Sugoroku 3

Problem Statement

There are $N$ squares called Square $1$ though Square $N$. You start on Square $1$.

Each of the squares from Square $1$ through Square $N-1$ has a die on it. The die on Square $i$ is labeled with the integers from $0$ through $A_i$, each occurring with equal probability. (Die rolls are independent of each other.)

Until you reach Square $N$, you will repeat rolling a die on the square you are on. Here, if the die on Square $x$ rolls the integer $y$, you go to Square $x+y$.

Find the expected value, modulo $998244353$, of the number of times you roll a die.

Notes

It can be proved that the sought expected value is always a rational number. Additionally, if that value is represented $\frac{P}{Q}$ using two coprime integers $P$ and $Q$, there is a unique integer $R$ such that $R \times Q \equiv P\pmod{998244353}$ and $0 \leq R \lt 998244353$. Find this $R$.

Constraints

• $2 \le N \le 2 \times 10^5$
• $1 \le A_i \le N-i(1 \le i \le N-1)$
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

$N$
$A_1$ $A_2$ $\dots$ $A_{N-1}$

Output

Print the answer.

Sample Input 1

3
1 1

Sample Output 1

4

The sought expected value is $4$, so $4$ should be printed.

Here is one possible scenario until reaching Square $N$:

• Roll $1$ on Square $1$, and go to Square $2$.
• Roll $0$ on Square $2$, and stay there.
• Roll $1$ on Square $2$, and go to Square $3$.

This scenario occurs with probability $\frac{1}{8}$.

Sample Input 2

5
3 1 2 1

发现正着定义状态很难定义。定义 \(dp_i\) 表示从第 \(i\) 个点到达第 \(n\) 个点的期望步数。
那么可以列出 $$dp_i=\frac{1}{a_i+1} dp_i+\frac{1}{a_i+1} dp_{i+1}\cdots+\frac{1}{a_i+1} dp_{i+a_i}+1$$

对dp数组维护后缀和即可。

#include<cstdio>
const int N=2e5+5,P=998244353;
int dp[N],a[N],n,add;
long long s[N],ret;
int pow(int x,int y)
{
	if(!y)
		return 1;
	int t=pow(x,y>>1);
	return y&1? 1LL*t*t%P*x%P:1LL*t*t%P;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
		scanf("%d",a+i);
	for(int i=n-1;i>=1;i--)
	{
//		scanf("%d",a+i);
		dp[i]=1LL*(s[i+1]-s[i+a[i]+1]+P)%P*pow(a[i],P-2)%P+1LL*(a[i]+1)*pow(a[i],P-2)%P ;
		dp[i]%=P;
//		printf("%",dp[i]);
		s[i]=s[i+1]+dp[i];
		s[i]%=P;
	}
	printf("%d",dp[1]);
}