hay look at my recursive pow


#include <stdio.h>
#include <stdlib.h>

unsigned long long rpow(int n, int p) {
  return p ? p==1 ? n : n * rpow(n,p-1) : 1;
}

int main(int argc, char **argv) {
  if(argv[1])
    if(argv[2]) {
      int n = atoi(argv[1]);
      int p = atoi(argv[2]);
      printf("%llu\n",rpow(n,p));
    }
  return 0;
}

>>140
so you're
Stopped reading right there sorry.
Disregard that, I didn't but I will point it out anyway so that I win the argument even if I am wrong. On to more important matters:

If you want to be an absolute pedant about it, you could of inferred that I was meani... oh wait, no you couldn't. I said the time complexity of a function pow() that performs a lookup on a table is constant. Giving the user a constant time pow() function. And even if I had meant overall amortized time of pow calculations, then your amortized time complexity is wrong anyway. The worst case sequence of events therein would be that the table is loaded, and then pow() is called a limitless number of times. This gives the cost of the load operation as 1/∞ along with the constant time cost of the lookups, so we get O(1) again.