快速幂

double myPow(double x, int n) {
  double ans= 1;
  double xx = x;
  long long nn = abs(n);
  while (nn != 0 && xx!=1.0f) {
    ans *= (nn & 0x1) == 0x1 ? xx : 1;
    nn >>= 1;
    xx *= xx;
  }
  return n>=0?ans:1/ans;
}