Problem Description
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.
Input
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.
Output
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.
Sample Input
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
Source
华东区大学生程序设计邀请赛_热身赛
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哔哩哔哩上大佬写的代码，以后作为模板使用，，，KMP刚开始接触，理解得还不是很透彻，，，
后记？！：
如何理解KMP 链接在此，还不点？！这位大佬是我看的多个讲解博客和视频中讲得最清晰的一个（大佬的其他算法的讲解也不错哦）。
//kmp裸题
#include<bits/stdc++.h>using namespace std;
const int maxn=1e6+10;
int nxt[maxn]; //存储真前缀和真后缀相匹配的最大长度，所谓真前缀，就是长度小于字符串的长度，例如，abcd中最长真前缀是abc，真后缀同理
char str[maxn],mo[maxn];void getNext(){int i=0,j=-1,len=strlen(mo); /*nxt[0]默认为-1，nxt[0]规定为0，当然也可以理解成j为真前缀和真后缀相匹配的最大长度，而真前缀长度肯定比字符 */nxt[0]=-1;while(i<len){if(j==-1||mo[i]==mo[j]) nxt[++i]=++j;else j=nxt[j];}
}int kmp(){getNext();int i=0,j=0,len1=strlen(str),len2=strlen(mo);int ans=0;while(i<len1){//cout<<"***j="<<j<<endl;if(j==-1||mo[j]==str[i]) i++,j++;else j=nxt[j];if(j==len2) ans++;//cout<<"j="<<j<<endl;}return ans;
}int main(){int t;scanf("%d",&t);while(t--){scanf("%s%s",mo,str);/*for(int i=0;i<=10;i++)cout<<nxt[i]<<endl;*/printf("%d\n",kmp());}return 0;
}