南阳5--Binary String Matching(Kmp)

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

 kmp  裸题；

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
char a[15], b[1010]; int p[1010], lena, lenb, ans;
void Getp()
{
    int  i = 0, j = -1;
    p[i] = j;
    while(i != lena)
    {
        if(j == -1 || a[i] == a[j])
        {
            i++, j++;
            p[i] = j; 
        } 
        else
            j = p[j];
    }
}
void Kmp()
{
    Getp();
    int i = 0, j = 0;
    while(i != lenb)
    {
        if(j == -1 || b[i] == a[j])    
            i++, j++;
        else 
            j = p[j];
        if(j == lena)
            ans++; 
    }  
    printf("%d
", ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        ans = 0;
        scanf("%s%s", a, b);
        lena = strlen(a);
        lenb = strlen(b);
        Kmp();
    }
    return 0;
}

暴力： 竟然也是 0 ms; 

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
char a[15], b[1010]; int lena, lenb; 
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int i, j, sum = 0;
        scanf("%s %s", a, b);
        int lena = strlen(a), lenb = strlen(b);
        for(i = 0; i <= lenb - lena; i++)
        {
            if(b[i] == a[0])
            {
                for(j = 1; j < lena; j++)
                    if(b[i+j] != a[j])
                        break;
                if(j == lena)
                    sum++;
            } 
        }
        printf("%d
", sum);
    } 
    return 0;    
}