50. Pow(x, n)

double myPow(double x, int n) {
    if (x == 0)
        return 0;
    long long val = n;
    double res = 1.0;
    if (val < 0)
        x = 1 / x, val = -val;
    while (val > 0) {
        if ((val & 1) == 1)
            res *= x;
        x *= x;
        val >>= 1;
    }
    return res;
}