传送门
思路：
- 题意：对于一个[1，n]的集合，每次删除最小元素后再随机删除一个元素知道集合只剩一个元素。输出每个元素的遗留概率。
- 官方题解：
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{
1, 0}, {
-1, 0}, {
0, 1}, {
0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 998244353;
const int N = 5e6 + 5;int t, n, f[N];int qmi(int a, int k)
{
int res = 1;while(k){
if(k & 1) res = (ll)res * a % mod;k >>= 1;a = (ll)a * a % mod;}return res;
}void Inint(){
f[1] = 1;for(int i = 2; i < N; i ++) f[i] = f[i-1]*i%mod;
}signed main(){
IOS;Inint();cin >> t;while(t --){
cin >> n;if(n == 1){
cout << 1 << endl;continue;}for(int i = 0; i < n/2; i ++) cout << 0 << " ";int m = f[n/2], d = qmi(2, n/2) * f[n/2]%mod;d = qmi(d, mod-2);cout << m*d%mod << " ";for(int i = 1; i < n/2; i ++){
m = m*(n/2+i)%mod;m = m*qmi(i*2, mod-2)%mod;cout << m*d%mod << " ";}cout << m*d%mod << endl;}return 0;
}