本題考查庫函式的實現原理，特別注意用O(logn)時間求a的n次方的優化演算法。

C++版

#include <iostream>
#include <cmath>
using namespace std;

bool g_InvalidInput = false;

double powerWithUnsignedExponent(double base, unsigned int exponent){
    if(exponent == 0)
        return 1;
    if(exponent == 1)
        return base;
    double result = powerWithUnsignedExponent(base, exponent >> 1);
    result *= result;
    if((exponent & 0x1) == 1)
        result *= base;
    return result;
}

double Power(double base, int exponent){
    g_InvalidInput = false;
    if(base == 0.0 && exponent < 0){
        g_InvalidInput = true;
        return 0.0;
    }
    unsigned int absExponent = (unsigned int)(exponent);
    if(exponent < 0)
        absExponent = -(unsigned int)(exponent);
    double result = powerWithUnsignedExponent(base, absExponent);
    if(exponent < 0)
        result = 1.0/result;
    return result;
}

int main()
{
    cout << "Hello world!" << endl;
    cout<<Power(2,3)<<endl;
    return 0;
}
 

  用O(logn)時間求a的n次方的優化演算法。

double powerWithUnsignedExponent(double base, unsigned int exponent){
    if(exponent == 0)
        return 1;
    if(exponent == 1)
        return base;
    double result = powerWithUnsignedExponent(base, exponent >> 1);
    result *= result;
    if((exponent & 0x1) == 1)
        result *= base;
    return result;
}