nyoj5 字符串匹配

Binary String Matching

时间限制： 3000 ms  |  内存限制： 65535 KB

 

 

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

 

算法实现：暴力的字符串匹配 你 O (n*m)

View Code

#include
#include<string>
using namespace std;

int work(string A,string B)
{
int len1=A.length(),len2=B.length();
int i,j,ans=0;
for(i=0;i<=len2-len1;i++)
    {
for(j=0;j        {
if(A[j]!=B[i+j]) break;
        }
if(j==len1) ans++;
    }
return ans;
}

int main()
{
int ntest;
    cin>>ntest;
string A,B;
while(ntest--)
    {
        cin>>A>>B;
        cout<    }
return 0;
}  

 

 

转载于:https://www.cnblogs.com/zhourongqing/archive/2011/09/29/2195877.html