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
解题思路:计算模式串在主串中出现的次数。
AC代码:
 #include<cstdio>
#include<string.h>
const int maxn=1e4+;
const int maxm=1e6+;
char text[maxm],pattern[maxn];
int prefix[maxn],lena,lenb,num,t;
void get_prefix_table(){//处理模式串前缀表
int j=,pos=-;
prefix[]=-;
while(j<lenb){
if(pos==-||pattern[pos]==pattern[j])prefix[++j]=++pos;
else pos=prefix[pos];
}
}
void kmp_search(){
int i=,j=;
while(i<lena){
if(j==-||text[i]==pattern[j])i++,j++;
else j=prefix[j];
if(j==lenb)num++,j=prefix[j];//此时的j应退回到j前面子串的最长公共前后缀长度的位置进行下一次匹配,可以有相交的模式串
}
}
int main(){
while(~scanf("%d",&t)){
while(t--){
scanf("%s%s",pattern,text);
memset(prefix,,sizeof(prefix));
num=,lena=strlen(text),lenb=strlen(pattern);
get_prefix_table();
kmp_search();
printf("%d\n",num);
}
}
return ;
}

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