本文实例演示了C语言求幂计算的高效解法。很有实用价值。分享给大家供大家参考。具体方法如下：

题目如下：

给定base，求base的幂exp

只考虑基本功能，不做任何边界条件的判定，可以得到如下代码：

#include

using namespace std;

int cacExp(int base, int exp)

{

int result = 1;

int theBase = 1;

while (exp)

{

if (exp & 0x01)

result = result * base;

base = base * base;

exp = exp >> 1;

}

return result;

}

int getRecurExp(int base, int exp)

{

if (exp == 0)

{

return 1;

}

if (exp == 1)

{

return base;

}

int result = getRecurExp(base, exp >> 1);

result *= result;

if (exp & 0x01)

result *= base;

return result;

}

int main()

{

for (int i = 1; i < 10; i++)

{

int result = cacExp(2, i);

//int result = getRecurExp(2, i);

cout << "result: " << result << endl;

}

return 0;

}

再来看看数值的整数次方求解方法：

#include

using namespace std;

bool equalZero(double number)

{

if (number < 0.000001 && number > -0.000001)

return true;

else

return false;

}

double _myPow(double base, int exp)

{

if (exp == 0)

return 1;

if (exp == 1)

return base;

double result = _myPow(base, exp >> 1);

result *= result;

if (exp & 0x01)

result *= base;

return result;

}

double _myPow2(double base, int exp)

{

if (exp == 0)

return 1;

double result = 1;

while (exp)

{

if (exp & 0x01)

result *= base;

base *= base;

exp = exp >> 1;

}

return result;

}

double myPow(double base, int exp)

{

if (equalZero(base))

return 0;

if (exp == 0)

return 1;

bool flag = false;

if (exp < 0)

{

flag = true;

exp = -exp;

}

double result = _myPow2(base, exp);

if (flag)

{

result = 1 / result;

}

return result;

}

void main()

{

double base = 2.0;

int exp = -5;

double result = myPow(base, exp);

cout << "result: " << result << endl;

}

相信本文所述对大家C程序算法设计的学习有一定的借鉴价值。