i=0; i<k; i++) powerk *=10; v *= powerk; //now print 1st k digits v