Power(int base, int exponent) 函數實現

 這個是個高效的演算法，時間複雜度為 O(logn)

原理：

a的n次方：

#include<iostream>#include<cmath>using namespace std;double PowerWithUnisgnedExponent(double base ,unsigned int exponent){if(exponent == 0)return 1;if(exponent == 1)return base;double result = PowerWithUnisgnedExponent(base, exponent >> 1);result *= result;if(exponent & 0x1 == 1)result *=base;return result;}double power(double base, int exponent){double result = PowerWithUnisgnedExponent(base,abs(exponent));if(exponent < 0)return 1.0/result;elsereturn result;}int main(){int base = 2;int exponent = 10;cout << power(base,exponent);return 0;}