程序代码：

#include<stdio.h>
#define N 10
void main()
{ 
    double x(int n);
    double s(int n);
    double n,sum;
    sum=s(n);
    printf("前10项和为: %lf\n ",sum);
    printf("\n");
}
double s(int n)
{  
   double x(int n);
   double i,sum=3;
   for(i=3;i<=10;i++)
       sum+=x(n);
    return sum;
}
double x(int n)
{   
     double x1=1,x2=2,x3;
         if(n=1)
             return x3=1;
         else if(n=2)
             return x3=2;
         else 
         {   x(n);
             x3=x1*x2;
             return x(n--);
         }
         return x3;
}
怎么总有警告啊，哪出问题了呢
就算没警告，结果也不对

程序代码：

#include<stdio.h>
void main()
{ 
    double x(double n);
    double s(double n);
    double n=10,sum;
    sum=s(n);
    printf("前10项和为: %lf\n ",sum);
    printf("\n");
}
double s(double n)
{  
   double x(double n);
   double i,sum=3;
   for(i=3;i<=10;i++)
       sum+=x(n);
    return sum;
}
double x(double n)
{   
     double x1=1,x2=2,x3;
         if(n=1)
             return x3=1;
         else if(n=2)
             return x3=2;
         else 
         {   x(n--);
             x3=x1*x2;
             return x(n--);
         }
}

程序代码：

#include<stdio.h>
void main()
{ 
    double x(double n);
    double s(double n);
    double n=10,sum;
    sum=s(n);
    printf("前10项和为: %lf\n ",sum);
    printf("\n");
}
double s(double n)
{  
   double x(double n);
   double i,sum=1;
   for(i=3;i<=10;i++)
       sum+=x(n);
       return sum;
}
double x(double n)
{   
         double y;
         if(n=1)
             return y=1;
         else if(n=2)
             return y=2;
         else 
         {   y=x(n)*(n-1);
             return y;
         }
}
你说的是这个？本质上不是一样的吗？改成这个结果还是一样错误

程序代码：

#include<stdio.h>
void main()
{ 
    double x(double n);
    double s(double n);
    double n,sum;
    sum=s(10);
    printf("前10项和为: %.0lf\n ",sum);
    printf("\n");
}
double s(double n)
{  
   double x(double n);
   double sum=3;
   for(n=10;n>=3;n--)
       sum+=x(n);
       return sum;
}
double x(double n)
{   
         double y;
         if(n==1)
             return y=1;
         else if(n==2)
             return y=2;
         else 
         {   y=x(n-1)*x(n-2);
             return y;
         }
}
这样运行才完全正确

程序代码：

#include<stdio.h>
void main()
{ 
    double x(double n);
    double s(double n);
    double n,sum;
    sum=s(10);
    printf("前10项和为: %.0lf\n ",sum);
    printf("\n");
}
double s(double n)
{  
   double x(double n);
   double sum=3;
   for(n=10;n>=3;n--)
       sum+=x(n);
       return sum;
}
double x(double n)
{   
         double y;
         if(n==1)
             return y=1;
         else if(n==2)
             return y=2;
         else 
         {   y=x(n-1)*x(n-2);
             return y;
         }
}
这样才对