CF 954H Path Counting

You are given a rooted tree. Let's denote d(x) as depth of node x: depth of the root is 1, depth of any other node x is d(y) + 1, where yis a parent of x.

The tree has the following property: every node x with d(x) = i has exactly a_i children. Maximum possible depth of a node is n, and a_n = 0.

We define f_k as the number of unordered pairs of vertices in the tree such that the number of edges on the simple path between them is equal to k.

Calculate f_k modulo 10⁹ + 7 for every 1 ≤ k ≤ 2n - 2.

Input

The first line of input contains an integer n (2 ≤ n ≤ 5 000) — the maximum depth of a node.

The second line of input contains n - 1 integers a₁, a₂, ..., a_n - 1 (2 ≤ a_i ≤ 10⁹), where a_i is the number of children of every node xsuch that d(x) = i. Since a_n = 0, it is not given in the input.

Output

Print 2n - 2 numbers. The k-th of these numbers must be equal to f_k modulo 10⁹ + 7.

Examples
input
Copy
4
2 2 2

output
Copy
14 19 20 20 16 16 
input
Copy
3
2 3

output
Copy
8 13 6 9 

Note

This the tree from the first sample:

$\text{[math]}$

【题意】

给出一棵深度为n的树，其中每个深度为i的节点都有a[i]个儿子。问对于每个k，有多少条简单路径满足其长度恰好为k。

n<=5000

【分析】

考虑枚举路径的端点。

设d[i,j]表示从某个深度为i的节点开始，只往下走且长度为j的路径条数。那么d[i,j]显然等于i的子树中深度为i+j的点数。

设u[i,j]表示从某个深度为i的节点开始，第一步必须往上走，且路径长度为j的方案。

有两种转移，一种是走到父亲后继续往上，贡献就等于u[i-1,j-1]。另一种转移是走到父亲后就开始往下走，贡献就等于u[i,j-2]*(a[i-1]-1)

综上，f[i][j]= f[i+1][j-1]*a[i]+

f[i-1][j-1]+

f[i][j-2]*(a[i-1]-1);

【代码】

#include<cstdio>
#include<cstring>
#include<iostream>
#define debug(x) cerr<<#x<<" "<<x<<'\n';
using namespace std;
typedef long long ll;
const int N=5005;
const ll mod=1e9+7;
const ll rev=5e8+4;
int n,a[N],p[N],f[N][N<<1],ans[N<<1];
int main(){
	scanf("%d",&n);p[0]=1;
	for(int i=1;i<n;i++) scanf("%d",&a[i]),p[i]=(ll)p[i-1]*a[i]%mod;
	for(int i=n;i;i--){
		f[i][0]=1;
		for(int j=1;j<=n-i;j++){
			f[i][j]=(ll)f[i+1][j-1]*a[i]%mod;
			ans[j]=((ll)ans[j]+(ll)f[i][j]*p[i-1])%mod;
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=2*n-2;j>=1;j--){
			f[i][j]=f[i-1][j-1];
			if(i>1&&j>1&&j-2<n&&j<=i+n-2) f[i][j]=((ll)f[i][j]+(ll)f[i][j-2]*(a[i-1]-1))%mod;
			ans[j]=((ll)ans[j]+(ll)f[i][j]*p[i-1])%mod;
		}
	}
	for(int i=1;i<=2*n-2;i++) printf("%I64d ",(ll)ans[i]*rev%mod);
	return 0;
}

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/10424160.html
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