ACM Binary String Matching

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 


程序代码:

#include<iostream> 
#include<string> 
using namespace std; 
int main() { 
    int N;
    cin>>N;
    while(N--) {
        string s,a,b;
        cin>>a>>b;
        int at =0,i=0,len;
        while((i=b.find(a,i))!=(string::npos)) { 
            at++;
            i++; 
        } 
        cout<<at<<endl; 
    } 
}

注解:
i=b.find(a,i)   //在字符串b中查找字符串a从b字符中第i个元素开始查找,返回为int型的数值重新赋值给i,string::npos是标准库的string容器属性。返回字符存放位置。 这个东西是一个容器，它将字符串分成一个一个来存储。
while((i=b.find(a,i))!=(string::npos)) { 
   at++;   //计算器
   i++;     
  }  //对b中字符串逐次与a比较直到结束