HDU 1686Oulipo ———————Hash or KMP

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## Oulipo ##

# Oulipo #

**Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 22302 Accepted Submission(s): 8608**

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

Source

华东区大学生程序设计邀请赛\_热身赛

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Hash\_code:

Problem : 1686 ( Oulipo )     Judge Status : Accepted
    RunId : 25951470    Language : G++    Author : 311709001223
    Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    const int MAXN=1e6+7;
    const int MAXM=1e4+7;
    const int base=233;
    char str1[MAXM],str2[MAXN];
    ull Hash[MAXN];
    int t;
    int id[30],idx;
    int main()
    {
        scanf("%d",&t);
        while(t--)
        {
            scanf("%s %s",str1,str2);
            int ans=0;
            int len1=strlen(str1);
            int len2=strlen(str2);
            if(len1>len2)
            {
                printf("%d\n",ans);
                continue;
            }
            ull sh1=0;//字串的hash值
            ull sh2=0;//母串的hash值,先计算前len1 长度的hash值
            ull hb=1;
            for(int i=0;i<len1;i++) sh1=sh1*base+str1[i]-'A';
            for(int i=0;i<len1;i++) sh2=sh2*base+str2[i]-'A';
            for(int i=0;i<len1;i++) hb*=base;
            for(int i=0;i+len1<=len2;i++)
            {
                if(sh1==sh2)    ans++;
                sh2=sh2*base -(str2[i]-'A')*hb + (str2[i+len1]-'A');
            }
            printf("%d\n",ans);
        }
        return 0;
    }

--------------------

KMP\_code

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int MAXN=1e6+7;
    int slen,plen;
    char s[MAXN],p[MAXN];
    int Next[MAXN];
    void preKMP(char x[],int m,int kmpNext[])
    {
        int i,j;
        j=kmpNext[0]=-1;
        i=0;
        while(i<m)
        {
            while(-1!=j && x[i]!=x[j])  j=kmpNext[j];
            if(x[++i]==x[++j])  kmpNext[i]=kmpNext[j];
            else    kmpNext[i]=j;
        }
    }
    int KMP(char x[],int m,char y[],int n)
    {
        int i,j;
        int ans=0;
        preKMP(x,m,Next);
        i=j=0;
        while(i<n)
        {
            while(-1!=j && y[i]!=x[j])  j=Next[j];
            i++;j++;
            if(j>=m)
            {
                ans++;
                j=Next[j];
            }
        }
        return ans;
    }
    
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%s %s",p,s);
            plen=strlen(p);
            slen=strlen(s);
            printf("%d\n",KMP(p,plen,s,slen));
        }
        return 0;
    }