字符串的应用

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<iostream>
#include<stack>
#include<string>
#include<vector>
using namespace std;
void main()
{
int n,i=0,k=0,t;
cout<<"请输入比较次数"<<endl;
cin>>n;
while(n--)
{
string s,s1;
cout<<"请输入字符串"<<endl;
cin>>s;
//cout<<s<<endl;
cout<<"请输入子串"<<endl;
cin>>s1;
//cout<<s1<<endl;;
while(s.find(s1,i)!=string::npos)
{
t=s.find(s1,i);
//cout<<"位置为 "<<t<<endl;
i=t+1;
k++;
}
cout<<k<<endl;
k=0;
i=0;
}
}