用C语言求一个整数的幂的最有效方法是什么?
// 2^3
pow(2,3) == 8
// 5^5
pow(5,5) == 3125
当前回答
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。
其他回答
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)
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);
}
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。
power()函数只适用于整数
int power(int base, unsigned int exp){
if (exp == 0)
return 1;
int temp = power(base, exp/2);
if (exp%2 == 0)
return temp*temp;
else
return base*temp*temp;
}
复杂度= O(exp)
Power()函数为负exp和浮点基数工作。
float power(float base, int exp) {
if( exp == 0)
return 1;
float temp = power(base, exp/2);
if (exp%2 == 0)
return temp*temp;
else {
if(exp > 0)
return base*temp*temp;
else
return (temp*temp)/base; //negative exponent computation
}
}
复杂度= O(exp)
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).
