8.11 J

                                           J - Oulipo

The French authorGeorges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. Aquote from the book:

Tout avait Pairnormal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puissurgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulaitl’association qui l’unissait au roman : stir son tapis, assaillant à toutinstant 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 commandanttout, où s’abolissait la raison : tout avait l’air normal mais…

Perec wouldprobably have scored high (or rather, low) in the following contest. People areasked to write a perhaps even meaningful text on some subject with as fewoccurrences of a given “word” as possible. Our task is to provide the jury witha program that counts these occurrences, in order to obtain a ranking of thecompetitors. These competitors often write very long texts with nonsensemeaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.

So we want toquickly find out how often a word, i.e., a given string, occurs in a text. Moreformally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finitestrings over that alphabet, a word W and a text T,count the number of occurrences of W in T. All theconsecutive characters of W must exactly match consecutive characters of T.Occurrences may overlap.

Input

The first line ofthe 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 testcase in the input file, the output should contain a single number, on a singleline: 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

 

题意：输入T组样例，每组样例为两个字符串，判断第一个字符串在第二个字符串中出现几次，例如AZA，

AZAZAZA，可以把第二个字符串拆成AZA,AZA,AZA(可以将字符串重复拆分)，所以第一个字符串在第二个字符串中出现的3次，输出3。

思路：现将第一个字符串转换成next数组，然后进行KMP比较，循环条件由第二个字符串控制，以确保将第二个字符串完全跑完，跑的时候记录相等字符串长度等于第一个字符串长度的次数。

#include<stdio.h>#include<string.h>char s1[1110000],s2[1110000];int next[1110000],extend[1110000];int main(){int i,j,k,T,m,n,max,f,c;while(scanf("%d",&T)!=EOF){while(T--){c=0;scanf("%s%s",s2,s1);memset(next,0,sizeof(next));memset(extend,0,sizeof(extend));m=strlen(s1);n=strlen(s2);j=0;for(i=1;i<n;){if(s2[i]==s2[j]){next[i]=j+1;i++;j++;}else if(j==0&&s2[i]!=s2[j]){i++;}else if(j>0&&s2[i]!=s2[j]){j=next[j-1];}}/*for(i=0;i<n;i++){printf("%d ",next[i]);}printf("\n");*/i=0;j=0;while(i<m)//KMP计算{if(s1[i]==s2[j]){i++;j++;}else if(j==0&&s1[i]!=s2[j]){i++;}else if(j>0&&s1[i]!=s2[j]){j=next[j-1];}if(j==n)//j每次记录的是两个字符串相同的长度{j=next[j-1];//printf("j=====%d\n",next[j]);c++;//统计出现次数}}printf("%d\n",c);}}return 0;}