HDU 6825 Set1 2020杭电多校（数学推导）

原题题面

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 $i\in [1,n]$, determine the probability of $i$ being left in the $S$.
It can be shown that the answers can be represented by $\frac{P}{Q}$, where $P$ and $Q$ are coprime integers, and print the value of $P\times Q^{-1}mod998244353$.

输入格式

The first line containing the only integer $T(T\in [1,40])$ denoting the number of test cases.
For each test case:
The first line contains a integer n .
It guarantees that: $\sum n\in [1,5\times 10^6]$.

输出格式

For each test case, you should output $n$ integers, $i-th$ of them means the probability of $i$ being left in the $S$.

输入样例

1
3

输出样例

0 499122177 499122177

题面分析

给一个包含$1$~$n$的$n$元集合$S$(保证$n$是奇数)，每次先删除最小的数字，再随机删除剩下的一个数字,经过若干轮删除后会留下一个数$i$，求每个$i$留下来的概率。
易知[1,$\lfloor \frac{n}{2}\rfloor$]留下来概率是0，因为在$\frac{n}{2}$轮的删除中它们必定被删除。
故接下来只要考虑[$\lfloor \frac{n}{2}\rfloor$+1,n]。
稍加打表可以发现
第一项是$\frac{(\lfloor \frac{n}{2}\rfloor )!}{2^{\lfloor \frac{n}{2}\rfloor }\ast \lfloor \frac{n}{2}\rfloor !}$
第二项是$\frac{(\lfloor \frac{n}{2}\rfloor +1)!}{2\ast 2^{\lfloor \frac{n}{2}\rfloor }\ast \lfloor \frac{n}{2}\rfloor !}$
第三项是$\frac{(\lfloor \frac{n}{2}\rfloor +2)!}{2\ast 4\ast 2^{\lfloor \frac{n}{2}\rfloor }\ast \lfloor \frac{n}{2}\rfloor !}$
故第$i$项为$\frac{(\lfloor \frac{n}{2}\rfloor +(i-1))!}{2^{i-1}\ast (i-1)!\ast 2^{\lfloor \frac{n}{2}\rfloor }\ast \lfloor \frac{n}{2}\rfloor !}$
化简得$$\frac{(\lfloor \frac{n}{2}\rfloor +(i-1))!}{2^{i-1}\ast (i-1)!\ast 2^{\lfloor \frac{n}{2}\rfloor }\ast \lfloor \frac{n}{2}\rfloor !}(n\ge 2)$$
证明不会，见出题人题解

AC代码（2074ms）

#include 
using namespace std;
const long long mod=998244353;
const long long maxn=5e6;
long long factor[maxn+10];
long long invn[maxn+10];
long long invFactor[maxn+10];
long long invpow2[maxn+10];
void init ()
{
    factor[0]=invFactor[0]=invn[0]=factor[1]=invFactor[1]=invn[1]=1;
    for (int i=2;i<=maxn;i++)
    {
        factor[i]=(long long)factor[i-1]*i%mod;
        invn[i]=(long long) (mod-mod/i)*invn[mod%i]%mod;
        invFactor[i]=(long long)invFactor[i-1]*invn[i]%mod;
    }
    invpow2[0]=1;
    for (int i=1;i<=maxn;i++)
    {
        invpow2[i]=(long long) invpow2[i-1]*invn[2]%mod;
    }
}
void solve1 ()
{
    init ();
    int t;
    scanf ("%d",&t);
    while (t--)
    {
        long long n;
        scanf ("%lld",&n);
        if(n==1)
        {
            printf ("1\n");
            continue;
        }
        for (int i=1;i<=n/2;i++)
        {
            printf ("0 ");
        }
        long long num=(long long) invFactor[n/2]*invpow2[n/2]%mod;
        for (long long i=1;i<=n/2+1;i++)
        {
            long long ans=factor[n/2+i-1];
            long long sumb=(long long)invpow2[i-1]*invFactor[i-1]%mod;
            ans=(long long)ans*num%mod;
            if(i!=1)
                ans=(long long)sumb*ans%mod;
            printf ("%lld",ans);
            if(i!=n/2+1)
                printf (" ");
        }
        printf ("\n");
    }
}
int main ()
{
//    ios_base::sync_with_stdio(false);
//    cin.tie(0);
//    cout.tie(0);
#ifdef ACM_LOCAL
    freopen ("in.txt","r",stdin);
    freopen ("out.txt","w",stdout);
    long long test_index_for_debug=1;
    char acm_local_for_debug;
    while (cin>>acm_local_for_debug)
    {
        cin.putback (acm_local_for_debug);
        if (test_index_for_debug>100)
        {
            throw runtime_error ("Check the stdin!!!");
        }
        auto start_clock_for_debug=clock ();
        solve1 ();
        auto end_clock_for_debug=clock ();
        cout<<"\nTest "<<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;
    }
#else
    solve1();
#endif
    return 0;
}

后记

赛前唯唯诺诺，赛后五分钟推出答案…
题面里的$[1,5\times 10^6]$一直看成$1.5\ast 10^6$，RE穿了…
DrGilbert 2020.8.4