【2020年杭电暑假第五场】6825 Set1

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=6825

题意

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

给出一个$n$，有$[1,n]$中$n$个数。
每次有如下操作：先删去最小的元素，再随机删掉一个元素。
求每个数留下来的概率期望。

思路

方法一：组合数学+数学推导

考虑元素$i$留下来的方案数，那么他前面有$i-1$个数，后面有$n-i$个数。
我们易得当$n-i\ge i-1$时，i才会被留下来，即$i\le \frac{n}{2}$时都是$0$，往后才有值。所以我们可以直接输出前$\frac{n}{2}$个$0$。

然后在处理后面的数字。
我们设每次删除最小元素为操作一，随机删除元素为操作二。
即$i$后面的$n-i$个数一定是操作二删除的，所以第一步，我们让后面$n-i$个数与前面$i-1$中的$n-i$个一一对应删除就行。（在$i-1$中选$n-i$个数出来在给他排序）
设第$i$个数被留下来的方案数为$cnt[i]$，则
$$cnt[i]=C_{i-1}^{n-i}\ast A_{n-i}^{n-i}=\frac{(i-1)!}{(2i-n-1)!}$$

然后在考虑剩下的$(i-1)-(n-i)=(2i-n-1)$个数，根据前面，在其中选择两两删除。
$(2i-n-1)$两两选择有$$C_{2i-n-1}^2\ast C_{2i-n-3}^2...C_4^2\ast C_2^2$$
$$\frac{(2i-n-1)!}{2!\ast (2i-n-3)!}\ast \frac{(2i-n-3)!}{2!\ast (2i-n-5)!}\ast ...\ast \frac{2!}{2!\ast 0!}$$
$$\frac{(2i-n-1)!}{(2!{)}^{\frac{2i-n-1}{2}}}$$

到这里还没有结束，然后我想了好久，是我太捞了。
比如$[1,6]$，我们选择的时候会重复选择多次，比如
$(1,2)(3,4)(5,6)$
$(1,2)(5,6)(3,4)$
$(3,4)(1,2)(5,6)$
$(3,4)(5,6)(1,2)$
$(5,6)(1,2)(3,4)$
$(5,6)(3,4)(1,2)$
这些都是上面包括的情况，因为选的先后次序不同，所以会出现重复的情况，一共出现相同的组合$(\frac{n}{2})!$，所以我们要在上面的基础上再除以$(\frac{2i-n-1}{2})!$
那么剩下数$(2i-n-1)$中两两选择的方案数为
$$\frac{(2i-n-1)!}{(2!{)}^{\frac{2i-n-1}{2}}\ast (\frac{2i-n-1}{2})!}$$
最后整合上述所有情况，第$i$个数被留下的总方案数为
$$cnt[i]=\frac{(i-1)!}{(2i-n-1)!}\ast \frac{(2i-n-1)!}{(2!{)}^{\frac{2i-n-1}{2}}\ast (\frac{2i-n-1}{2})!}$$
化简得
$$cnt[i]=\frac{(i-1)!}{(2!{)}^{\frac{2i-n-1}{2}}\ast (\frac{2i-n-1}{2})!}$$

总方案数为$sum={\sum }_{i=1}^{n}cnt[i]$
第$i$个数被留下的概率期望为$p[[i]=\frac{cnt[i]}{sum}$
时间复杂度为$O(n)$

事先预处理所有的阶乘和阶乘逆元。

方法二：dp

参考：https://www.cnblogs.com/luyouqi233/p/13434815.html
dp方法我也不太会，哭了。

Code(3026MS)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pdd;

#define INF 0x7f7f7f
#define mem(a, b) memset(a , b , sizeof(a))
#define FOR(i, x, n) for(int i = x;i <= n; i++)

const ll mod = 998244353;
// const int maxn = 1e5 + 10;
// const double eps = 1e-6;

const int N = 5e6 + 5;

ll F[N];
ll invn[N];
ll invF[N];
ll cnt[N];

ll quick_pow(ll a, ll b)
{
    
    
    ll ans = 1;
    while(b)
    {
    
    
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans % mod;
}

void Init()
{
    
    
    F[0] = F[1] = invn[0] = invn[1] = invF[0] = invF[1] = 1;
    for(int i = 2;i < N; i++){
    
    
        F[i] = F[i - 1] * i % mod;
        invn[i] = (mod - mod / i) * invn[mod % i] % mod;
        invF[i] = invF[i - 1] * invn[i] % mod;
    }
}

void solve() {
    
    

    Init();

    int T;
    scanf("%d",&T);

    ll inv2 = 499122177;

    while(T--)
    {
    
    
        int n;
        scanf("%d",&n);
        if(n == 1) printf("1\n");
        else{
    
    
            printf("0");
            for(int i = 2;i <= n / 2; i++)
                printf(" 0");

            ll ans = 0;

            for(int i = n / 2 + 1;i <= n; i++) {
    
    
                cnt[i] = F[i - 1] * invF[(2 * i - n - 1) / 2] % mod * quick_pow(inv2, (2 * i - n - 1) / 2) % mod;
                ans = (ans + cnt[i]) % mod;
            }

            ll sum = quick_pow(ans, mod - 2);

            for(int i = n / 2 + 1;i <= n; i++) {
    
    
                printf("% lld",sum * cnt[i] % mod);
            }

            printf("\n");

        }
    }
}


signed main() {
    
    
    ios_base::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr);
#ifdef FZT_ACM_LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
    
    
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
    solve();
#endif
    return 0;
}