codeforces Gym 100187D Holidays(排列组合)

最新推荐文章于 2019-08-15 11:41:23 发布

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本文介绍了一个数学问题的解决方法，该问题是计算在一个特定场景下所有可能的礼物交换组合数量。具体来说，在一个由n个不同种族组成的银河帝国中，我们需要找出所有可能的方式，即一部分种族给另一部分种族赠送礼物的不同组合。

Everyone knows that the battle of Endor is just a myth fabled by George Lucas for promotion of his movie. Actually, no battle of Endor has happened and the First Galactic Empire prospers to this day.

There are creatures of n races living in the First Galactic Empire. In order to demonstrate their freedom, equality and brotherhood the Emperor commanded to introduce the holidays. During each of these holidays creatures of one non-empty subset of races should give gifts to creatures of another non-empty subset of races, not intersecting the first one.

The Emperor's stuff is not very strong in maths so you should calculate how many such holidays can be introduced. Two holidays are considered different if they differ in the subset of races which give gifts or in the subset of races which receive gifts.

Input

The input contains the only integer n (1 ≤ n ≤ 200000) — the number of races living in the First Galactic Empire.

Output

Find the number of holidays the Emperor commanded to introduce. This number can be very large, so output the reminder of division of this number by 109 + 9.

Examples

input
2

output
2

input
3

output
12

input
5

output
180

input
200000

output
82096552

题意：给你n个人，从中取出若干个分成两个子集，不能为空，一个子集给另一个送礼物，问有多少种组合。

当n=5是：第一个子集从5个中选取一个，第二个就可以从4个中选1个(或者2个，3个，4个);

第一个子集从5个中选取两个，第二个就可以从3个中选1个(或者2个，3个);

第一个子集从5个中选取三个，第二个就可以从2个中选1个(或者2个);

第一个子集从5个中选取四个，第二个就可以从4个中选1个

所以可推出∑ni=2Cin∗(2n-i −1),

#include<cstdio> 
#include<cstring>
#include<algorithm>
#define N 200010
#define mod 1000000009
#define LL long long
using namespace std;
LL f[N],s[N];
int nn;
LL pow_quick(LL n,LL m)
{
	LL ans=1;
	while(m)
	{
		if(m&1)
		   ans=ans*n%mod;
		m>>=1;
		n=n*n%mod;
	}
	return ans;
}
LL C(int n,int m)
{
	return f[n]*pow_quick(f[m],mod-2)%mod*pow_quick(f[n-m],mod-2)%mod;   //逆元 a*a^(p-2) ≡ 1 (mod p)。
}
int main()
{
	scanf("%d",&nn);
	s[0]=1;   f[0]=1;
	for(int i=1;i<=nn;i++)
	{
		f[i]=f[i-1]*i%mod;
		s[i]=s[i-1]*2%mod;
	}
	LL sum=0;
	for(int i=1;i<nn;i++)
		sum+=C(nn,i)*(s[nn-i]-1)%mod;
	printf("%lld\n",sum%mod);
	return 0;
}