C语言动态链表-程序

题目：把k的值插入到b前面，b是程序里面的数，程序里没有b，k插入到最后。
问题： 如果b是程序里第一个数，要求插入到第一个数前面，但是老是插到最后，求改正，其他插入没错。

#include<stdio.h>
#include<malloc.h>
#include<stdlib.h>
struct student
{
    int data;
    struct student *next;
}a;
struct student *p,*q,*head,*l,*m,*t;
int main()
{
    l=q=head=&a;
    int n,i,c=0,k,b;
    printf("请输入n:\n");
    scanf("%d",&n);
    q->next=NULL;
    printf("输入链表：\n");
    for(i=0;i<n;i++)
    {
        p = (struct student*)malloc(sizeof(struct student));
        c++;
        if(c==1) t = head = l = p;
        scanf("%d",&p->data);
        q->next = p;
        q = q->next;
    }
    p->next=NULL;
    printf("请输入b,k的值：\n");
    scanf("%d%d",&b,&k);
    m = (struct student*)malloc(sizeof(struct student));
    m->data = k;
    c=0;
    printf("把值插入链表中为：\n");
    l = head->next;
    q = &a;
    for(i=0;i<n;i++)
    {
        c++;
        if(l->data == b&&c==1)
        {
            q->data = k;
            break;
        }
        else if(l->data == b)
        {
            head->next = m;
            m->next = l;
            break;
        }
        if(c==n-1)
        {
            p->next = m;
            m->next=NULL;
        }
        l = l->next;
        head = head->next;
    }
    for(i=0;i<n+2;i++)
    {
        printf("%d\n",t->data);
        t = t->next;
    }
    return 0;
}

#include<stdio.h>
#include<malloc.h>
#include<stdlib.h>
struct student
{
    int data;
    struct student* next;
}a;
struct student* p, * q, * head, * l, * m, * t;
int main()
{
    l = q = head = &a;
    int n, i, c = 0, k, b;
    printf("请输入n:\n");
    scanf("%d", &n);
    q->next = NULL;
    printf("输入链表：\n");
    for (i = 0;i < n;i++)
    {
        p = (struct student*)malloc(sizeof(struct student));
        c++;
        if (c == 1) t = head = l = p;
        scanf("%d", &p->data);
        q->next = p;
        q = q->next;
    }
    p->next = NULL;
    printf("请输入b,k的值：\n");
    scanf("%d%d", &b, &k);
    m = (struct student*)malloc(sizeof(struct student));
    m->data = k;
    c = 0;
    printf("把值插入链表中为：\n");
    l = head;
    q = &a;
    for (i = 0;i < n;i++)
    {
        c++;
        if (l->data == b && c == 1)
        {
            head = m;
            m->next = l;
            break;
        }
        else if (l->data == b)
        {
            q->next = m;
            m->next = l;
            break;
        }
        
        //用q表示l的前一个
        q = l;
        l = l->next;
    }
    if (c == n)
    {
        p->next = m;
        m->next = NULL;
    }
    t = head;
    for (i = 0;i < n + 1;i++)
    {
        printf("%d\n", t->data);
        t = t->next;
    }
    return 0;
}