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‘

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

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
題意描述：
輸入兩個串s1和s2
計算並輸出s1在s2中能夠匹配多少次
解題思路：
KMP模板題，加深對next數組的理解，主要問題是更新i或者j的位置，其實不用更新，你會發現其實到最後j是會自動返回的。
代碼實現:

 1 #include<stdio.h>
 2 #include<string.h>
 3 char s[1000010],t[10010];
 4 void get_next(char t[],int next[],int l2);
 5 int kmp(char s[],char t[]);
 6 int main()
 7 {
 8     int T;
 9     scanf("%d",&T);
10     while(T--)
11     {
12         scanf("%s%s",t,s);
13         printf("%d\n",kmp(s,t)); 
14     }
15     return 0;
16 }
17 int kmp(char s[],char t[])
18 {
19     int i,j,l1,l2,c;
20     
21     int next[10010];//next[]數組中存的是左上匹配串前後綴的相似度，從0到l2 
22     l1=strlen(s);
23     l2=strlen(t);
24     get_next(t,next,l2);
25     
26     c=0;
27     i=0;
28     j=0;
29     while(i < l1)
30     {
31         if(j==-1 || s[i] == t[j])
32         {
33             i++;
34             j++;
35         }
36         else//少了個else 
37             j=next[j];
38         if(j==l2)//i和j均不用改動
39             c++;
40     }
41     return c;
42 }
43 void get_next(char t[],int next[],int l2)
44 {
45     int i,j;
46     i=0;
47     j=-1;
48     next[0]=-1;
49     while(i < l2)
50     {
51         if(j==-1 || t[i] == t[j])
52         {
53             i++;
54             j++;
55             if(t[i] != t[j])
56             next[i]=j;
57             else
58             next[i]=next[j];
59         }
60         else//少了個else 
61             j=next[j];
62     }
63     /*for(i=0;i<=l2;i++)
64         printf("%d ",next[i]);
65     printf("\n");*/
66 }

易錯分析：

1、next數組存儲的結果是錯位的，正好能夠被下次利用。

2、代碼能力，註意細節，像少個else，找了半天錯。

POJ 3461 Oulipo