学步园

// 此方法能减少乘法次数， O（1）

#include <iostream>
#include <cassert>

using namespace std;

// do not consider doulbe overflow

double pow(double base, int exponent)
{
	// initialize
	int i = 0;

	assert(base!=0 && exponent>=0);// if the condition is true, go ahead, else break;
	if (exponent==0) return 1.0;

	double power = 1.0;
	unsigned int mask = 1;
	double ret = 1.0;

	for (i=0; i<32; ++i)
	{
		if (i==0) power = base;
		else power = power*power;

		if ( ((mask<<i) & (unsigned int)exponent) != 0)
		{
			ret = ret * power;
		}
	}
	
	return ret;
}

int main(int argc, char* argv[])
{
	cout<<pow(2.5,2)<<endl;
	cout<<pow(-2.5,2)<<endl;
	cout<<pow(1,0)<<endl;
	cout<<pow(100,3)<<endl;
	cout<<pow(-100,4)<<endl;
	return 0;
}