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
思路：这个只是简单匹配，要是有时间限制的话，可以用KMP算法处理
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdio> using namespace std; int main(){ int n;
cin>>n;
string a,b;
while (n--)
{
cin>>a>>b;
int count = ;
for (int i = ; i < b.length(); i++)
{
int k = i;
int j = ;
while(k<b.length() && j<a.length()){ if (j==a.length()- && b[k]==a[j])
{
count++;
}
if (b[k]==a[j])
{
k++;j++;
}else{
break;
} } }
cout<<count<<endl;
} return ;
}