POJ——T 3461 Oulipo

Posted 2020-10-02

http://poj.org/problem?id=3461

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42698		Accepted: 17154

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

Source

BAPC 2006 Qualification

 

 

①：hash 搞   对于y的任意子串x   hash(x,y)=hash[y]-hash[x]*p^(y-x+1)

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 
 6 using namespace std;
 7 
 8 #define p 147
 9 #define ull unsigned long long
10 char s1[10005],s2[1000005];
11 ull n,ss1,ss2,P;
12 ull lens,hs[1000005],ans;
13 
14 inline void Get_hash()
15 {
16     lens=0;
17     for(ull i=1;i<=ss1;i++)
18         lens=lens*p+s1[i]-‘A‘;
19 }
20 inline void Get_hs()
21 {
22     for(ull i=1;i<=ss2;i++)
23         hs[i]=hs[i-1]*p+s2[i]-‘A‘;
24 }
25 inline ull Q_pow(ull a,ull b)
26 {
27     ull ret=1,base=a;
28     for(;b;b>>=1,base*=base)
29         if(b&1) ret*=base;
30     return ret;
31 }
32 inline ull Check(ull x,ull y)
33 {
34     return hs[y]-hs[x-1]*P;
35 }
36 
37 int main()
38 {
39     for(cin>>n;n--;ans=0)
40     {
41         scanf("%s%s",s1+1,s2+1);
42         ss1=strlen(s1+1),ss2=strlen(s2+1);
43         Get_hash(); Get_hs(); P=Q_pow(p,ss1);
44         for(ull i=ss1;i<=ss2;i++)
45             if(lens==Check(i-ss1+1,i)) ans++;
46         cout<<ans<<endl;
47     }
48     return 0;
49 }

 

    ② kmp

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 
 6 using namespace std;
 7 
 8 int n,ss1,ss2;
 9 char s1[10005],s2[1000005];
10 int lens,next[1000005],ans;
11 
12 inline void Get_next()
13 {
14     int j=0;next[1]=0;
15     for(int i=2;i<=ss1;i++)
16     {
17         for(;j>0&&s1[i]!=s1[j+1];) j=next[j];
18         if(s1[i]==s1[j+1]) j++;
19         next[i]=j;
20     }
21 }
22 
23 inline void kmp()
24 {
25     int j=0;
26     for(int i=1;i<=ss2;i++)
27     {
28         for(;j>0&&s2[i]!=s1[j+1];) j=next[j];
29         if(s1[j+1]==s2[i]) j++;
30         if(j==ss1) ans++,j=next[j];
31     }
32 }
33 
34 int main()
35 {
36     for(cin>>n;n--;ans=0)
37     {
38         scanf("%s%s",s1+1,s2+1);
39         ss1=strlen(s1+1),ss2=strlen(s2+1);
40         Get_next(); kmp();
41         printf("%d\n",ans);
42     }
43     return 0;
44 }