求长度为nn且∀i∈[1,n],ai−1<ai>ai+1 or ai−1>ai<ai+1\forall i\in [1,n], a_{i-1}<a_i>a_{i+1} ~ or ~a_{i-1}>a_i<a_{i+1}的排列数量(对PP取模)

3≤n≤4200,P≤1093\le n\le 4200, P \le 10^9

Luogu P2467

Solution

首先很容易想到设dp[i][j][0/1]dp[i][j][0/1]表示到第ii位,第ii位放jj这个数,且这个数为谷/峰的方案数

但是显然这个会算重,考虑把jj换一个定义:在前ii位中排第jj位的数,那么就很好转移了

注意开个滚动数组优化空间

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
inline int read ()
{
	int sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

const int Maxn = 4200 + 100;

int N, Mod, Dp[2][Maxn][2];

int main()
{
#ifdef hk_cnyali
	freopen("A.in", "r", stdin);
	freopen("A.out", "w", stdout);
#endif
	N = read(), Mod = read();
	Dp[1][1][0] = Dp[1][1][1] = 1;
	int now = 1;
	for (int i = 2; i <= N; ++i)
	{
		now ^= 1;
		int sum = 0;
		for (int j = 1; j <= i; ++j)
		{
			(Dp[now][j][0] = sum) %= Mod;
			(sum += Dp[now ^ 1][j][1]) %= Mod;
		}
		sum = 0;
		for (int j = i; j >= 1; --j)
		{
			(sum += Dp[now ^ 1][j][0]) %= Mod;
			(Dp[now][j][1] = sum) %= Mod;
		}
	}
	int ans = 0;
	for (int i = 1; i <= N; ++i)
		(ans += Dp[now][i][0] + Dp[now][i][1]) %= Mod;
	cout<<ans<<endl;
	return 0;
}