实现函数 doubule Power(double base, int exponent)，求base的e

实现函数 doubule ?Power(double base,int exponent)，求base的exponent次方，不得使用库函数，同时不需要考虑大数问题。

解法一：直接写出函数，而未考虑base为0，且exponent为负的情况。

double MyPower1(double dBase,int iExporent)
{
	double dRes = 1.0;
	for (int i=1; i<=iExporent; i++)
	{
		dRes *= dBase;
	}

	return dRes;
}

解法二：较完整的解法，未考虑base为0，且exponent为负的情况。程序中添加了g_InvalidInput用来表示错误类型。添加了equal()函数，用于确定double类型的数值是否相等，并肩iExporent转换成了无符号整数。

bool g_InvalidInput = false;
double MyPower2(double dBase,int iExporent)
{
	g_InvalidInput = false;

	if (equal(dBase,0) && iExporent<0){
		g_InvalidInput = true;
		return 0.0;
	}

	unsigned int absExporent = (unsigned int )iExporent;
	if (iExporent < 0){
		absExporent = (unsigned int )(-iExporent);
	}
	

	double dRes = PowerWithUnsignedExpoent(dBase,absExporent);
	if (iExporent < 0){
		dRes = 1.0/dRes;
	}

	return dRes;
}

double PowerWithUnsignedExpoent(double dBase,unsigned int iExporent)
{
	double dRes = 1.0;
	for (unsigned int i=1; i<=iExporent; i++)
	{
		dRes *= dBase;
	}
	
	return dRes;
}

bool equal(double dA,double dB)
{
	double dSub = dA-dB;
	if (dSub < 0.0000001 && dSub > -0.0000001){
		return true;
	}else{
		return false;
	}
}

通过上述公式，可以将计算指数的次数减少到O(logN)次，同时可以使用递归。下面代码中的另外两处优化的地方是，将除以2使用右移替代；将取模运算，用和0x1进行与运算，提高了程序的运行效率。

double MyPower3(double dBase,int iExporent)
{
	g_InvalidInput = false;
	
	if (equal(dBase,0) && iExporent<0){
		g_InvalidInput = true;
		return 0.0;
	}
	
	unsigned int absExporent = (unsigned int )iExporent;
	if (iExporent < 0){
		absExporent = (unsigned int )(-iExporent);
	}
	
	double dRes = PowerWithUnsignedExpoent_Recursive(dBase,absExporent);
	if (iExporent < 0){
		dRes = 1.0/dRes;
	}
	
	return dRes;
}

double PowerWithUnsignedExpoent_Recursive(double dBase,unsigned int iExporent)
{
	if (iExporent == 0)
		return 1;
	if (iExporent == 1)
		return dBase;
	
	double dRes = PowerWithUnsignedExpoent_Recursive(dBase,iExporent >> 1);
	dRes *= dRes;
	
	if (iExporent & 0x1 == 1){
		dRes *= dBase;
	}
	
	return dRes;
}