HDOJ 4869 Turn the pokers

最后的结果中正面向上的奇偶性是一定的，计算出正面向上的范围low，up

结果即为 C(m，low）+ C(m，low+2) +.... + C(m,up) ，用逆元取模

Turn the pokers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 828    Accepted Submission(s): 302

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) 
    
    
    
    
     
   
   
   
   

 

Author

FZU

 

Source

2014 Multi-University Training Contest 1

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

const int maxn=110000;
const LL mod=1000000009LL;

LL a[maxn],n,m;

void ex_gcd(LL a,LL b,LL &d,LL &x,LL &y)
{
	if(!b)
	{
		d=a; x=1; y=0;
	}
	else
	{
		ex_gcd(b,a%b,d,y,x);
		y-=a/b*x;
	}	
}

LL lower,upper;

void cal(LL lower,LL upper)
{
    LL cur=1LL,t=0,ret=0;
    while(t<=upper)    
    {
        if(t>=lower&&t<=upper)
            if((t-lower)%2==0)
                ret=(ret+cur)%mod;    
        t++;
        cur=(cur*(m-t+1))%mod;
        LL d,x,y;
        ex_gcd(t,mod,d,x,y);
        x=(x+mod)%mod;
        cur=(cur*x)%mod;
    }
    cout<<ret<<endl;
}

int main()
{
while(cin>>n>>m)
{
    lower=upper=0LL;
    for(int i=0;i<n;i++)
        cin>>a[i];
    lower=upper=a[0];
    for(int i=1;i<n;i++)
    {
        LL last_low=lower,last_up=upper;

        ///get low bound
        if(a[i]<=last_low)
            lower-=a[i];
        else if(a[i]<=last_up)
            lower=0+((a[i]+last_up)%2==1);
        else
            lower=a[i]-last_up;

        ///get upper bound
        if(a[i]+last_up<=m)    
            upper=a[i]+last_up;
        else if(a[i]+last_low<=m)
            upper=m-((a[i]+last_low)%2==1);
        else
            upper=m-(a[i]+last_low-m);
    }
    cal(lower,upper);
}
    return 0;
}