[ABC299F] Square Subsequence

Problem Statement

You are given a string $S$ consisting of lowercase English letters.
Print the number of non-empty strings $T$ that satisfy the following condition, modulo $998244353$.

The concatenation $TT$ of two copies of $T$ is a subsequence of $S$ (not necessarily contiguous).

Constraints

• $S$ is a string consisting of lowercase English letters whose length is between $1$ and $100$, inclusive.

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Input

The input is given from Standard Input in the following format:

$S$

Output

Print the answer.

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Sample Input 1

ababbaba

Sample Output 1

8

The eight strings satisfying the condition are a, aa, ab, aba, b, ba, bab, and bb.

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Sample Input 2

zzz

Sample Output 2

1

The only string satisfying the condition is z.
Note that this string contributes to the answer just once, although there are three ways to extract the subsequence zz from $S = S_1S_2S_3 = $ zzz: $S_1S_2 = $ zz, $S_1S_3 = $ zz, and $S_2S_3 = $ zz.

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Sample Input 3

ppppqqppqqqpqpqppqpqqqqpppqppq

Sample Output 3

580

考虑枚举两个串的开头，就设为 \(a\) 和 \(b\)，满足 \(s_a=s_b\),然后分别跳子序列自动机。

为了使所有跳的操作不重复，我们应该强制第一个串的开头就是某个字母的第一次出现。

然后进行dp，定义 \(dp_{i,j}\) 为第一个串目前跳的到了点 \(i\)，第二个串跳到了点 \(j\) 的情况。当然，要满足 \(i<a\)

理论上，要把所有的 \(dp_{i,j}\) 全部计入答案。但是我们发现如果这样会算重。比如串 ababab，那么枚举第一个 a时，会把 ab 这个串数两次。去一下重就可以了。如果 \(i\) 的下一个 \(s_a\) 的出现地方时 \(b\)，那么我们才计入答案。

复杂度：枚举两个串开头是 \(O(n)\) 的，dp \(O(|\Sigma|n^2)\)，总复杂度 \(O(n^3|\Sigma|)\)

#include<bits/stdc++.h>
using namespace std;
const int N=105,P=998244353;
int n,nx[N][26],dp[N][N],ans;
char s[N];
void add(int&x,int y)
{
	x=x+y>=P? x+y-P:x+y;
}
int main()
{
	scanf("%s",s+1),n=strlen(s+1);
	for(int i=n-1;i>=1;i--)
	{
		memcpy(nx[i],nx[i+1],sizeof(nx[i]));
		nx[i][s[i+1]-'a']=i+1;
	}
//	printf("%d\n",nx[1][0]);
	for(int i=0;i<26;i++)
	{
		int st=0;
		for(int j=1;j<=n;j++)
		{
			if(s[j]-'a'^i)
				continue;
			if(!st)
				st=j;
			else
			{
//				printf("hjh:%d %d\n",st,j);
				memset(dp,0,sizeof(dp));
				dp[st][j]=1;
				for(int a=st;a<j;a++)
				{
					for(int b=j;b<=n;b++)
					{
//						if(dp[a][b])
//							printf("%d %d\n",a,b);
						if(nx[a][i]==j)
							add(ans,dp[a][b]);
						for(int c=0;c<26;c++)
							if(nx[a][c]<j&&nx[a][c]&&nx[b][c])
								add(dp[nx[a][c]][nx[b][c]],dp[a][b]);
					}
				}
			}
		}
//		add(ans,(bool)st);
	}
	printf("%d",ans);
}