Codeforces 1000D Yet Another Problem On a Subsequence

题目：点击打开链接
题意：定义一个数列是“好的”：第一个数字a[0]为数列长度+1。定义一个数列的子序列是“好的”：这个子序列能分割成几个“好的”数列。给出一个数列，求“好的”子序列的数目。
分析：定义dp[i]为以a[i]为开头的好序列数目，第一种情况很好求，第二种情况有点难想，应该从后往前遍历，先计算第一种情况，然后累加之后的情况。（dp[i]代表以i开始的符合要求的序列数，j是我们列举出的i之后一个字符串的开始地址，这里的C是组合数，dp[i] += C(j - i - 1, A[i] )* dp[j];）
代码：

#pragma comment(linker, "/STACK:102400000,102400000")///手动扩栈
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define mod 998244353
#define PI acos(-1.0)
const int N = 1e6+10;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};ll n,a[N],dp[N],s[N],ans,fac[N];ll qp(ll x,ll y) {ll s=1;while(y) {if(y&1) s=s*x%mod;x=x*x%mod;y>>=1;}return s;
}ll C(ll x,ll y) {if(x<0||y<0) return 0;else return fac[x]*qp(fac[y],mod-2)%mod*qp(fac[x-y],mod-2)%mod;
}int main() {///重定向语句：ifsteam cin("C:\\xxx\\xxx\\xxx.xxx");ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);#ifndef debugfreopen("C:\\Users\\Administrator\\Desktop\\data.in","r",stdin);#endiffac[0]=1;for(int i=1;i<=1005;i++) fac[i]=fac[i-1]*i%mod;cin>>n;for(int i=1;i<=n;i++) cin>>a[i];for(int i=n-1;i>=1;i--) {if(a[i]>0&&( n-i >=a[i])) {dp[i] += C(n-i,a[i]);for(int j=i+a[i]+1;j<=n;j++)dp[i] = (dp[i]+C(j-i-1,a[i])*dp[j]%mod)%mod;}}for(int i=1;i<=n;i++)ans = (ans + dp[i])%mod;cout<<ans<<endl;return 0;
}

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