刷题小分队-剑指offer(二)

11. 数值的整数次方

 1 #include <iostream>
 2
 3 int error = 0;
 4 double Power(double base, int exponent);
 5
 6 int main(int argc, char *argv[]) {
 7   double base;
 8   int exponent;
 9   std::cin >> base >> exponent;
10
11   double result = Power(base, exponent);
12
13   std::cout << error << std::endl;
14   std::cout << result << std::endl;
15 }
16
17 double Power(double base, int exponent) {
18   if(0 == exponent)
19     return 0;
20   if(exponent > 0) {  // 指数为正
21     if(1 == exponent)
22       return base;
23     if(exponent % 2 == 0) { // 偶数
24       double tmp = Power(base, exponent/2);
25       return tmp*tmp;
26     } else {  // 奇数
27       double tmp = Power(base, (exponent-1)/2);
28       return tmp*tmp*base;
29     }
30   } else {  // 指数为负
31     if(base == 0) {
32       error = -1;
33       return 0;
34     }
35     if(-1 == exponent)
36       return 1/base;
37     if(exponent % 2 == 0) {  // 偶数
38       double tmp = Power(base , exponent/2);
39       return tmp*tmp;
40     } else {  // 奇数
41       double tmp = Power(base , (exponent+1)/2);
42       return tmp*tmp*(1/base);
43     }
44   }
45 }

*book:

 1 #include <iostream>
 2
 3 bool error = false;
 4 double Power(double base, int exponent);
 5 double PositivePower(double base, unsigned int exponent);
 6 bool equal(double m, double n);
 7
 8 int main(int argc, char *argv[]) {
 9   double base;
10   int exponent;
11   std::cin >> base >> exponent;
12
13   double result = Power(base, exponent);
14
15   std::cout << error << std::endl;
16   std::cout << result << std::endl;
17 }
18
19 double Power(double base, int exponent) {
20   // 底数为0, 指数为负数/0/正数
21   if(equal(base, 0.0)) {
22     if(exponent < 0) {
23       error = true;
24       return 0;
25     } else
26       return 0;
27   }
28
29   //指数为正数/0
30   unsigned int absexponent = (unsigned int) exponent;
31   if(exponent < 0) {
32     absexponent = (unsigned int)(-exponent);
33   }
34   double result = PositivePower(base, absexponent);
35   return (exponent < 0) ? 1/result : result;
36 }
37
38 bool equal(double m, double n) {
39   if(m-n < 0.0000001 && m-n > -0.0000001)
40     return true;
41   return false;
42 }
43
44 double PositivePower(double base, unsigned int exponent) {
45   if(exponent == 1)
46     return base;
47   if(exponent == 0)
48     return 0;
49   double result = PositivePower(base, exponent >> 1);
50   result *= result;
51   if(exponent & 0x1 == 1)  // 奇数
52     result *= base;
53   return result;
54 }

思路：利用乘方的性质

本题看似简单，但完美的写出程序需要思考全面：

1）底数/指数分别为0，负数，正数的情形；（a的n次幂是a的-n次幂的倒数，利用此点可简化程序）

2）底数判断是否为0的方法（double base== 0错误）；