PE427.n-sequences

题目大意

定义一个串的值为最长相同子串长度，求所有长度为n的每项为[1,n]的所有串的值之和模1e9+9

题解

推完就对了很奇妙

先求至多为m的答案

(ans_m=[x^n]nsum_{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)

设(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没有影响可以去掉

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

分母展开成无穷级数，枚举(-(n-1)x^{m+1})的次数来计算，(nx)的次数就用(n)减一下得到

因为实际上是个((a+b)^i)的形式，所以当b的次数和总次数确定时只有一项算到

枚举m，时间复杂度是调和级数的形式，预处理幂可以做到O(n log n)，跑6s左右

最后的(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
",(ans+mod)%mod*n%mod*qpower(n-1,Mod)%mod);
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}