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

分析：本题是简单的字符串匹配问题，运用的是kmp思想

代码：

#include<stdio.h>
#include<string.h>
int p1,p2;
char a[11],b[1010];
int next[1010];
void getnext()
{
int i,j;
next[0]=-1;
i=-1;j=0;
while(j<p1)
{
if(i==-1||a[i]==a[j])
{
i++;j++;
if(a[i]!=a[j])
{
next[j]=i;
}
else next[j]=next[i];
   }
else   i=next[i];
}
}
void kmp()
{
int i,j;
int count=0;
i=j=0;
getnext();
while(i<p2)
{
if(j==-1||a[j]==b[i])
{
i++;j++;
}
else j=next[j];
if(j==p1)
{
count++;
j=next[j];
}
}
printf("%d\n",count);

}
int main()
{
int n,i,j;
scanf("%d",&n);
while(n--)
{
scanf("%s%s",a,b);
p1=strlen(a);
p2=strlen(b);
kmp();
}
return 0;
}