c - 2/1, 3/2, 5/3, 8/5, 13/8...前20项的和

 double
 pres(const int n) {
     double numerator = ;    //分子.
     double denominator = ;    //分母.
     double sum = ;
     double tmp;

     for(int i = ; i <= n; i++) {
         sum += (numerator / denominator);

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

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

     return sum;
 }
 //result = 32.6603(前20项)