题目链接：hdu 4869 Turn the pokers

题目大意：给定n和m，表示有n次翻牌的机会，m张牌，一开始所有的牌均背面朝上，每次翻牌可以选择xi张牌同时翻转。问说最后有多少种能。

解题思路：只要确定最后正面朝上的牌的个数即可。所以在读入xi的时候维护上下限即可。

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

using namespace std;
typedef long long ll;
//typedef __int64 ll;
const ll mod = 1e9+9;
const int maxn = 1e5+10;
int n, m, l, r;
ll c[maxn];

ll pow_mod (ll x, int k) {
    ll ans = 1;
    while (k) {
        if (k&1)
            ans = ans * x % mod;

        x = x * x % mod;
        k /= 2;
    }
    return ans;
}

void init () {
    int x, p, q;
    l = r = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", &x);

        if (l >= x)
            p = l - x;
        else if (r >= x)
            p = ( (l&1) == (x&1) ? 0 : 1);
        else
            p = x - r;

        if (r + x <= m)
            q = r + x;
        else if (l + x <= m)
            q = ( ((l+x)&1) == (m&1) ? m : m-1);
        else
            q = 2 * m - l - x;
        l = p;
        r = q;
    }
}

ll solve () {
    c[0] = 1;
    ll ans = 0;

    for (int i = 0; i <= r; i++) {

        if (i) {
            if (m - i < i)
                c[i] = c[m-i];
            else
                //c[i] = c[i-1] * (m-i+1) % mod * pow_mod(i, mod-2)  % mod;
                c[i] = c[i-1] * ( (ll)(m-i+1) * pow_mod(i, mod-2)  % mod) % mod;
        }

        if (i >= l && (i&1) == (l&1))
            ans = (ans + c[i]) % mod;
    }
    return ans;
}

int main () {
    while (scanf("%d%d", &n, &m) == 2) {
        init();
        printf("%lld\n", solve());
    }
    return 0;
}