急呀 C语言

有一线性表示存储在一个带头结点的循环单链表L中,写出计算线性表元素个数的算法 可不可帮我用C语言编下这道题,要完全的程序,就是可以直接可以运行的... 帮帮忙了,真的很急需

已测试运行通过：

#include <stdio.h>
#include <stdlib.h>

typedef struct _list {
int val;
struct _list* next;
} *node, list;

node Insert( node head, node pos, int val )
{
node tmp;
tmp = ( node )malloc( sizeof( list ) );
tmp->val = val;
tmp->next = pos ? pos->next : head->next;
if ( pos ) {
pos->next = tmp;
} else {
head->next = tmp;
}
return tmp;
}

void Create( node head, int* beg, int* end )
{
node t;
head->next = t = NULL;
head->val = end - beg;
while ( beg != end ) {
t = Insert( head, t, *beg++ );
}
t->next = head;
}

int Count( node head )
{
int n = 0;
node h = head;
while( head->next != h ) {
head = head->next;
++n;
}
return n;
}

void Print( char* msg, node head )
{
node h = head;
printf( msg );
head = head->next;
while ( head != h ) {
printf( "%d ", head->val );
head = head->next;
}
putchar( '\n' );
}


int main()
{
int a[] = { 1,3,5,7,9 };
int n;
list head;
Create( &head, a, a + 5 );
n = Count( &head );
Print( "list:", &head );
printf( "COUNT:%d\n", n );
getchar();
return 0;
}

p = head ;

int n = 0 ; //计数器

while(p->next)

{

p=p->next;

n ++ ; //往后走一步，计数器自加一，

}

cout<<"结点个数为： "<<n<<endl;

遍历。。。。设头节点为head

p = head ;

int n = 0 ; //计数器

while(p->next)

{

p=p->next;

n ++ ; //往后走一步，计数器自加一，

}

cout<<"结点个数为： "<<n<<endl;