c - 2/1, 3/2, 5/3, 8/5, 13/8...前20项的和

 1 double

 2 pres(const int n) {

 3     double numerator = 2;    //分子.

 4     double denominator = 1;    //分母.

 5     double sum = 0;

 6     double tmp;

 7     

 8     for(int i = 1; i <= n; i++) {

 9         sum += (numerator / denominator);

10         

11         tmp = numerator;    //临时保存当前分子(以下一个数使用).

12         

13         numerator += denominator;    //下一个分子为当前(分子分母的和).

14         denominator = tmp;    //下一个分母为当前分子.

15     }

16     

17     return sum;

18 }

19 //result = 32.6603(前20项)