这个是个高效的算法，时间复杂度为 O(logn)

原理：

a的n次方：

#include<iostream>

#include<cmath>

using namespace std;

double PowerWithUnisgnedExponent(double base ,unsigned int exponent)

{

	if(exponent == 0)

		return 1;

	if(exponent == 1)

		return base;

	double result = PowerWithUnisgnedExponent(base, exponent >> 1);

	result *= result;

	if(exponent & 0x1 == 1)

		result *=base;

	return result;

}

double power(double base, int exponent)

{

	double result = PowerWithUnisgnedExponent(base,abs(exponent));

	if(exponent < 0)

		return 1.0/result;

	else

		return result;

}

int main()

{

	int base = 2;

	int exponent = 10;

	cout << power(base,exponent);

	return 0;

}