Binary String Matching

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

源代码：

基于数据结构模式匹配BP算法

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char s[1000],t[10];
void index(char t[10],char s[1000])
{
int i=0,j=0;
int count=0;
while(i<strlen(s))
{
if(s[i] == t[j])
{
i++;
j++;
}
else
{
i=i-j+1;
j=0;
}
if(j>=strlen(t))
{

count++;
i=i-j+1;
j=0;
}
}
printf("%d\n",count);
}
int main()
{
int n;
scanf("%d",&n);
// getchar();
while(n--)
{
scanf("%s%s",t,s);
index(t,s);
}
system("pause");
return 0;
}

基于模式匹配KMP算法

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char p[1000], s[1000];
int next[1000];
void getnext()
{
int i = 0, j = -1, len = strlen(p);
next[0] = -1;
while (i < len - 1)
{
if (j == -1 || p[j] == p[i])
{
j++;
i++;
next[i] = j;
}
else
j = next[j];
}
}
int kmp()
{
int i = -1, j = -1, lenp = strlen(p), lens = strlen(s);
getnext();
int num = 0;
while (i < lens)
{
if (j == -1 || s[i] == p[j])
{
++i;
++j;
}
else
j = next[j];
if (j == lenp)
{
i-=j;
j = -1;
num++;
}
}
return num;
}
int main(void)
{
int n;
while (~scanf("%d", &n))
{
while (n--)
{
scanf("%s%s", p, s);
printf("%d\n", kmp());
}
}
system("pause");
return 0;
}