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)