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 wordW and a textT, count the number of occurrences ofW inT. All the consecutive characters of W must exactly match consecutive characters ofT. 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:

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 wordW in the textT.

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIANSample Output
130KMP問題。。。。不過和KMP稍有不同，這個題要求計算次數而不是匹配成功就可以，所以next數組多求一位。AC代碼：
#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;int lent,lenp,next[10005];char p[10005],t[1000005];void getnext(){    int i=0,j=-1;    next[0]=-1;    while(i!=lenp)    {        if(j==-1||p[i]==p[j])        {            next[++i]=++j;        }        else        {            j=next[j];        }    }}int KMP(){    int i=0,j=0,c=0;    while(i!=lent&&j!=lenp)    {        if(t[i]==p[j]||j==-1)        {            ++i,++j;        }        else        {            j=next[j];        }        if(j==lenp)        {            c++;            j=next[j];        }    }    return c;}int main(){    int n,i,k,j,ans;    scanf("%d",&n);    getchar();    while(n--)    {        scanf("%s%s",p,t);        lent=strlen(t);        lenp=strlen(p);        getnext();        ans=KMP();        printf("%d/n",ans);    }    return 0;}