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

题意：

求模式串在子串出现的次数。

跑个KMP就好了，，

#include <stdio.h>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
using namespace std;
#define mod 1000000007
#define inf 0x7f7f7f7f
int Next[10005];
char w[10005];
char t[1000005];
int lenw,lent;
int ans;
void GetNext()
{
    int i=0;
    int j=-1;
    Next[i]=j;
    while(i<lenw)
    {
        if(j==-1 || w[i]==w[j])
        {
            i++;
            j++;
            Next[i]=j;
        }
        else
            j=Next[j];
    }
}
void kmp()
{
    int i=0;
    int j=0;
    while(i<lent)
    {
        if(j==-1 ||t[i]==w[j])
        {
            i++;
            j++;   //匹配一个j+1；
        }
        else
            j=Next[j];
        if(j>=lenw)    //直到匹配lenw个。。
        { 
            ans++;   //出现的次数加一。。
            j=Next[j];
        }
    }
}
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        ans=0;
        scanf("%s%s",w,t);
        lenw=strlen(w);
        lent=strlen(t);
        if(lenw>lent)
        {
            cout<<0<<endl;
            continue;
        }
        GetNext();
        kmp();
        cout<<ans<<endl;
    }
    return 0;
}