NYOJ 5 Binary String Mathing (substr函数)

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 

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
	int T;
	string a;
	string b;
	cin>>T;
	while(T--)
	{
	cin>>a;
	cin>>b;
	int count=0;
	int len1=a.size();
	int len2=b.size();
	for(int i=0;i<len2;i++)
	{
        if(b.substr(i,len1)==a.substr(0,len1))
        count++;
	}	
	cout<<count<<endl;	
	} 
	return 0;
 }