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
代码如下：
#include<iostream>
#include<string.h>
using namespace std;
char a[15],b[1005];
int main()
{
int n;
cin>>n;
while(n--)
{
int ans=0;
cin>>a>>b;
int m=strlen(a);
int l=strlen(b);
for(int i=0;i<l;i++)
{
if(strncmp(a,&b[i],m)==0)//strncmp函数
ans++;
}
cout<<ans<<endl;
}
return 0;
}