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

31110011101101011100100100100011010110100010101011 

样例输出

303 

来源

网络

上传者

naonao

此题如果用数组做的话，是一道水体，但是作为训练数据结构的话，还是有点小难度的-----至少个人是这么认为的

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


free(head);
         printf("%d\n",count);

}
return 0;
}