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

輸入
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.
輸出
For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
樣例輸入

3
11
1001110110
101
110010010010001
1010
110100010101011 

樣例輸出

3
0
3 

簡單來說就是給你兩個字串，讓你求出s1在s2 中出現的次數。一般想到的是KMP，畢竟都知道字串的題大多數都能用KMP。這裡講的是string的用法。

#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;
    }
}
/*
s2.find(s1,start);
表示的是s2字串上,從start開始s1在s2處的下標
如果找不到的話就返回一個常量，也就是 string::npos  4294967295
*/