E - Sugoroku 3(期望dp)

E - Sugoroku 3(期望dp)

$d{p}_i=\frac{{\sum }_{j=i}^{i+a_i}d{p}_j}{a_i+1}+1$

右边的$d{p}_i$ 移项得到：

$\frac{a_i}{a_i+1}d{p}_i=\frac{{\sum }_{j=i+1}^{i+a_i}d{p}_j}{a_i+1}+1$

$d{p}_i=\frac{{\sum }_{j=i+1}^{i+a_i}d{p}_j}{a_i}+\frac{a_i+1}{a_i}=\frac{({\sum }_{j=i+1}^{i+a_i}d{p}_j)+a_i+1}{a_i}$

取模用逆元递推即可。

时间复杂度：$O(n)$

// Problem: E - Sugoroku 3
// Contest: AtCoder - LINE Verda Programming Contest（AtCoder Beginner Contest 263）
// URL: https://atcoder.jp/contests/abc263/tasks/abc263_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// Date: 2022-08-09 12:44:54
// --------by Herio--------

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=2e5+5,M=2e4+5,inf=0x3f3f3f3f,mod=998244353;
const int hashmod[4] = {
    402653189,805306457,1610612741,998244353};
#define mst(a,b) memset(a,b,sizeof a)
#define db double
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n){
    
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\n",a[n]); 
}
template <typename T>		//x=max(x,y) x=min(x,y)
void cmx(T &x,T y){
    
	if(x<y) x=y;
}
template <typename T>
void cmn(T &x,T y){
    
	if(x>y) x=y;
}
int n;
ll dp[N];
ll suf[N];
ll a[N];
ll inv[N];
ll p=mod;
void init(int n){
    
	inv[1]=1;
	for(int i=2;i<=n;i++) 
	inv[i]=(p-p/i)*inv[p%i]%p;
}
int main(){
    
	ll q = 1;
	cin>>n;
	rep(i,1,n-1) cin>>a[i];
	init(n);
	dp[n] = 0;
	for(int i=n-1;i;i--){
    
		dp[i] = (suf[i+1]-suf[i+a[i]+1]+a[i]+1)*inv[a[i]]%mod;
		suf[i] = (suf[i+1] + dp[i])%mod;
	}
	printf("%lld",(dp[1]%mod+mod)%mod);
	return 0;
}