/**

* O(logN)的时间复杂度

* Note:

* 需要记住的是，double类型和float类型是不能直接比较大小的，需要近似比较大小。

*/

#include <iostream>

#include <cassert>

double power(double base, int exponent);

bool eq(double a, double b)

{

if(a-b <0.0000001 | a-b > -0.0000001 )

return true;

return false;

}

double power(double base, int exponent)

{

if(eq(base, 0.0) && exponent >= 0)

return 0.0;

else if(eq(base, 0.0) && exponent < 0)

{

//std::cout<<"The base is zero, and the exponent shouldn't be nagetive interger!"<<std::endl;

return -1;

}

else if(!eq(base, 0.0) && exponent < 0)

{

return 1.0/power(base,-exponent);

}

else

{

double result = 1.0;

while(exponent > 0)

{

std::cout << exponent << std::endl;

if(exponent & 0x01)

{

result *= base;

}

base *= base;

exponent=exponent>>1;

}

return result;

}

}

int main(int argc, char const *argv[])

{

assert(power(0.0, -1) < 0);

assert(eq(power(2.0, -1),0.5));

assert(eq(power(2.0, 2),4));

assert(eq(power(0.0, 3),0) );

assert(eq(power(0.0, 0),0 ));

return 0;

}