递归求简单交错幂级数的部分和

 1 double fn(double x, int n)
 2 {
 3     double ret;
 4     double item, flag;
 5 
 6     if (n == 1)
 7     {
 8         ret = x;
 9     }
10     else
11     {
12         if (n % 2)
13         {
14             flag = 1;
15         }
16         else
17         {
18             flag = -1;
19         }
20         item = 1;
21         for (int i = 1; i <= n; i++)
22         {
23             item = item * x;
24         }
25         item = item * flag;
26         ret = fn(x, n - 1) + item;
27     }
28 
29     return ret;
30 }