NYOJ 5 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 
资料参考strstr函数：http://www.cplusplus.com/reference/cstring/strstr/（c语言版）
      <string>:find函数http://www.cplusplus.com/reference/string/string/find/（c++）
代码参考：

 
#include<iostream>
#include<string>
#include<algorithm>
 using namespace std;
 int main() {
 int n;
 cin >> n;
 while (n--) {
 int flag, s = 0;
 string ch, str;
 cin >> ch >> str;
 flag = str.find(ch, 0);
 while (flag != string::npos) {
        //The position of the first character of the first match.
        //If no matches were found, the function returns string::npos.
 s++;
 flag = str.find(ch, flag + 1);
 }
 cout << s << endl;
 }
 }