HDU - 4821(95/600)字符串水题_en a string s and two integers l and m, we conside-优快云博客

本文链接：https://blog.youkuaiyun.com/NineFailure/article/details/77262227

本文探讨了如何计算给定字符串中满足特定长度和多样性的子串数量。通过哈希算法实现高效计算，介绍了算法的设计思路和实现代码。

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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
2

随便哈希一下

#include<bits/stdc++.h>
using namespace std;
#define unsigned long long  int
int bs=233;
int bte[100001],haxi[100001];
main()
{
    bte[0]=1;
    for(int a=1;a<=100000;a++)bte[a]=bte[a-1]*bs;
    int n,m;
    string q;
    while(cin>>n>>m)
    {
        cin>>q;
        haxi[0]=0;
        memset(haxi,0,sizeof(haxi));
        for(int a=0;a<q.size();a++)haxi[a+1]=haxi[a]*bs+q[a];
        int dan=0;
        int cc=q.size();
        for(int a=1;a<=m;a++)
        {
            map<int,int>mp;
            int er=0;
            for(int b=1,lj=1,sdk=1;a+b*m-1<=cc;b++,lj++)
            {
                int zs=haxi[a+b*m-1]-haxi[a+(b-1)*m-1]*bte[m];
        //      cout<<q[a+b*m-1-1]<<" ";
                if(mp[zs]==1)er++;
                mp[zs]++;
                if(lj<n)continue;
                if(er==0)dan++;
                int sdsd=haxi[a+sdk*m-1]-haxi[a+(sdk-1)*m-1]*bte[m];
                sdk++;
                mp[sdsd]--;
                if(mp[sdsd]==1)er--;
            }
    //      cout<<endl;
        }
        cout<<dan<<endl;
    }
}