LOJ6268拆分数

/*

相当于每种物品都有无限个的背包

毕竟考场上写exp是个比较危险的行为

对数据进行根号分治是个比较好的方法

对于小于等于根号的部分暴力背包转移
对于大于根号的 最多只会拿根号个 dp一下就好了

*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#include<cmath>
#define ll long long
#define M 100010
#define mmp make_pair
using namespace std;
const int mod = 998244353, g = 3;
void add(int &a, int b) {
	a += b;
	a -= a >= mod ? mod : 0;
	a += a < 0  ? mod : 0;

}

int poww(int a, int b) {
	int tmp = a, ans = 1;
	for(; b; b >>= 1, tmp = 1ll * tmp * tmp % mod) if(b & 1) ans = 1ll * ans * tmp % mod;
	return ans;
}

int read() {
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
int a[M * 4], b[M * 4], biao, f[350][M], n, pw[M * 4], iw[M * 4];

void fft(int *a, int n, int dft) {
	for(int i = 0, j = 0; i < n; i++) {
		if(i < j) swap(a[i], a[j]);
		for(int l = n >> 1; (j ^= l) < l; l >>= 1);
	}
	for(int step = 1; step < n; step <<= 1) {
		int wn = (dft == 1) ? pw[step] : iw[step];
		for(int i = 0; i < n; i += step << 1) {
			int wnk = 1;
			for(int j = i; j < i + step; j++) {
				int x = a[j], y = 1ll * wnk * a[j + step] % mod;
				a[j] = (x + y) % mod;
				a[j + step] = (x - y + mod) % mod;
				wnk = 1ll * wnk * wn % mod;
			}
		}
	}
	if(dft == -1) {
		int inv = poww(n, mod - 2);
		for(int i = 0; i < n; i++) a[i] = 1ll * a[i] * inv % mod;
	}
}

int main() {
	n = read();
	a[0] = 1;
	biao = sqrt(n) + 1;
	for(int i = 1; i < biao; i++) {
		for(int j = i; j <= n; j++) {
			add(a[j], a[j - i]);
		}
	}
	f[1][biao] = 1;
	for(int i = 1; i <= biao; i++) {
		for(int j = 0; j <= n; j++) {
			if(f[i][j]) {
				if(j + i <= n) add(f[i][j + i], f[i][j]);
				if(j + biao <= n) add(f[i + 1][j + biao], f[i][j]);
			}
			add(b[j], f[i][j]);
		}
	}
	add(b[0], 1);
	for(int i = 1; i < 4 * M; i <<= 1) pw[i] = poww(g, (mod - 1) / 2 / i), iw[i] = poww(pw[i], mod - 2);
	n++;
	int up = (n << 1) - 1;
	while(up - (up & -up)) up += (up & -up);
	fft(a, up, 1), fft(b, up, 1);
	for(int i = 0; i < up; i++) a[i] = 1ll * a[i] * b[i] % mod;
	fft(a, up, -1);
	for(int i = 1; i < n; i++) cout << a[i] << '\n';
	return 0;
}