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HDOJ 4869 Turn the pokers

2024-07-13 17:21:27   222人阅读


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;
}


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