南阳OJ----Binary String Matching

Binary String Matching

时间限制：3000 ms | 内存限制：65535 KB

难度：3

描述Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

输入The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.输出For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011

样例输出

3
0
3

来源网络上传者naonao

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct node
{
  char a;
  struct node *next;
}Node;
int i;
bool flag=true;
int main( void )
{
    Node *head,*p1,*p2;
    int t,n,count;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        head=NULL;
        count=n=0;
        p1=p2=( Node * )malloc( sizeof(Node) );
        char str[10]={''};
        gets(str);
        //getchar();
        while(p1->a=getchar(),p1->a!='n')
        {
            if(n++==0)
                      head=p1;
            else 
                 p2->next=p1;
                 p2=p1;
            p1=(Node*)malloc(sizeof(Node));
        }
        p2->next=NULL;
          p1=head;
    /*      while(p1!=NULL)
          {
              putchar(p1->a);
              p1=p1->next;
          }
          puts("");
    */
         while(p1!=NULL)
         {
         
              while(p1!=NULL&&p1->a!=*str)   //找到第一个位置
                 {
                   p1=p1->next;
                 }
              if(p1!=NULL)                //防止万一没有找到
              {
                  p2=p1->next;

                for( ::i=0; str[i]!=''; i++ )
                {
                  if(str[i]==p1->a)
                  {
                      p1=p1->next;
                  }
                  else 
                  {
                      ::flag=false;
                           p1=p2;
                             break;
                  }
                  
                  if((p1==NULL&&str[i+1]!=''))
                      {
                         ::flag=false ;
                          break;
                      }
                }
                    
                 if(::flag)
                 {
                          p1=p2 ;
                          count++;
                 }
                else 
                    ::flag=true;
              }
            }

         free(head);
         printf("%dn",count);
        
         }
    return 0;
}