[ABC284G] Only Once

Problem Statement

For a sequence of length $N$, $A = (A_1,A_2,\dots,A_N)$, consisting of integers between $1$ and $N$, inclusive, and an integer $i\ (1\leq i \leq N)$, let us define a sequence of length $10^{100}$, $B_i=(B_{i,1},B_{i,2},\dots,B_{i,10^{100}})$, as follows.

• $B_{i,1}=i$.
• $B_{i,j+1}=A_{B_{i,j}}\ (1\leq j<10^{100})$.

Additionally, let us define $S_i$ as the number of distinct integers that occur exactly once in the sequence $B_i$.
More formally, $S_i$ is the number of values $k$ such that exactly one index $j\ (1\leq j\leq 10^{100})$ satisfies $B_{i,j}=k$.

You are given an integer $N$. There are $N^N$ sequences that can be $A$. Find the sum of $\displaystyle \sum_{i=1}^{N} S_i$ over all of them, modulo $M$.

Constraints

• $1\leq N \leq 2\times 10^5$
• $10^8\leq M \leq 10^9$
• $N$ and $M$ are integers.

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Input

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

$N$ $M$

Output

Print the answer as an integer.

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Sample Input 1

4 100000000

Sample Output 1

624

As an example, let us consider the case $A=(2,3,3,4)$.

• For $i=1$: we have $B_1=(1,2,3,3,3,\dots)$, where two integers, $1$ and $2$, occur exactly once, so $S_1=2$.
• For $i=2$: we have $B_2=(2,3,3,3,\dots)$, where one integer, $2$, occurs exactly once, so $S_2=1$.
• For $i=3$: we have $B_3=(3,3,3,\dots)$, where no integers occur exactly once, so $S_3=0$.
• For $i=4$: we have $B_4=(4,4,4,\dots)$, where no integers occur exactly once, so $S_4=0$.

Thus, we have $\displaystyle \sum_{i=1}^{N} S_i=2+1+0+0=3$.

If we similarly compute $\displaystyle \sum_{i=1}^{N} S_i$ for the other $255$ sequences, the sum of this value over all $256$ sequences is $624$.

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Sample Input 2

7 1000000000

Sample Output 2

5817084

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Sample Input 3

2023 998244353

Sample Output 3

737481389

Print the sum modulo $M$.

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Sample Input 4

100000 353442899

Sample Output 4

271798911

题目讲的很复杂的样子。但是如果我们将 \(a_i\) 看作从 \(i\) 向 \(a_i\) 连一条边，那么将会连出来一棵基环树。题目要求的就是所有带标号基环树中所有非环上节点的深度和。

这题看似dp，却难以dp。考虑从数学角度入手。

枚举深度，尝试求出在所有的基环树中有多少个深度为 \(i\) 的节点。首先可以选出这个点到环上的 \(i\) 个节点组成的链，共 \(C_{n}^i\times i!\)，然后选出这个环，枚举环上有 \(j\) 个节点，众所周知，一个 \(j\) 元环有 \((j-1)!\) 种排列方法，但是还要枚举这条链接到换上的那个节点，再乘个 \(j\)，有 \(j!\) 种方法。剩下的点随便连，共 \(n^{n-i-j}\) 种。

最终答案为

\(\begin{aligned}
&\sum\limits_{i=1}^ni\sum\limits_{j=1}^{n-i}C_{n}^i\times i!\times C_{n-i}^j\times j!\times n^{n-i-j}\\&=\sum\limits_{i=1}^ni\sum\limits_{j=1}^{n-i}\frac{n!}{i!(n-i)!}\times i!\times \frac{(n-i)!}{j!(n-i-j)!}\times j!\times n^{n-i-j}\\&=\sum\limits_{i=1}^ni\sum\limits_{j=1}^{n-i}\frac{n!}{(n-i-j)!}\times n^{n-i-j}
\end{aligned}\)

会发现此时如果能调换一下求和顺序，会顺眼很多。但是后面即和 \(i\) 又和 \(j\) 有关，但是也之和 \(i+j\) 有关。令 \(T=i+j\)

\(\begin{aligned}
&=\sum\limits_{i=1}^ni\sum\limits_{j=1}^{n-i}\frac{n!}{(n-i-j)!}\times n^{n-i-j}\\&=\sum\limits_{T=2}^n\frac{n!}{(n-T)!}n^{n-T}\sum\limits_{i=1}^{T-1}i\\&=\sum\limits_{T=2}^n\frac{n!}{(n-T)!}n^{n-T}\frac 12 T(T-1)
\end{aligned}\)

#include<cstdio>
const int N=2e5+5;
int n,P,f[N],pw[N],ans;
int main()
{
	scanf("%d%d",&n,&P);
	for(int i=pw[0]=1;i<=n;i++)
		pw[i]=pw[i-1]*1LL*n%P;
	f[n+1]=1;
	for(int i=n;i>=1;i--)
		f[i]=f[i+1]*1LL*i%P;
	for(int i=2;i<=n;i++)
		(ans+=1LL*f[n-i+1]*pw[n-i]%P*(i*(i-1LL)/2%P)%P)%=P;
	printf("%d",ans);
}