public static double myPow(double x,int n)
{
if (n < 0) return 1.0 / power(x, -n);
else return power(x, n);
}
static double power (double x,int n)
{
if (n == 0) return 1;
double v = power(x, n / 2);
if (n % 2 == 0) return v * v;
else return v * v * x;
}
分治法-pow(x,n)
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转载自blog.csdn.net/smj20170417/article/details/80609666
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