poj3461 Oulipo

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

昨天去做cf的时候B题写KMP的时候调了半天调不出来真是打脸……

所以现在开始疯狂刷KMP（当复习吧）

这题当模板用

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void write(LL a)
{
	if (a<0){printf("-");a=-a;}
	if (a>=10)write(a/10);
	putchar(a%10+'0');
}
inline void writeln(LL a){write(a);printf("\n");}
int next[100010];
char s[10010],p[1000010];
int ls,lp;
inline void pre()
{
	memset(next,0,sizeof(next));
	int j=0;
	for (int i=2;i<=ls;i++)
	{
		while (j>0 && s[j+1]!=s[i])j=next[j];
		if (s[j+1]==s[i])j++;
		next[i]=j;
	}
}
inline void KMP()
{
	int j=0,ans=0;
	for (int i=1;i<=lp;i++)
	{
		while (j>0 && s[j+1]!=p[i])j=next[j];
		if (s[j+1]==p[i])j++;
		if (j==ls)
		{
			ans++;
			j=next[j];
		}
	}
	printf("%d\n",ans);
}
inline void work()
{
	scanf("%s",s+1);
	scanf("%s",p+1);
	ls=strlen(s+1);lp=strlen(p+1);
	pre();
	KMP();
}
int main()
{
	int T=read();
	while (T--)work();
}