c语言求幂运算（C语言幂运算）

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：

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 #include <iostream> 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：

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 #include <iostream> 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; }