【南理oj】5 - Binary String Matching（水）

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

31110011101101011100100100100011010110100010101011 

样例输出

303 

来源

网络

上传者

naonao

 

 

挺简单的。

代码如下：

#include <cstdio>#include <cstring>char base[14];char pr[1111];int l1,l2;bool check(int n){for (int i=0;i<l1;i++){if (base[i]!=pr[n++])return false;}return true;}int main(){int u;int l;int ans;scanf ("%d",&u);while (u--){scanf ("%s",base);scanf ("%s",pr);l1=strlen(base);l2=strlen(pr);l=l2-l1;ans=0;for (int i=0;i<=l;i++){if (check(i))ans++;}printf ("%d\n",ans);}return 0;}