Counting of Trees AtCoder - 5633

Counting of Trees

Problem Statement

 

Given is an integer sequence D1,…,DN of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition:

• For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is Di.

Notes

 

• A tree of N vertices is a connected undirected graph with N vertices and N−1 edges, and the distance between two vertices are the number of edges in the shortest path between them.
• Two trees are considered different if and only if there are two vertices x and y such that there is an edge between x and y in one of those trees and not in the other.

Constraints

 

• 1≤N≤105
• 0≤Di≤N−1

Input

Input is given from Standard Input in the following format:

N
D1 D2 … DN

Output

 

Print the answer.

Sample Input 1

 

4
0 1 1 2

Sample Output 1

 

2

For example, a tree with edges (1,2), (1,3), and (2,4) satisfies the condition.

Sample Input 2

 

4
1 1 1 1

Sample Output 2

 

0

Sample Input 3

 

7
0 3 2 1 2 2 1

Sample Output 3

 

24

题意：就是给出一串序列，每个数字代表到第一棵树的距离，求出树的所有可能的排列个数。

思路：首先判定树是否合法：第一个数字一定是0，并且所有树的距离，中间不能有中断，必须1-2-3-4-5-6....

肯定是树的每层的可能的排列数的乘积。

但是每层树的排列数我一开始错误的认为比如样例3中第三层应该是A34（3   4的全排列），但是A34的结果是6，实际排列数应该是8，2的3次方emmmmm（我一开始默认为二叉树的排列了。。）可以存在一棵树与两棵以上的树相连的。

所以就是每层的树的下一层树的次幂的乘积。

2000ms的题我补题都能tle。。。加了O2优化才过的（不晓得AtCoder能不能用嗷）

 

#pragma GCC optimize(2) 
#include

using namespace std;
const int mod=998244353;
long long a[100010];
long long n,x;
long long ksm(long long a,long long b){
	long long res=1;
	while(b){
		if(b&1){
			res*=a;
			res%=mod;
		}
		a*=a;
		a%=mod;
		b>>=1;
	}
	return res;
}

int main(){
//	ios::sync_with_stdio(0);
//	cin.tie(0),cout.tie(0);
	scanf("%lld",&n);
	int flag=0;
	scanf("%lld",&x);
	int ans=x;
	if(x!=0) flag=1;
	for(int i=1;ians){
			ans=x;
		}
	}
	
	if(flag){
		printf("0\n");
	}
	else{
	long long sum=1;
		for(int i=1;i