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最新推荐文章于 2019-01-14 08:43:16 发布

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最新推荐文章于 2019-01-14 08:43:16 发布

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文章标签：人工智能

原文链接：http://www.cnblogs.com/NaCl/p/9580189.html

本文介绍了一个有趣的编程挑战：在一个给定的文本中找到特定大小的不同子串的数量，并将其视为一个隐藏的质数。通过使用hash算法而非传统的set集合，有效地解决了这一难题。

Description

Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a given text. Such number could be the number of different substrings of a given size that exist in the text. As you soon will discover, you really need the help of a computer and a good algorithm to solve such a puzzle.
Your task is to write a program that given the size, N, of the substring, the number of different characters that may occur in the text, NC, and the text itself, determines the number of different substrings of size N that appear in the text.

As an example, consider N=3, NC=4 and the text “daababac”. The different substrings of size 3 that can be found in this text are: “daa”; “aab”; “aba”; “bab”; “bac”. Therefore, the answer should be 5.

Input

The first line of input consists of two numbers, N and NC, separated by exactly one space. This is followed by the text where the search takes place. You may assume that the maximum number of substrings formed by the possible set of characters does not exceed 16 Millions.

Output

The program should output just an integer corresponding to the number of different substrings of size N found in the given text.

Sample Input
3 4
daababac

Sample Output
5
一开始用STL的set来做果断超时了，然后就想用hash做，这道hash有点意思，他给出的NC
值就是关键字，然后给出的字符串是文本，注意是文本，而不是单单是字母，所以在求每个字符的hash值有点变化。

#include<stdio.h>
#include<string.h> 
#include<iostream> 
#include<algorithm>
#include<string> 
#include<set>
#include<queue>
#define maxn 18000000+50
using namespace std;
char s[maxn]; 
bool hash[maxn*10];
int a[300];
int n,N;
int main()
{
    set<string>S;
    while(cin>>n>>N)
    {
    int count=0;
    memset(hash,0,sizeof(hash));
    memset(a,0,sizeof(a));
    scanf("%s",s);
    int k=strlen(s);
    for(int i=0;i<k;i++)
    {
        if(a[s[i]-' ']==0) 
        { 
        a[s[i]-' ']=1;
        }
    } 
    int num=0;
    for(int i=0;i<=256;i++)
      if(a[i]==1)
      {
       a[i]=num++;
      }
    for(int i=0;i<k-n+1;i++)
          {
            long long h=0;
            for(int j=i;j<i+n;j++)
            h=h*N+a[s[j]-' '];
            if(!hash[h])
            {
            count++;
            hash[h]=1;
           }
           }
           cout<<count<<endl;
     }
    return 0;
}

转载于:https://www.cnblogs.com/NaCl/p/9580189.html