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)