妙手代表的状态即为 $\{z,y+z,x+y+z\}$，不向包含这个状态的 $S$ 转移即可，复杂度为 $\mathcal O(nV2^{X+Y+Z})$。
```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int p = 998244353;
int n, m, x, y, z, e, ans = 1, f[41][1 << 17];
int main() {
cin >> n >> x >> y >> z, m = (1 << (x + y + z)) - 1, e = (1 << (z - 1)) | (1 << (y + z - 1)) | (1 << (x + y + z - 1));
f[0][0] = 1;
for (int i = 1; i <= n; i++, ans = ans * 10ll % p) {
for (int j = 0; j <= m; j++) {
for (int k = 1; k <= 10; k++) {
int s = ((j << k) | (1 << (k - 1))) & m;
if ((s | e) != s) f[i][s] = (f[i][s] + f[i - 1][j]) % p;
}
}
}
for (int i = 0; i <= m; i++) ans = (ans + p - f[n][i]) % p;
cout << ans << '\n';
}
```