nyoj 5 Binary String Matching<水过>

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 
来源
网络
上传者

naonao

代码：

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;char ck[15],ch[1050];int main(){int t;scanf("%d",&t);while (t--){scanf("%s%s",ck,ch);int ii=strlen(ch);int jj=strlen(ck);int s=0;for (int i=0;i<=ii-jj;i++){int kp=0;for (int j=0;j<jj;j++){if (ch[i+j]==ck[j])kp++;}if (kp==jj)s++;}printf("%d\n",s);}return 0;}