快速幂(迭代法)

迭代法求快速幂 pow(x, n).

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class Solution { public:     double multiply(double x, long long n) {         double ans = 1.0;         double base_x = x;         while (n > 0) {             if (n % 2 == 1) {                 ans *= base_x;             }             base_x *= base_x;             n /= 2;         }         return ans;     }     double myPow(double x, int n) {         long long N = n;         return N >= 0 ? multiply(x, N): (1.0 / multiply(x, -(long long)N));     } };