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)

#include<iostream>
#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<len1;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<<work(A,B)<<endl;
    }
    return 0;
}