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.
InputThe 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.
OutputFor 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 http://blog.csdn.net/v_july_v/article/details/7041827（kmp的算法详解）
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 using namespace std;
6
7 int n;
8 int nxt[1000005];
9 char p[10005],s[1000005];
10
11 void getNext(char *p){
12 nxt[0]=-1;
13 int len=strlen(p);
14 for(int i=0;i<len;i++){
15 int k=nxt[i];
16 while(k!=-1&&p[i]!=p[k]) k=nxt[k];
17 nxt[i+1]=k+1;
18 }
19 }
20
21 int kmp(char *s,char *p){
22 getNext(p);
23 int ans=0,i=0,j=0,lens=strlen(s),lenp=strlen(p);
24 while(i<lens){
25 if(j==-1||s[i]==p[j]){
26 i++;
27 j++;
28 }
29 else j=nxt[j];
30 if(j==lenp){
31 ans++;
32 j=nxt[j];
33 }
34 }
35 return ans;
36 }
37
38 int main()
39 { scanf("%d",&n);
40 while(n--){
41 scanf("%s%s",p,s);
42 printf("%d\n",kmp(s,p));
43 }
44 }