PE427.n-sequences

題目大意

定義一個串的值爲最長相同子串長度，求所有長度爲n的每項爲[1,n]的所有串的值之和模1e9+9

題解

推完就對了很奇妙

先求至多爲m的答案

\(ans_m=[x^n]n\sum_{i=0}^{\infty} (n-1)^i(\sum_{j=1}^m x^j)^{i+1}\)

\(=[x^n]\frac{n}{n-1}\sum_{i=1}^{\infty} (n-1)^i(\sum_{j=1}^m x^j)^i\)

設\(F=(n-1)\sum_{i=1}^m x^i=(n-1)\frac{x^{m+1}-x}{x-1}\)

\(ans_m=[x^n]\frac{n}{n-1}\sum_{i=1}^{\infty} F^i\)

\(=[x^n]\frac{n}{n-1}*\frac{F}{1-F}\)

\(=[x^n]\frac{n}{n-1}*(\frac{1}{1-F}-1)\)

-1沒有影響可以去掉

\(=[x^n]\frac{n}{n-1}*\frac{1-x}{1-(nx-(n-1)x^{m+1})}\)

分母展開成無窮級數，枚舉\(-(n-1)x^{m+1}\)的次數來計算，\(nx\)的次數就用\(n\)減一下得到

因爲實際上是個\((a+b)^i\)的形式，所以當b的次數和總次數確定時只有一項算到

枚舉m，時間複雜度是調和級數的形式，預處理冪可以做到O(n log n)，跑6s左右

最後的\(Ans=\sum_i i*(ans_i-ans_{i-1})\)

code

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define C(n,m) (jc[n]*Jc[m]%mod*Jc[(n)-(m)]%mod)
#define add(a,b) a=((a)+(b))%mod
#define mod 1000000009
#define Mod 1000000007
#define ll long long
#define file
using namespace std;

ll jc[7500001],Jc[7500001],Ans[7500001],p1[7500001],p2[7500001],ans;
int n,i,j,k,l;

ll qpower(ll a,int b) {ll ans=1; while (b) {if (b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;} return ans;}

int main()
{
	#ifdef file
	freopen("PE427.in","r",stdin);
	#endif
	
	scanf("%d",&n);
	jc[0]=1;
	fo(i,1,n) jc[i]=jc[i-1]*i%mod;
	Jc[n]=qpower(jc[n],Mod);
	fd(i,n-1,0) Jc[i]=Jc[i+1]*(i+1)%mod;
	
	p1[0]=p2[0]=1;
	fo(i,1,n) p1[i]=p1[i-1]*(-(n-1))%mod,p2[i]=p2[i-1]*n%mod;
	
	fo(i,1,n)
	{
		fo(j,0,n)
		if ((i+1)*j<=n)
		{
			add(Ans[i],p1[j]*p2[n-(i+1)*j]%mod*C(j+(n-(i+1)*j),j));
			if ((i+1)*j<n)
			add(Ans[i],-p1[j]*p2[n-(i+1)*j-1]%mod*C(j+(n-(i+1)*j-1),j));
		}
		else
		break;
	}
	fo(i,1,n) add(ans,(Ans[i]-Ans[i-1])*i);
	
	printf("%lld\n",(ans+mod)%mod*n%mod*qpower(n-1,Mod)%mod);
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}