题解 T1出了个大阴间题

传送门

确实是个大阴间题
是个挺显然的状压，但就是没想到，看到 \(n\leqslant 18\) 想到meet in middle那边去了……

• 对于第二维取值范围非常有限，且这些取值都与某个标准值相差不大的情况，也可以把第二维记录为与这个标准值的差值

先令 \(f[s][i]\) 为集合 \(s\) 中 \(a\) 值为 \(i\) 的总和
发现 \(i\) 的合法取值只有 \(max_{a_i}\) 和 \(max_{a_i}+1\ (i\in s)\) 两种
所以令 \(f[s][0]\) 和 \(f[s][1]\) 分别对应这两种
转移就比较显然了

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define fir first
#define sec second
#define make make_pair
//#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 (f=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int a[N];
ll k;
const ll mod=1e9+7;
inline void md(ll& a, ll b) {a+=b; a=a>=mod?a-mod:a;}

namespace force{
	int p[N];
	ll maxn, ans;
	pair<ll, ll> tem[N];
	void solve() {
		for (int i=1; i<=n; ++i) p[i]=i;
		do {
			for (int i=1; i<=n; ++i) tem[i]=make(a[p[i]], 0);
			pair<ll, ll> now=tem[1]; ll sum=0;
			for (int i=2; i<=n; ++i) {
				sum=(sum+((now.fir==tem[i].fir)?(now.fir+1):max(now.fir, tem[i].fir))%mod*k%mod+now.sec+tem[i].sec)%mod;
				now=make((now.fir==tem[i].fir)?(now.fir+1):max(now.fir, tem[i].fir), (2ll*max(now.sec, tem[i].sec)+1)%mod);
			}
			if (now.fir>maxn) {
				maxn=now.fir; ans=sum;
			}
			else if (now.fir==maxn) ans=(ans+sum)%mod;
		} while (next_permutation(p+1, p+n+1));
		printf("%lld %lld\n", maxn, ans);
		exit(0);
	}
}

namespace task{
	int maxn[1<<18], siz[1<<18];
	ll f[1<<18][2], g[1<<18][2], dlt[20];
	void solve() {
		int lim=1<<n;
		for (int s=1; s<lim; ++s)
			for (int i=0; i<n; ++i) if (s&(1<<i))
				maxn[s]=max(maxn[s], a[i+1]), ++siz[s];
		for (int i=2; i<=n; ++i) dlt[i]=2*dlt[i-1]+1;
		for (int i=0; i<n; ++i) g[1<<i][0]=1;
		for (int s=1; s<lim; ++s) {
			for (int i=0; i<n; ++i) if (!(s&(1<<i))) {
				if (g[s][0]) {
					if (maxn[s]==a[i+1]) md(f[s|(1<<i)][1], (f[s][0]+g[s][0]%mod*(k*(a[i+1]+1)%mod+dlt[siz[s]])%mod)%mod);
					else md(f[s|(1<<i)][0], (f[s][0]+g[s][0]%mod*(k*max(maxn[s], a[i+1])%mod+dlt[siz[s]])%mod)%mod);
					g[s|(1<<i)][maxn[s]==a[i+1]]+=g[s][0];
				}
				if (g[s][1]) {
					if (max(maxn[s]+1, a[i+1])==maxn[s|(1<<i)]+1||maxn[s]+1==a[i+1]) md(f[s|(1<<i)][1], (f[s][1]+g[s][1]%mod*(k*max(maxn[s]+1, a[i+1]+1)%mod+dlt[siz[s]])%mod)%mod);
					else md(f[s|(1<<i)][0], (f[s][1]+g[s][1]%mod*(k*max(maxn[s]+1, a[i+1])%mod+dlt[siz[s]])%mod)%mod);
					g[s|(1<<i)][max(maxn[s]+1, a[i+1])==maxn[s|(1<<i)]+1||maxn[s]+1==a[i+1]]+=g[s][1];
				}
			}
		}
		if (g[lim-1][1]) printf("%d %lld\n", maxn[lim-1]+1, f[lim-1][1]);
		else printf("%d %lld\n", maxn[lim-1], f[lim-1][0]);
		exit(0);
	}
}

signed main()
{
	freopen("repair.in", "r", stdin);
	freopen("repair.out", "w", stdout);

	n=read(); k=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	// force::solve();
	task::solve();

	return 0;
}