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.

题意：找一个串在另一个串中出现了多少次

思路：kmp统计一下即可

#include <cstdio>

#include <map>

#include <iostream>

#include<cstring>

#include<bits/stdc++.h>

#define ll long long int

#define M 6

using namespace std;

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

int moth[]={,,,,,,,,,,,,};

int dir[][]={, ,, ,-, ,,-};

int dirs[][]={, ,, ,-, ,,-, -,- ,-, ,,- ,,};

const int inf=0x3f3f3f3f;

const ll mod=1e9+;

int nextt[];

void get_next(string s){

    nextt[]=;

    int len=s.length();

    for(int i=,j=;i<=len;i++){

        while(j>&&s[j]!=s[i-]) j=nextt[j];

        if(s[j]==s[i-]) j++;

        nextt[i]=j;

    }

}

int main(){

    ios::sync_with_stdio(false);

    int t;

    cin>>t;

    while(t--){

        string a,b;

        cin>>a>>b;

        get_next(a);

        int ans=;

        int l1,l2;

        l1=a.length(); l2=b.length();

        for(int i=,j=;i<=l2;i++){

            while(j>&&(j==l1||b[i-]!=a[j])) j=nextt[j];

            if(b[i-]==a[j]) j++;if(j==l1) ans++;

        }

        cout<<ans<<endl;

    }

    return ;

}