int pow( int base, int exponent)

{   // Does not work for negative exponents. (But that would be leaving the range of int) 
    if (exponent == 0) return 1;  // base case;
    int temp = pow(base, exponent/2);
    if (exponent % 2 == 0)
        return temp * temp; 
    else
        return (base * temp * temp);
}

int pow( int base, int exponent)

{   // Does not work for negative exponents. (But that would be leaving the range of int) 
    if (exponent == 0) return 1;  // base case;
    int temp = pow(base, exponent/2);
    if (exponent % 2 == 0)
        return temp * temp; 
    else
        return (base * temp * temp);
}

如果要取2的a次方。最快的方法是按幂位移位。

2 ** 3 == 1 << 3 == 8
2 ** 30 == 1 << 30 == 1073741824 (A Gigabyte)

平方求幂。

int ipow(int base, int exp)
{
    int result = 1;
    for (;;)
    {
        if (exp & 1)
            result *= base;
        exp >>= 1;
        if (!exp)
            break;
        base *= base;
    }

    return result;
}

这是在非对称密码学中对大数进行模求幂的标准方法。

如果你想得到一个整数的2的幂，最好使用shift选项:

Pow(2,5)可以替换为1<<5

这样效率更高。

请注意，平方求幂并不是最优的方法。这可能是一种适用于所有指数值的通用方法，但对于特定的指数值，可能有更好的序列，需要更少的乘法。

例如，如果你想计算x^15，用平方求幂的方法会给你:

x^15 = (x^7)*(x^7)*x 
x^7 = (x^3)*(x^3)*x 
x^3 = x*x*x

这一共有6次乘法。

事实证明，这可以通过“仅仅”5次加法链幂运算来完成。

n*n = n^2
n^2*n = n^3
n^3*n^3 = n^6
n^6*n^6 = n^12
n^12*n^3 = n^15

没有有效的算法来找到这个最优的乘法序列。从维基百科:

The problem of finding the shortest addition chain cannot be solved by dynamic programming, because it does not satisfy the assumption of optimal substructure. That is, it is not sufficient to decompose the power into smaller powers, each of which is computed minimally, since the addition chains for the smaller powers may be related (to share computations). For example, in the shortest addition chain for a¹⁵ above, the subproblem for a⁶ must be computed as (a³)² since a³ is re-used (as opposed to, say, a⁶ = a²(a²)², which also requires three multiplies).