hdu 4869 Turn the pokers

Turn the pokers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1213    Accepted Submission(s): 449

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

3 4

3 2 3

3 3

3 2 3

 

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：把m个0，经过n次翻转，统计得到的情况总数。
1.对于第x个0来说，如果翻转偶数次，就是等于没有翻转。
首先，1出现的次数是连续的。
如果1的个数满足[ L , R ]的范围。 那么在L --》 R的范围也是合理的。

2.最后出现1的结果的个数 一定和原来翻转的个数和奇偶性相同。由1可以证明。
比如翻转 3 ，2 那么必然只会有奇数个1存在。

3.求出满足的最小1的个数，和最大1的个数，然后sum(  C(m,i)  i &1 == sum&1 ,sum =所以数的和  );

C(n,m)%p，由于p是素数用Lucas定理来求.
**/
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef __int64 LL;

const int maxn = 100000+1;
const LL p = 1000000009 ;
LL fac[maxn];
void init()
{
    fac[0] = 1;
    for(int i=1;i<maxn;i++)
        fac[i]=(fac[i-1]*i)%p;
}
LL pow_mod(LL a,LL b)
{
    LL ans =1;
    while(b)
    {
        if(b&1) ans=(ans*a)%p;
        b=b>>1;
        a=(a*a)%p;
    }
    return ans;
}
LL C(LL n,LL m)
{
    if(m==0 || n==m) return 1;
    LL sum1 = fac[n];
    LL sum2 =(fac[m]*fac[n-m])%p;
    sum1 = (sum1*pow_mod(sum2,p-2))%p;
    return sum1;
}
int main()
{
    init();
    LL n,m;
    LL Min,Max,k,ans,pp,q;
    while(scanf("%I64d%I64d",&n,&m)>0)
    {
        Min = Max = 0;
        for(k=1;k<=n;k++)
        {
            scanf("%I64d",&ans);
            //最少的1
            if(Min>=ans) pp = Min - ans;/**尽可能的让1变成0**/
            else if(Max>=ans) pp =  (  (Max&1) ==(ans&1) )? 0:1;
            /** 区间[ Min ,Max ] 包含ans，判断奇偶性能否取到就行 **/
            else pp = ans -Max;
            /**比Max都大，只能把所以1变成0，部分0变成1了。**/

            //最多的1
            if(Max+ans<=m) q = Max+ans;/**尽可能的让0变成1**/
            else if(Min+ans<=m) q = ( ((Min+ans)&1)==(m&1) ) ? m:m-1;
            /** Min+ans<=m<=Max+ans  所以，m可能取到，但是要判断奇偶性 **/
            /**这里用Max+ans   或者 Min+ans都一样的**/
            else q = 2*m -(Min+ans);
            /**  Min+ans>m  ，那么只能2*m-(Min+ans) **/

            Min =pp;
            Max =q;
        }
        LL sum = 0;
        for(LL i=Min;i<=Max;i=i+2)
        {
            sum = (sum+C(m,i))%p;
        }
        printf("%I64d
",sum);
    }
    return 0;
}