字符匹配

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.

#include<iostream>
#include<string>
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 << num << endl;
}
}

查找(find)
语法: 
  size_type find( const basic_string &str, size_type index );
  size_type find( const char *str, size_type index );
  size_type find( const char *str, size_type index, size_type length );
  size_type find( char ch, size_type index );

find()函数: 

返回str在字符串中第一次出现的位置（从index开始查找）。如果没找到则返回string::npos, 
返回str在字符串中第一次出现的位置（从index开始查找，长度为length）。如果没找到就返回string::npos, 
返回字符ch在字符串中第一次出现的位置（从index开始查找）。如果没找到就返回string::npos 
例如, 

    string str1( "Alpha Beta Gamma Delta" );
    unsigned int loc = str1.find( "Omega", 0 );
    if( loc != string::npos )
      cout << "Found Omega at " << loc << endl;
    else
      cout << "Didn't find Omega" << endl;