Binary String Matching

最新推荐文章于 2023-11-06 16:56:22 发布

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本博客介绍了一个算法，用于计算一个字符串作为另一个字符串中子串出现的次数，适用于只包含二进制字符的场景。

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

#include<stdio.h>
#include<string.h>
int main()
{
	int n,t;
	char a[10],b[1000],*p;
	scanf("%d",&n);
	while(n--)
	{
		t=0;
		scanf("%s%s",a,b);
		p=b;
		while( (p=strstr(p,a)) !=NULL )
		{
			t++;
			p++;
		}
		printf("%d\n",t);
	}
	return 0;
}
//其它人写的，差不多
#include<iostream>
#include<string>
#include<string.h>
using namespace std;
int main()
{
	int n;
	 char A[11];
	 char B[1100];
	 char *p;
	cin>>n;
	while(n--)
	{
		int count=0;
      cin>>A>>B;
      p=strstr(B,A);
	  while(p)
	  {
		  count++;
		  p=p+1;
		  p=strstr(p,A);
	  }
	  cout<<count<<endl;
	}
	return 0;
}