hdu 4821String（字符串hash）_hdu 字符串hash-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_37164003/article/details/77407332

本文介绍了一种使用字符串Hash解决特定子串问题的方法。该问题要求找出所有由M个不同的长度为L的子串拼接而成的长度为M*L的可恢复子串。通过构建高效的Hash函数，可以在大型字符串中快速定位并计数这些特殊子串。

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String

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 0 Accepted Submission(s) : 0

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Problem Description

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

Source

2013 Asia Regional Changchun

想法：字符串hash

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<string>
typedef unsigned long long ull;
using namespace std;
const int maxn= 100050;
const ull base =31;
ull nbase[maxn],hash[maxn];
int m,l;
map<ull, int> mp;
int main()
{
	ull tmp;
	nbase[0] = 1;
	for (int i = 1;i<maxn; i++)
    {
        nbase[i]=nbase[i-1]*base;
    }
	while(~scanf("%d%d",&m,&l))
	 {
	     string s;
		cin>>s;
		int slen=s.size();
		hash[slen] = 0;
		for (int i = slen-1; i >= 0; i--)
			hash[i] = hash[i+1]*base+s[i]-'a'+1;           //关键1
		int ans = 0;
		for (int i = 0; i<l&&i+m*l<=slen; i++)
		{
			mp.clear();
			for (int j = i; j<i+m*l; j += l)
			{
				tmp = hash[j] - hash[j+l]*nbase[l];
				mp[tmp]++;
			}
			if (mp.size() ==m)
				ans++;
			for (int j=i+m*l; j+l<=slen; j +=l)
			 {
				tmp = hash[j-m*l] - hash[j-(m-1)*l]*nbase[l];
				mp[tmp]--;
				if (mp[tmp] == 0)
                                mp.erase(tmp);
				tmp = hash[j] - hash[j+l]*nbase[l];
				mp[tmp]++;
				if (mp.size() == m)
					ans++;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}