组合数学--Set1 hdu-6825

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 $i\in [1,n]$, 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 $P\times Q^{-1}$ $mod$ $998244353$.

思路

• 对于每次数据 $n$,设 $num=\frac{n}{2}$， 可知 $\forall i\in [1,num]$的期望值是0
• 设$\forall i\in [num+1,n]$,每次最后剩余他的次数 $cnt[i]$
  1. 首先对于 $i$ 后面的 $n-i$ 个数，需要前面的 $i-1$ 个数来达到消去的效果，所以这里的总次数是
     $$C_{i-1}^{n-i}\ast (n-i)!$$
  2. 使后面 $n-i$ 个数消去后， $i$ 前面还剩余 $(i-1)-(n-i)=2i-n-1$ 个数，而这里两两匹配的总个数为
     $$\prod _{j=1}^{2i-n-1}j,(j\in 2k+1|k\in N)$$
  3. 综上可得$cnt[i]=C_{i-1}^{n-i}\ast (n-i)!\ast {\prod }_{j=1}^{2i-n-1}j,(j\in 2k+1|k\in N)$，即 $j$ 为奇数
• 对于每个$n$,总数 $sum={\sum }_{i=1}^{i=n}cnt[i]$,可知
  $su{m}_1=1=2^0\ast 0!$ ,$(0!=1)$
  $su{m}_3=2=2^1\ast 1!$
  $su{m}_5=8=2^2\ast 2!$
  $su{m}_7=48=2^3\ast 3!$
  $su{m}_9=384=2^4\ast 4!$
  所以可得知 $num=n/2,su{m}_n=2^{num}\ast (num)!$
• 所以总的期望值 $$p[i]=\frac{C_{i-1}^{n-i}\ast (n-i)!\ast {\prod }_{j=1}^{2i-n-1}j}{2^{num}\ast (num)!}，(j\in 2k+1|k\in N)$$
• 下面的代码中$fac[i]=i!,ifac[i]=fa{c}^{-1}[i],mfac[\frac{2i-n-1}{2}]={\prod }_{j=1}^{2i-n-1}j$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template <typename T>
inline void read(T &x)
{
    
    
    int tmp = 1;char c = getchar();x = 0;
    while (c > '9' || c < '0'){
    
    if (c == '-')tmp = -1;c = getchar();}
    while (c >= '0' && c <= '9'){
    
    x = x * 10 + c - '0';c = getchar();}
    x *= tmp;
}
const int inf=0x3f3f3f3f;
const int mod=998244353;
const int N=5e6+10;
const int M=1e7+10;
ll fp(ll x,ll y){
    
    
	ll ans=1;
	while(y){
    
    
		if(y&1) ans=ans*x%mod;
		x=x*x%mod;
		y>>=1;
	}
	return ans;
}
ll fac[N],ifac[N],mfac[N];
inline ll C(int n,int k){
    
    
	return fac[n]*ifac[n-k]%mod*ifac[k]%mod;
}
int main(){
    
    
	ios::sync_with_stdio(false);cin.tie(nullptr); cout.tie(nullptr);
	fac[0]=1;
	for(int i=1;i<=5e6;++i) fac[i]=fac[i-1]*i%mod;
	ifac[N-10]=fp(fac[N-10],mod-2);
	for(int i=5e6-1;i>=0;--i) ifac[i]=ifac[i+1]*(i+1)%mod;
	mfac[0]=mfac[1]=1;
	for(int i=1;i<=3e6;++i) mfac[i+1]=mfac[i]*(i<<1|1)%mod;
	int T;read(T);
	while(T--){
    
    
		int n;read(n);
		if(n==1){
    
    
			puts("1");
			continue;
		}
		int num=n/2;
		ll down=fp(2,num)*fac[num]%mod;
		down=fp(down,mod-2);
		for(int i=1;i<=num;++i){
    
    
			if(i==1) printf("0");
			else printf(" 0");
		}
		for(int i=num+1;i<=n;++i){
    
    
			printf(" %lld",C(i-1,n-i)*fac[n-i]%mod*mfac[(2*i-n-1)/2]%mod*down%mod);
		}
		puts("");
	}
	return 0;
}