F - Variety of Digits

https://atcoder.jp/contests/abc235/editorial/3259
#include<iostream>
#include<cstring>
using namespace std;

const int N = 1e3 + 10;
const int mod = 998244353;
long long dp[5][5][(1 << 10) + 10];

string s;
int n, m;
int main () {
    cin >>  s;
	n = s.size();
   int a[n + 10];
    for (int i = 0; i < n ; i ++)
    	a[i] = s[n - 1 - i] - '0';
    
    int mask = 0;
    cin >> m;
    for (int i = 0; i < m; i ++) {
    	int t; cin >> t;
    	mask |= 1 << t;
	}
	
	for (int i = n -1; i >= 0; i --) {
		long long ndp[5][5][(1 << 10) + 10];
		memset (ndp, 0, sizeof ndp);
		for (int x = 1; x <= (i == n - 1?a[i] : 9); x ++) 
			ndp[0][i < n -1 || x < a[i]][1<<x] += 1, ndp[1][i < n - 1 || x < a[i]][1<<x] += x;
		
		for (int l = 0; l < 2; l ++) {
			for (int j = 0; j < (1 << 10); j ++) {
				for (int x = 0; x <= (l?9:a[i]); x ++)
				{
					(ndp[0][l || x < a[i]][j | 1 << x] += dp[0][l][j]) %= mod;
					(ndp[1][l || x < a[i]][j | 1 << x] += dp[1][l][j] * 10 % mod + dp[0][l][j] * x % mod) %= mod; 
				}
			}
		}
		memcpy(dp, ndp, sizeof dp);
	}
	long long ans = 0;
	for (int l = 0; l < 2; l ++)
	for (int i = 0; i < (1 << 10); i ++)
		if ((i & mask) == mask)
			(ans += dp[1][l][i]) %= mod;
	cout <<ans << endl;
    return 0;
}

数位dp dp[i][S]i表示现在是否小于n，1表示小于， 0表示等于S表示当前出现的数字的集合如果和mask一样，那么就有贡献。否则无