2020杭电多校第五场 Set1（组合数学，概率）
Problem Description
You are given a set S={1…n}. It guarantees that n is odd. You have to do the following operations until there is only 1 element in the set:
Firstly, delete the smallest element of S. Then randomly delete another element from S.
For each i∈[1,n], determine the probability of i being left in the S.
It can be shown that the answers can be represented by PQ, where P and Q are coprime integers, and print the value of P×Q−1 mod 998244353.
Input
The first line containing the only integer T(T∈[1,40]) denoting the number of test cases.
For each test case:
The first line contains a integer n .
It guarantees that: ∑n∈[1,5×106].
Output
For each test case, you should output n integers, i-th of them means the probability of i being left in the S.
Sample Input
1
3
Sample Output
0 499122177 499122177
Source
2020 Multi-University Training Contest 5
这场全是推公式题，恶心心
感谢arc3102大佬指出一些错误
题意：
n个数，每次选一个最小的数扔掉，再仍以选择一个数扔掉。求最后剩下的数是i的概率。
思路：
假设剩下的数字是 i i i，那么前面有 i − 1 i-1 i−1个数，后面有 n − i n-i n−i个数。
要求剩下的是 i i i，则需要满足 i − 1 ≥ n − i i-1≥n-i i−1≥n−i。则后面 n − i n-i n−i个数肯定是和前面的数一一对应被删掉。前面剩下 2 ∗ i − n − 1 2*i-n-1 2∗i−n−1个数最后两两配对被删掉。
则 c n t [ i ] = ( i − 1 n − i ) ∗ ( n − i ) ! ∗ ( 2 ∗ i − n − 1 ) ! ( 2 ! ) ( 2 ∗ i − n − 1 ) 2 ∗ ( 2 ∗ i − n − 1 ) 2 ! cnt[i]=\tbinom{i-1}{n-i}*(n-i)!*\frac{(2*i-n-1)!}{(2!)^{\frac{(2*i-n-1)}{2}}*\frac{(2*i-n-1)}{2}!} cnt[i]=(n−ii−1)∗(n−i)!∗(2!)2(2∗i−n−1)∗2(2∗i−n−1)!(2∗i−n−1)!
最后要除这个 ( 2 ∗ i − n − 1 ) 2 ! \frac{(2*i-n-1)}{2}! 2(2∗i−n−1)!，是因为选出来的 ( 2 ∗ i − 1 − n ) 2 \frac{(2*i-1-n)}{2} 2(2∗i−1−n)个二元组中，第一个数字必须递增，所以第一个数字仍以排列的情况要除去。
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