题解 排队

传送门

怎么连套路 DP 都不会了啊/kk

貌似是个很套路的 DP
考虑按从小到大的顺序加入下一个数
于是每次加入的数一定大于已有的所有数
所以加到一个峰的旁边峰数不变，否则峰数+1
令 \(f_{i, j}\) 为前 \(i\) 个数，有 \(j\) 个刺头的方案数
有

\[f_{i, j}=f_{i-1, j}*2j+f_{i-1,j-1}*(i-2(j-1)) \]

于是发现在 \(\bmod p\) 意义下不同的 \(i\) 只有 p 个，预处理后矩阵快速幂
剩下的边角暴力矩乘即可
复杂度 \(O(m^3p+m^3\log n)\)

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
// #define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m; ll mod;
inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}

namespace force{
	int a[N], ans;
	void solve() {
		for (int i=1; i<=n; ++i) a[i]=i;
		do {
			int cnt=0;
			if (a[1]>a[2]) ++cnt;
			if (a[n]>a[n-1]) ++cnt;
			for (int i=2; i<n; ++i) if (a[i]>a[i-1] && a[i]>a[i+1]) ++cnt;
			if (cnt==m) ++ans;
		} while (next_permutation(a+1, a+n+1));
		cout<<ans%mod<<endl;
	}
}

namespace task1{
	ll f[N][15];
	void solve() {
		f[0][0]=1;
		for (int i=1; i<=n; ++i)
			for (int j=1; j<=m; ++j)
				f[i][j]=(f[i-1][j]*2*j+f[i-1][j-1]*(i-2*j+2))%mod;
		cout<<f[n][m]<<endl;
	}
}

namespace task{
	struct matrix{
		int n, m;
		ll a[11][11];
		matrix(){memset(a, 0, sizeof(a));}
		matrix(int x, int y){n=x; m=y; memset(a, 0, sizeof(a));}
		void resize(int x, int y) {n=x; m=y; memset(a, 0, sizeof(a));}
		inline ll* operator [] (int t) {return a[t];}
		void put() {for (int i=0; i<=n; ++i) {for (int j=0; j<=m; ++j) cout<<setw(2)<<a[i][j]<<' '; cout<<endl;}cout<<endl;}
		inline matrix operator * (matrix b) {
			matrix ans(n, b.m);
			for (int i=0; i<=n; ++i)
				for (int k=0; k<=m; ++k)
					for (int j=0; j<=b.m; ++j)
						ans[i][j]=(ans[i][j]+a[i][k]*b[k][j])%mod;
			return ans;
		}
	}mat, tr[1010], link, I;
	inline matrix qpow(matrix a, ll b) {matrix ans=I; for (; b; a=a*a,b>>=1) if (b&1) ans=ans*a; return ans;}
	void solve() {
		mat.resize(1, m); I.resize(m, m);
		mat[0][0]=1;
		for (int i=0; i<=m; ++i) I[i][i]=1;
		for (int i=1; i<=mod; ++i) {
			tr[i].resize(m, m);
			for (int j=0; j<=m; ++j) {
				tr[i][j][j]=2*j;
				if (j>0) tr[i][j-1][j]=i-2*j+2;
			}
		}
		link=tr[1];
		for (int i=2; i<=mod; ++i) link=link*tr[i];
		link=qpow(link, n/mod);
		for (int i=1; i<=n%mod; ++i) link=link*tr[i];
		mat=mat*link;
		printf("%lld\n", (mat[0][m]%mod+mod)%mod);
	}
}

signed main()
{
	freopen("queue.in", "r", stdin);
	freopen("queue.out", "w", stdout);
	
	n=read(); m=read(); mod=read();
	// if (m==1) cout<<qpow(2, n-1)<<endl;
	// else force::solve();
	task::solve();
	
	return 0;
}