The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter ‘e‘. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive ‘T‘s is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {‘A‘, ‘B‘, ‘C‘, …, ‘Z‘} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format: One line with the word W, a string over {‘A‘, ‘B‘, ‘C‘, …, ‘Z‘}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W). One line with the text T, a string over {‘A‘, ‘B‘, ‘C‘, …, ‘Z‘}, with |W| ≤ |T| ≤ 1,000,000.

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

1
3
0
注意哪个是len1哪个是len2....
AC-code:
#include<cstdio>
#include<cstring>
const int N=10010;
char str[N],buf[N*100];
int p[N],len1,len2,cnt;
void getp()
{
	int i=0,j=-1;
	p[i]=j;
	while(i<len1)
	{
		if(j==-1||str[i]==str[j])
		{
			i++,j++;
			if(str[i]==str[j])
				p[i]=p[j];
			else
				p[i]=j;
		}
		else
			j=p[j];
	}
}
void kmp()
{
	getp();
	int i=0,j=0;
	while(i<len2)
	{
		if(j==-1||buf[i]==str[j])
		{
			i++,j++;
			if(j==len1)
				cnt++;
		}
		else
			j=p[j];	
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	getchar();
	while(n--)
	{
		scanf("%s%s",str,buf);
		len1=strlen(str);
		len2=strlen(buf);
		cnt=0;
		kmp();
		printf("%d\n",cnt);
	}
	return 0;
}