C语言：计算并输出S=1+(1+2^0.5)+(1+2^0.5+3^0.5)...+(1+2^0.5+3^0.5+...+n^0.5)

//计算并输出S=1+(1+2^0.5)+(1+2^0.5+3^0.5)...+(1+2^0.5+3^0.5+...+n^0.5)

 #include<math.h>
 #include<stdio.h>
 double fun(int n)
 {
     double s=,S=;
     for (int i = ; i <= n+; i++)//这里判断条件为n+1.
     {
         S += s;
         s += sqrt(i);//2^0.5=2^(1/2)=根号下2=sqrt(2)
         printf("%f\n", S);
         printf("%f\n", s);
     }
     return S;
 }
 void main()
 {
   FILE *wf;
   int n;
   double s;
   printf("\n\nInput n: ");
   scanf("%d",&n);
   s=fun(n);
   printf("\n\ns=%f\n\n",s);
 /******************************/
   wf=fopen("out.dat","w");
   fprintf(wf,"%f",fun());
   fclose(wf);
 /*****************************/
 }

//使用递归解决：

 double f(int n)
 {
     if (n == ) return ;
     else return sqrt(n) + f(n - );
 }
 double fun(int n)
 {
     double S=;
     for (int i = ; i<=n; i++)
     {
         S += f(i);
     }
     return S;
 }

//也可以使用pow函数。功能：求x的y次方。用法：pow(x,y)。