剑指Offer系列-面试题11：数值的整数次方

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题目：实现函数double 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。

#include <iostream>using namespace std;bool g_InvalidInput = false;    /// 输入是否非法的标识/// 判断两个double类型的数是否相等bool equals(double num1, double num2){    if(-0.0000001 < (num1 - num2) && (num1 - num2) < 0.0000001)    {        return true;    }    else    {        return false;    }}/// 计算幂乘double calcWithUnsignedExponent(double base, unsigned int exponent){    double result = 1.0;    for(int i = 1; i <= exponent; ++i)        result *= base;    return result;}double Power(double base, int exponent){    g_InvalidInput = false;    if(equals(base, 0.0) && exponent < 0)   /// 如果底数为0而且指数小于0    {        g_InvalidInput = true;        return 0.0;    }    unsigned int absExponent = (unsigned int)(exponent);    if(exponent < 0)        absExponent = (unsigned int)(-exponent);    double result = calcWithUnsignedExponent(base, absExponent);    if(exponent < 0)        result = 1.0 / result;    return result;}int main(){    double base;    int exponent;    cout << "Input base and exponent:" << endl;    cin >> base >> exponent;    double result = Power(base, exponent);    if(equals(result, 0.0))    {        if(g_InvalidInput)            cout << "Input error!" << endl;        else            cout << "0.0" << endl;    }    else        cout << result << endl;    return 0;}

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