The sequence of integers a1,a2,…,aka1,a2,…,ak is called a good array if a1=k−1a1=k−1 and a1>0a1>0. For example, the sequences [3,−1,44,0],[1,−99][3,−1,44,0],[1,−99] are good arrays, and the sequences [3,7,8],[2,5,4,1],[0][3,7,8],[2,5,4,1],[0] — are not.

A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences [2,−3,0,1,4][2,−3,0,1,4], [1,2,3,−3,−9,4][1,2,3,−3,−9,4] are good, and the sequences [2,−3,0,1][2,−3,0,1], [1,2,3,−3−9,4,1][1,2,3,−3−9,4,1] — are not.

For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353.

The first line contains the number n (1≤n≤103)n (1≤n≤103) — the length of the initial sequence. The following line contains nn integers a1,a2,…,an (−109≤ai≤109)a1,a2,…,an (−109≤ai≤109) — the sequence itself.

In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353.

In the first test case, two good subsequences — [a1,a2,a3][a1,a2,a3] and [a2,a3][a2,a3].

In the second test case, seven good subsequences — [a1,a2,a3,a4],[a1,a2],[a1,a3],[a1,a4],[a2,a3],[a2,a4][a1,a2,a3,a4],[a1,a2],[a1,a3],[a1,a4],[a2,a3],[a2,a4] and [a3,a4][a3,a4]

今天是我做出来一道动态规划题大喜的日子，这个题开了一个一维数组f[i]，表示从1开始到i结束的序列里有几个符合条件的序列。f[i]分两部分，一部分是f[i-1],另一部分是包含a[i]的序列。包含a[i]的序列又分两部分，一个good array，或者一个good array和前面的序列组合成的序列。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;long long n;
long long tr[1005][1005],a[1005];
long long f[1010];const long long mod=998244353;int main()
{scanf("%lld",&n);tr[0][0]=1; tr[1][0]=1; tr[1][1]=1;for (int i=2; i<=n; i++){tr[i][0]=1;for (int j=1; j<n; j++){tr[i][j]=(tr[i-1][j-1]+tr[i-1][j])%mod;}tr[i][i]=1;}for (int i=1; i<=n; i++){scanf("%lld",&a[i]);}memset(f,0,sizeof(f));for (int i=2; i<=n; i++){f[i]=(f[i]+f[i-1])%mod;for (int j=i-1; j>=1; j--){int st=a[j]-1;if (st<0 || st>i-j-1) continue;f[i]=(f[i]+tr[i-j-1][st])%mod;f[i]=(f[i]+tr[i-j-1][st]*f[j-1])%mod;}}cout<<f[n]<<endl;return 0;
}