ARG102E：Stop. Otherwise...

传送门

Sol

对于每个 (i) ，可以把 (k) 个数字分成 ((x,i-x)) 的若干组。
那么就是求每组只能其中选择一个且可以重复的方案数。
预处理 (f[i][j]) 表示从 (j) 个组内选 (i) 个，每个组必须选的方案数。
(f[i][j]=(f[i-1][j]+f[i-1][j-1] imes 2)) 注意组有序，而选出来是无序的，所以要强制选择最后一组
那么每个组可以不选的方案数 (g[i][j]) 就是

[g[i][j]=sum_{t=0}^{j}(^{j}_{t})f[i][t] ]

拆开组合数可以 (NTT) 优化。
对于询问分两种情况

1. (i) 为奇数
   那么 (x e i-x) 直接枚举多少个选组内的，其它的可重组合即可
2. (i) 为偶数
   存在一组 (x=i-x) 枚举是否选 (x) (因为只能选一个)，然后像奇数一样做即可
   (Theta(n^2logn)) 丝毫不虚

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

namespace IO {
	const int maxn((1 << 21) + 1);

	char ibuf[maxn], *iS, *iT, c;
	int f;

	inline char Getc() {
		return iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), iS == iT ? EOF : *iS++) : *iS++;
	}

	template <class Int> inline void In(Int &x) {
		for (f = 1, c = Getc(); c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;
		for (x = 0; c >= '0' && c <= '9'; c = Getc()) x = (x << 1) + (x << 3) + (c ^ 48);
		x *= f;
	}
}

using IO :: In;

const int maxn(4005);
const int mod(998244353);

int f[maxn][maxn], c[maxn][maxn], fac[maxn], ifac[maxn];

inline void Inc(int &x, int y) {
	x += y;
	if (x >= mod) x -= mod;
}

inline int Pow(ll x, int y) {
	ll ret = 1;
	for (; y; y >>= 1, x = x * x % mod)
		if (y & 1) ret = ret * x % mod;
	return ret;
}

inline int Calc(int n, int num, int res) {
	int ret = 0;
	if (!res) ret = f[n][num];
	else {
		for (int j = 0; j <= n; ++j)
			Inc(ret, 1LL * f[n - j][num] * c[j + res - 1][j] % mod);
	}
	return ret;
}

int a[maxn << 2], b[maxn << 2], len, l, r[maxn << 2], w[2][maxn << 2];

inline void NTT(int *p, int opt) {
	for (int i = 0; i < len; ++i) if (i < r[i]) swap(p[i], p[r[i]]);
	for (int i = 1; i < len; i <<= 1)
		for (int j = 0, t = i << 1; j < len; j += t)
			for (int k = 0; k < i; ++k) {
				int nw = w[opt == -1 ? 1 : 0][len / i * k];
				int x = p[j + k], y = 1LL * p[j + k + i] * nw % mod;
				p[j + k] = (x + y) % mod, p[j + k + i] = (x - y + mod) % mod;
			}
	if (opt == -1) {
		int inv = Pow(len, mod - 2);
		for (int i = 0; i < len; ++i) p[i] = 1LL * p[i] * inv % mod;
	}
}

inline void Init() {
	f[0][0] = c[0][0] = ifac[0] = fac[0] = 1;
	for (int i = 1, j; i <= 4000; ++i)
		for (c[i][0] = j = 1; j <= i; ++j)
			c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
	for (int i = 1; i <= 2000; ++i)
		for (int j = 1; j <= 2000; ++j)
			f[i][j] = (f[i - 1][j] + 1LL * f[i - 1][j - 1] * 2 % mod) % mod;
	for (int i = 1; i <= 2000; ++i) fac[i] = 1LL * fac[i - 1] * i % mod;
	ifac[2000] = Pow(fac[2000], mod - 2);
	for (int i = 1999; i; --i) ifac[i] = 1LL * ifac[i + 1] * (i + 1) % mod;
	for (len = 1; len <= 4000; len <<= 1) ++l;
	for (int i = 0; i < len; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	for (int i = 1; i < len; i <<= 1) {
		int wn = Pow(3, (mod - 1) / (i << 1));
		for (int nw = 1, k = 0; k < i; ++k, nw = 1LL * nw * wn % mod)
			w[0][len / i * k] = nw, w[1][len / i * k] = Pow(nw, mod - 2);
	}
	for (int i = 0; i <= 2000; ++i) {
		for (int j = 0; j < len; ++j) a[j] = b[j] = 0;
		for (int j = 0; j <= 2000; ++j) a[j] = ifac[j], b[j] = 1LL * ifac[j] * f[i][j] % mod;
		NTT(a, 1), NTT(b, 1);
		for (int j = 0; j < len; ++j) a[j] = 1LL * a[j] * b[j] % mod;
		NTT(a, -1);
		for (int j = 0; j <= 2000; ++j) f[i][j] = 1LL * a[j] * fac[j] % mod;
	}
}

int n, k;

int main() {
	In(k), In(n), Init();
	int r = k << 1;
	for (int i = 2; i <= r; ++i) {
		if (i & 1) {
			int num = max(0, min(i >> 1, i - 1) - max(1, i - k) + 1);
			int res = k - (num << 1);
			printf("%d
", Calc(n, num, res));
		}
		else {
			int num = max(0, min(i >> 1, i - 1) - max(1, i - k) + 1);
			int res = k - (num << 1) + 1;
			if (num) --num;
			printf("%d
", (Calc(n, num, res) + Calc(n - 1, num, res)) % mod);
		}
	}
	return 0;
}