剑指offer 求一个数的n次方，不考虑大数问题

#include <stdio.h>
#define ESP 0.0000001

typedef enum bool{
false = 0,
true = 1
}bool;

bool value_num = false;   //定义一个全局变量用来区分正确时返回的0和错误时返回的0
double PowWithunsignedexp(double Base, unsigned int exp);
bool equal(double One, double Two);

double MyPow (double Base, int exp)
{
bool value_num = true;
double result;
unsigned int unsignedexp;
if ( equal(Base,0.0) && exp == 0)
{
value_num = false;
return 0.0;              //指数和底数都是0时返回0，value_num为false说明传值错误
}
if (exp == 0)
return 1.0;
else if (exp < 0)
unsignedexp = (unsigned int)(-exp);    //指数为负数时取绝对值
else
unsignedexp =  (unsigned int)(exp);

result = PowWithunsignedexp (Base,unsignedexp);

if(exp < 0)
return (1/result);
else
return result;
}

double PowWithunsignedexp(double Base, unsigned int exp)  //用递归求幂，因为指数为大于0的指数，有奇数和偶数两种情况
{
if (exp == 1)
return Base;
else if (exp%2 == 0)
return PowWithunsignedexp(Base,exp/2)*PowWithunsignedexp(Base,exp/2);
else
return PowWithunsignedexp(Base,exp/2)*PowWithunsignedexp(Base,(exp/2+1));

}

bool equal(double One, double Two)   //比较浮点数的大小
{
if ((One - Two) <= ESP && (Two - One) <= ESP)
return true;
return
false;
}