<模板>Hdu4869 Turn the pokers 组合数求余费马小定理

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#组合数取模

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Turn the pokers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1115 Accepted Submission(s): 418

Problem Description

During summer vacation,Alice stay at home for a long time, with nothing to do. She went out and bought m pokers, tending to play poker. But she hated the traditional gameplay. She wants to change. She puts these pokers face down, she decided to flip poker n times, and each time she can flip Xi pokers. She wanted to know how many the results does she get. Can you help her solve this problem?

Input

The input consists of multiple test cases.
Each test case begins with a line containing two non-negative integers n and m(0<n,m<=100000).
The next line contains n integers Xi(0<=Xi<=m).

Output

Output the required answer modulo 1000000009 for each test case, one per line.

Sample Input

Sample Output

  
   8
3


   
    
     Hint
    For the second example:
0 express face down,1 express face up
Initial state 000
The first result:000->111->001->110
The second result:000->111->100->011
The third result:000->111->010->101
So, there are three kinds of results(110,011,101)

题意：有M张牌，N次操作，每次翻转xi张牌,问最后有多少种情况。

题解：不论前面怎么翻转，我们只需要知道最后牌的状态是哪几种，然后求组合数就可以了，比如有6张牌，两张向上，或四张向上。这里有个很神奇的地方，比赛的时候没有想到，最后的结果只需要知道最少是多少张朝上，最多有多少张朝上，他们的状态都是同奇或者同偶。比如最少1张向上，那么他们其他状态一定是1+2i张向上的，所有我们只需要左右区间即可。不信你自己模拟翻一翻牌就知道了。然后就是我们得仔细考虑怎么求最小最大了。这个多YY多推敲，万一不行就实践试试总会搞出来的。so我们知道最后牌的几种状态k=[L,R]（位置可以不同），所以接下来就需要求这些状态的组合数了C（M,k）。这里数据好像很大的样子。我们可以考虑用这个不记得是费马小定理还是什么东西。(a/b)%M=a*(b^(M-2))%M。我们可以把组合数C(M,i)==C(M,i-1)*(M-i+1)/i 这个关系.套用左边那个公式。递推求得所有C(M,X)的值，然后找到区间[L,R]的组合数相加，即是最后的解。

增加一个链接:http://blog.youkuaiyun.com/modiz/article/details/38257837

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
using namespace std;
#define LL __int64
//#define DEBUG
const LL M=1000000009;
LL c[100110];
LL ksm(LL a,LL b)
{
	LL ans=1;
	while (b)
	{
		if (b&1)
			ans=ans* a % M;
		b>>=1;
		a=a*a%M;
	}
	return ans;
}
int main()
{
	LL n,m;
//	freopen("1009.in","r",stdin);
	while (~scanf("%I64d%I64d",&n,&m))
	{
		LL l=0,r=0;
		for (int i=1;i<=n;i++)
		{
			LL ll,rr,x;
			scanf("%I64d",&x);
			if (l>=x) ll=l-x;
			else if (r>=x) ll=(l+x)&1;
			else ll=x-r;
			if (r+x<=m) rr=r+x;
			else if (l+x<=m) rr=m-((m+l+x)&1);
			else rr=2*m-l-x;
			l=ll;
			r=rr;
		}
		c[0]=1;
	//	cout<<l<<" "<<r<<endl;
		for (int i=1;i<=m;i++)
			if (m-i<i) c[i]=c[m-i];
			else c[i]=((c[i-1]*(m-i+1))%M * ksm(i,M-2))%M;
		LL ans=0;
		for (int i=l;i<=r;i+=2)
			ans=(ans+c[i]) % M;
		cout<<ans<<endl;
	}
	return 0;
}