目录
B. Queue Sorting
应该还会再补几题
B. Queue Sorting
题解：
Dilworth定理:
【偏序关系与偏序集、Hasse图、极大元、极小元、全序关系、最大元、良序集/三小时讲不完离散数学之集合论/考研复试/期末复习考前冲刺/近世代数/抽象代数】https://www.bilibili.com/video/BV1FK4y1i7FP?vd_source=8dcdcdf1464ff87d2684660b142a2bfe
偏序集-Dilworth定理_偏序集的例子-CSDN博客
#define int long long//__int128 2^127-1(GCC)
#define PII pair<int,int>
const int inf = 0x3f3f3f3f3f3f3f3f, N = 1e5 + 5, mod = 998244353;
class modint {
public:int x;modint(int o = 0) { o %= mod; x = o >= 0 ? o : mod + o; }modint& operator = (int o) { o %= mod; return x = o >= 0 ? o : mod + o, *this; }modint& operator +=(modint o) { return x = x + o.x >= mod ? x + o.x - mod : x + o.x, *this; }modint& operator -=(modint o) { return x = x - o.x < 0 ? x - o.x + mod : x - o.x, * this; }modint& operator *=(modint o) { return x = x * o.x % mod, *this; }modint& operator ^=(int b) {modint a = *this, c = 1;for (; b; b >>= 1, a *= a)if (b & 1)c *= a;return x = c.x, *this;}modint& operator /=(modint o) { return *this *= o ^= mod - 2; }friend modint operator +(modint a, modint b) { return a += b; }friend modint operator -(modint a, modint b) { return a -= b; }friend modint operator *(modint a, modint b) { return a *= b; }friend modint operator /(modint a, modint b) { return a /= b; }friend modint operator ^(modint a, int b) { return a ^= b; }friend bool operator ==(modint a, int b) { return a.x == b; }friend bool operator !=(modint a, int b) { return a.x != b; }bool operator ! () { return !x; }modint operator - () { return x ? mod - x : 0; }bool operator <(const modint& b)const { return x < b.x; }bool operator >(const modint& b)const { return x > b.x; }
};
inline modint qpow(modint x, int y) { return x ^ y; }
int fpow(int x, int r)
{int result = 1;while (r){if (r & 1)result = result * x % mod;r >>= 1;x = x * x % mod;}return result;
}namespace binom {int fac[N], ifac[N];int __ = []{fac[0] = 1;for (int i = 1; i <= N - 5; i++)fac[i] = fac[i - 1] * i % mod;ifac[N - 5] = fpow(fac[N - 5], mod - 2);for (int i = N - 5; i; i--)ifac[i - 1] = ifac[i] * i % mod;return 0;}();inline int C(int n, int m){if (n < m || m < 0)return 0;return fac[n] * ifac[m] % mod * ifac[n - m] % mod;}inline int A(int n, int m){if (n < m || m < 0)return 0;return fac[n] * ifac[n - m] % mod;}
}
using namespace binom;
void add(int& x, const int& y) {(x += y) %= mod;
}
signed main()
{ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);int n;cin >> n;vector<int>a(n + 1), pre(n + 1);for (int i = 1; i <= n; i++) {cin >> a[i];pre[i] = pre[i - 1] + a[i];}vector<vector<modint>>dp(n + 1, vector<modint>(pre[n] + 2));dp[1][pre[1] + 1] = 1;for (int i = 1; i < n; i++) {for (int j = 1; j <= pre[i] + 1; j++) {for (int x = 0; x < a[i + 1]; x++) {for (int k = 1; k < j; k++) {dp[i + 1][x + k + 1] += dp[i][j] * C(j - k + a[i + 1] - x - 2, a[i + 1] - x - 1);}}dp[i + 1][a[i + 1] + j] += dp[i][j];}}modint ans = 0;for (int i = 1; i <= pre[n]+1; i++) {ans += dp[n][i];}cout << ans.x << "\n";}
