数据结构练习（33）数值的整数次方

http://zhedahht.blog.163.com/blog/static/254111742009101563242535/

思路：

见过斐波那契，然后这题直接想到了递归2分法求解。

方法一还是蛮不错的，利用了位运算。不过代码写起来比较冗长。

#include <iostream>

#include <bitset>

using namespace std;



double PowerWithExponent_1(double base, size_t exp)

{

    bitset<32> bits(exp);



    if (bits.none())

        return 1.0;



    double b[32];

    for (int i = 0; i < 32; ++i)

        b[i] = 1.0;



    double power = 1.0;

    size_t cnt = 0;

    for (int i = 0; i < 32 && cnt <= bits.count(); ++i)

    {

        if (i == 0)

            power = base;

        else

            power *= power;



        if (bits.at(i))

            b[i] = power, ++cnt;

    }



    power = 1.0;

    for (int i = 0; i < 32; ++i)

        if (bits.at(i))

            power *= b[i];



    return power;

}



double PowerWithExponentRecursively(double base, unsigned int exp)

{

    if (exp == 0)

        return 1.0;



    double power = PowerWithExponentRecursively(base, exp >> 1);

    power *= power;



    if (exp & 0x01)

        power *= base;



    return power;

}



int main()

{

    double power1 = PowerWithExponent_1(2.3, 3);

    double power2 = PowerWithExponentRecursively(2.3, 3);

    cout << power1 << " " << power2 << endl;

    return 0;

}