Binet's formula for n-th Fibonnaci number accuracy

#include <bits/stdc++.h>

// How good is Binet's formula for approximating the n-th Fibonacci number?
// Spoilers: it's good for N <= 72

typedef long long ll;
typedef long double ld;

const int MAXN = 1000;

const ld phi = 1.618033988749894848204586834365638117720309179805762862135;
const ld oneoversqrt5 = 0.447213595499957939281834733746255247088123671922305144854;

ll fibdp[MAXN];
ld phipow[MAXN]; // phi^n
ld phipowrec[MAXN]; // 1 / (phi^n)

// Regular Fibonacci formula:
// fib(0) = 0, fib(1) = 1, fib(n) = fib(n-1) + fib(n-2), for n >= 2
ll fibaccurate(int n) {
  return (fibdp[n] != -1) ? fibdp[n] : fibdp[n] = fibaccurate(n-1) + fibaccurate(n-2);
}

// Binet's formula:
// fib(n) = [ (phi^n - (-phi)^(-n)) / sqrt(5) ]
// where [x] = closest integer to x
ll fibapprox(int n) {
  return round( oneoversqrt5 * (phipow[n] - phipowrec[n]*(n&1?-1:1)) );
}


int main() {

  memset(fibdp, -1, sizeof(fibdp));
  fibdp[0] = 0;
  fibdp[1] = 1;
  phipow[0] = 1.0;
  for (int i = 1; i < MAXN; ++i) {
    phipow[i] = phipow[i-1] * phi;
    phipowrec[i] = 1.0 / phipow[i];
  }

  int n = 2;
  ll f1, f2;
  do {
    f1 = fibaccurate(n);
    f2 = fibapprox(n);
    ++n;
    cout << "n = " << setw(3) << n << " : " << setw(20) << f1 << " " << setw(20) << f2 << endl;
  } while (f1 == f2 && n < MAXN);

}