//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)
1 #include<math.h> 2 #include<stdio.h> 3 double fun(int n) 4 { 5 double s=1,S=0; 6 for (int i = 2; i <= n+1; i++)//这里判断条件为n+1. 7 { 8 S += s; 9 s += sqrt(i);//2^0.5=2^(1/2)=根号下2=sqrt(2) 10 printf("%f\n", S); 11 printf("%f\n", s); 12 } 13 return S; 14 } 15 void main() 16 { 17 FILE *wf; 18 int n; 19 double s; 20 printf("\n\nInput n: "); 21 scanf("%d",&n); 22 s=fun(n); 23 printf("\n\ns=%f\n\n",s); 24 /******************************/ 25 wf=fopen("out.dat","w"); 26 fprintf(wf,"%f",fun(20)); 27 fclose(wf); 28 /*****************************/ 29 }
//使用递归解决：
1 double f(int n) 2 { 3 if (n == 1) return 1; 4 else return sqrt(n) + f(n - 1); 5 } 6 double fun(int n) 7 { 8 double S=0; 9 for (int i = 1; i<=n; i++) 10 { 11 S += f(i); 12 } 13 return S; 14 }
//也可以使用pow函数。功能：求x的y次方。用法：pow(x,y)。