String （字符串哈希）

最新推荐文章于 2025-03-04 22:00:00 发布

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最新推荐文章于 2025-03-04 22:00:00 发布

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分类专栏：算法和数据结构

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本文探讨了在一个给定的字符串S中，寻找所有长度为M*L的可恢复子串的方法。这些子串由M个不同的，长度为L的子串拼接而成，且这M个子串在相同位置的字符不完全相同。文章提供了一个使用哈希算法解决此问题的示例代码。

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Given a string S and two integers L and M, we consider a substring of S as “recoverable” if and only if
(i) It is of length M*L;
(ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings are considered as “diversified” if they don’t have the same character for every position.
Two substrings of S are considered as “different” if they are cut from different part of S. For example, string "aa" has 3 different substrings "aa", "a" and "a".
Your task is to calculate the number of different “recoverable” substrings of S.

Input

The input contains multiple test cases, proceeding to the End of File.
The first line of each test case has two space-separated integers M and L.
The second ine of each test case has a string S, which consists of only lowercase letters.
The length of S is not larger than 10^5, and 1 ≤ M * L ≤ the length of S.

Output

For each test case, output the answer in a single line.

Sample Input

3 3
abcabcbcaabc

Sample Output

【代码】

#include <iostream>
#include <algorithm>
#include <cstring>
#include<map>
#define ull unsigned long long
using namespace std;
char s[100100];
ull hsh[100100],p[100100];
map<ull,int> mp;
int gethash(int l,int r)
{
    return hsh[r]-hsh[l-1]*p[r-l+1];
}
int main()
{
    int m,l;
    while(cin>>m>>l)
    {
        scanf("%s",s+1);
        int num=0;
        int len=strlen(s+1);
        p[0]=1;
        for(int i=1;i<=len;i++)
        {
            hsh[i]=hsh[i-1]*131+s[i]-'a'+1;
            p[i]=p[i-1]*131;
        }
        for(int i=1;i<=l&&i+m*l<=len;i++)
        {
            mp.clear();
            for(int j=i;j<i+m*l;j+=l)
            {
                ull x=gethash(j,j+l-1);
                mp[x]++;
            }
            if(mp.size()==m)
                num++;
            for(int j=i+m*l;j+l-1<=len;j+=l)
            {
                ull x=gethash(j,j+l-1);
                mp[x]++;
                ull y=gethash(j-m*l,j-m*l+l-1);
                mp[y]--;
                if(mp[y]==0) mp.erase(y);
                if(mp.size()==m) num++;
            }
        }
        cout<<num<<endl;
    }
    return 0;
}