hdu 4869

Turn the pokers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 676    Accepted Submission(s): 231

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?

 

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

using namespace std;

typedef long long ll;

const int MAX = 1e5 + 7;
const int MOD = 1e9 + 9;
int N, M;
int x[MAX];
ll c[MAX];

ll mod_pow(ll x, ll n, ll mod) {
        ll res = 1;
        while (n > 0) {
                if (n & 1) res = res * x % mod;
                x = x * x % mod;
                n >>= 1;
        }

        return res;
}

void solve() {
        int Min = 0, Max = 0;
        for (int i = 1; i <= N; ++i) {
                int t1, t2;
                if (Min >= x[i]) {
                        t1 = Min - x[i];
                } else if (x[i] <= Max) {
                        t1 = !((Min & 1) == (x[i] & 1));
                } else {
                        t1 = x[i] - Max;
                }

                if (Max + x[i] <= M) {
                        t2 = Max + x[i];
                } else if (Min + x[i] <= M) {
                        t2 = (((Min + x[i]) & 1) == (M & 1)) ? M : M - 1;
                } else {
                        t2 = 2 * M - (Min + x[i]);
                }

                Min = t1; Max = t2;
        }

        //printf("min = %d max = %d\n", Min, Max);
        ll ans = 0;
        c[0] = 1;
        if (Min == 0) ans += 1;
        for (int i = 1; i <= Max; ++i) {
                if (M - i < i) c[i] = c[M - i];
                else
                c[i] = c[i - 1] * (M - i + 1) % MOD * mod_pow(i, MOD - 2,MOD) % MOD;
                if (i >= Min && (i & 1) == (Min & 1)) ans = (ans + c[i]) % MOD;

        }

        printf("%I64d\n", ans);
}
int main()
{
    while (scanf("%d%d", &N, &M) != EOF) {
            for (int i = 1; i <= N; ++i) {
                    scanf("%d", &x[i]);
            }

            solve();
    }
    //cout << "Hello world!" << endl;
    return 0;
}

 

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

 

首先1个个数的奇偶性和翻牌次数的奇偶性相同，然后根据分类讨论找出出现的最少的1 和最多的1，最终的结果肯定是连续的奇数或者偶数

然后求出sum(c(M,i)) (   Min < i <= Max)连续的奇数或偶数

根据费马小定理可以知道若p为质数，gcd(a, p） = 1,则a ^ (p - 2)  = 1 / p （mod p） 根据这个求出组合数就可以了

 

 

 

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