一个含有 nn 个互异正整数的序列 cc

若一棵带点权的有根二叉树满足其所有顶点的权值都在集合 c[1],c[2],...,c[n]c[1],c[2],...,c[n] 中, 就是好的二叉树，一棵带点权的树的权值,是其所有顶点权值的总和。

给出一个整数 mm,对于任意的 s(1<=s<=m)s(1<=s<=m) 计算出权值为 ss 的好二叉树的个数。

两个二叉树不同当他们点数不同或者结构不同，答案对 998244353998244353 取模

1≤n≤105,1≤m≤105,Ci≤1051 \le n \le 10^5 , 1 \le m \le 10^5 , C_i \le 10^5

Solution

19-8-9-1

模板题，第一次写多项式求逆和开根

  • 求逆：B(x)=2B′(x)−A(x)B′(x)2B(x) = 2B'(x) - A(x)B'(x)^2
  • 开根：B(x)=A(x)+B′(x)22B′(x)B(x) = \frac{A(x) + B'(x)^2}{2B'(x)}，推导时构造完全平方数

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back
#define DEBUG(x) cout << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 1e6 + 100;
const int Mod = 998244353;
const int g = 3;

int N, M;
int C[Maxn];

inline void Add (int &a, int b) { if ((a += b) >= Mod) a -= Mod; }

inline int Pow (int a, int b)
{
	int ans = 1;
	for (int i = b; i; i >>= 1, a = (LL) a * a % Mod) if (i & 1) ans = (LL) ans * a % Mod;
	return ans;
}

namespace Poly
{
	int n, rev[Maxn]; 
	int _W[2][Maxn];

	inline void init (int maxn = 1e6)
	{
		for (int mid = 1; mid <= maxn; mid <<= 1)
		{
			_W[0][mid] = Pow (g, (Mod - 1) / (mid << 1));
			_W[1][mid] = Pow (_W[0][mid], Mod - 2);
		}
	}

	inline void dft (int *A, int fl)
	{
		for (int i = 0; i < n; ++i) if (rev[i] < i) swap (A[i], A[rev[i]]);
		for (int mid = 1; mid < n; mid <<= 1)
		{
			int W = _W[0][mid];
			if (fl == -1) W = _W[1][mid];
			for (int i = 0; i < n; i += (mid << 1))
			{
				int Wn = 1;
				for (int j = i; j < i + mid; ++j, Wn = (LL) Wn * W % Mod)
				{
					int x = A[j], y = (LL) Wn * A[j + mid] % Mod;
					A[j] = (x + y) % Mod;
					A[j + mid] = (x - y + Mod) % Mod;
				}
			}
		}
		if (fl == -1) {int inv = Pow (n, Mod - 2); for (int i = 0; i < n; ++i) A[i] = (LL) A[i] * inv % Mod; }
	}
 
	inline void mul (int *A, int N, int *B, int M, int *C)
	{
		static int F[Maxn], G[Maxn];
		n = 1; while (n <= N + M) n <<= 1;
		for (int i = 0; i < n; ++i) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) ? (n >> 1) : 0);
		for (int i = 0; i < n; ++i) F[i] = (i <= N) ? A[i] : 0;
		for (int i = 0; i < n; ++i) G[i] = (i <= M) ? B[i] : 0;

		dft (F, 1), dft (G, 1);
		for (int i = 0; i < n; ++i) F[i] = (LL) F[i] * G[i] % Mod;
		dft (F, -1);
		for (int i = 0; i < n; ++i) C[i] = (i <= N + M) ? C[i] : 0;
	}

	inline void get_inv (int *A, int *B, int N)
	{
		if (N == 1) { B[0] = Pow (A[0], Mod - 2); return ; }
		int len = (N + 1) >> 1;
		get_inv (A, B, len);

		static int F[Maxn], G[Maxn];
		n = 1; while (n <= 2 * N) n <<= 1;
		for (int i = 0; i < n; ++i) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) ? (n >> 1) : 0);
		for (int i = 0; i < n; ++i)
		{
			F[i] = (i < N) ? A[i] : 0;
			G[i] = (i < len) ? B[i] : 0;
		}

		dft (F, 1), dft (G, 1);
		for (int i = 0; i < n; ++i) F[i] = (2ll * G[i] % Mod - (LL) F[i] * G[i] % Mod * G[i] % Mod + Mod) % Mod;
		dft (F, -1);

		for (int i = 0; i < n; ++i) B[i] = (i < N) ? F[i] : 0;
	}

	inline void get_sqrt (int *A, int *B, int N)
	{
		if (N == 1) { B[0] = 1; return ; }
		int len = (N + 1) >> 1;
		get_sqrt (A, B, len);

		n = 1; while (n <= 2 * N) n <<= 1;
		for (int i = 0; i < n; ++i) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) ? (n >> 1) : 0);

		static int F[Maxn], G[Maxn], S[Maxn], H[Maxn];
		for (int i = 0; i < n; ++i)
		{
			F[i] = (i < N) ? A[i] : 0;
			G[i] = (i < len) ? B[i] : 0;
			S[i] = (i < len) ? 2ll * B[i] % Mod	: 0;
			H[i] = 0;
		}

		dft (F, 1), dft (G, 1); 
		get_inv (S, H, N), dft (H, 1);
		for (int i = 0; i < n; ++i) F[i] = (LL) (F[i] + (LL) G[i] * G[i] % Mod) % Mod * H[i] % Mod;
		dft (F, -1);

		for (int i = 0; i < n; ++i) B[i] = (i < N) ? F[i] : 0;
	}
}

int A[Maxn], B[Maxn];

inline void Solve ()
{
	C[0] = 1;
	for (int i = 1; i <= M; ++i) C[i] = (- 4ll * C[i] % Mod + Mod) % Mod;

	Poly :: init ();
	Poly :: get_sqrt (C, B, M + 1);
	(++B[0]) %= Mod;
	Poly :: get_inv (B, A, M + 1);
	for (int i = 0; i <= M; ++i) A[i] = 2ll * A[i] % Mod;
	for (int i = 1; i <= M; ++i) printf("%d\n", A[i]);
}

inline void Input ()
{
	N = read<int>(), M = read<int>();
	for (int i = 1; i <= N; ++i) C[read<int>()] = 1;
}

int main()
{

#ifndef ONLINE_JUDGE
	freopen("E.in", "r", stdin);
	freopen("E.out", "w", stdout);
#endif

	Input ();
	Solve ();

	return 0;
}

//一定注意：所有的F和G都要开static !!!!!!