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

循环AC

#include 
     
      #include 
      
        #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;#define mem(a) memset(a, 0, sizeof(a))char a[100], b[1100];int main() { int n; cin >> n; while (n --) { mem(a); mem(b); cin >> a >> b; int res = 0, p1 = strlen(a), p2 = strlen(b); int num = 0; while (num <= p2-p1) { int flag = 1; for (int i = num, j = 0; j
          
         
        
       
      
     

标程：#include中的find()函数的应用。另外，m!=string::npos 意思是：m不等于字符串的尾部。

#include
     
      #include
      
       using namespace std;int main(){    string s1,s2;    int n;    cin>>n;    while(n--)    {        cin>>s1>>s2;        unsigned int m=s2.find(s1,0);        int num=0;        while(m!=string::npos)        {            num++;            m=s2.find(s1,m+1);        }        cout<
       
        <