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.

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.

OutputFor 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.

Sample Input

3

BAPC

BAPC

AZA

AZAZAZA

VERDI

AVERDXIVYERDIAN

Sample Output

1

3

0

在主串中匹配到j>=plen时，j继续回溯。

 #include<stdio.h>

 #include<string.h>

 int Next[],n,m,_,tlen,plen;

 char t[],p[];

 void prekmp() {

     tlen=strlen(t);

     plen=strlen(p);

     int i,j;

     j=Next[]=-;

     i=;

     while(i<plen) {

         while(j!=-&&p[i]!=p[j]) j=Next[j];

         if(p[++i]==p[++j]) Next[i]=Next[j]; //这里判断的是模式串

         else Next[i]=j;

     }

 }

 int kmp() {

     prekmp();

     int i,j,ans=;

     i=j=;

     while(i<tlen) {

         while(j!=-&&t[i]!=p[j]) j=Next[j];

         i++;j++;

         if(j>=plen) {  //主要区别

             ans++;

             j=Next[j];

         }

     }

     return ans;

 }

 int main() {

     for(scanf("%d",&_);_;_--) {

         scanf("%s",p);

         scanf("%s",t);

         printf("%d\n",kmp());

     }

 }