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
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0
给出一个子串一个主串，求主串中包括多少个子串
KMP算法，要深入理解
2015,7,20
#include<stdio.h>
#include<string.h>
char ch[1001000],str[110000];
int next[110001];
void getnext(int len){
	int i,j;
	i=0;j=-1;
	next[0]=-1;
	while(i<len){
		if(j==-1||str[i]==str[j]){
			++i;++j;
		//	if(str[i]==str[j]) 这三行是优化。能匹配的更快 
			//	next[i]=next[j];
		//	else next[i]=j;
			next[i]=j;
		}
		else j=next[j];
	}
}
int kmp(int len1,int len2){
	int i,j,count=0;
	getnext(len2);
	i=j=0;
	while(i<len1){//从主串開始匹配到最后 
		if(j==-1||ch[i]==str[j]){
			++i;++j;
		}
		else j=next[j];
		if(j==len2){//假设子串匹配完那么就加一 
			count++;
		}
	}
	return count;
}
int main(){
	int t,len1,len2;
	scanf("%d",&t);
	while(t--){
		scanf("%s%s",str,ch);
		len1=strlen(ch);
		len2=strlen(str);
		printf("%d\n",kmp(len1,len2));
	}
	return 0;
}