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
就是简单的找子字符串：
#include<iostream>
#include<string>
using namespace std;
int main(){
int t,m,n,ans;
cin>>t;
while(t--){
string a,b;
ans=0;
cin>>b>>a;
m=a.length();
n=b.length();
for(int i=0;i<=m-n;i++){
if(b==a.substr(i,n)){
ans++;
}
}
cout<<ans<<endl;
}
return 0;
}