Binary String Matching

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本文介绍了一种解决二进制字符串匹配问题的算法。该算法通过在一个较长的二进制字符串B中查找较短的二进制字符串A出现的所有位置，并统计其出现次数。示例展示了如何使用C++实现这一算法，包括输入多个测试用例并输出匹配计数。

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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

string 中的find(string&,location) ，在一个字符串中查找指定的单个字符或字符组。如果找到，返回首次匹配的开始位置，如果没有找到匹配的内容，则返回 string::npos。,一般有2个输入参数，一个是待查询的字符串，一个是查询的起始位置，默认起始位置为0.

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

int main()
{
	int N;	
	
	cin >> N;
	while(N--)
	{
		string s, temp;
		cin >> temp >> s;
		int ans = s.find(temp);
		
		int num = 0;
		while(ans != string::npos)
		{
			++num;
			ans = s.find(temp, ans+1);
		}
		
		cout << num << endl;
	}
	
	return 0;
}