题解 [HNOI2012]集合选数

题目传送门

题目大意

直接看题面吧。

思路

感觉挺水的一道题啊？怎么评到紫色的啊?考试的时候LJS出了这个题的加强版我就只想出这个思路，然后就爆了。。。

不难发现，我们可以构造矩阵：

x 2x 4x 6x ...
3x 6x 12x 24x 48x ...
9x 18x 36x ...

然后实际上就相当于在这个矩阵中选出一些数使得两两不相邻。因为行数列数都是 \(\log\) 级别的，所以直接状压就好了。

\(\texttt{Code}\)

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define mod 1000000001
#define MAXN 1000005

template <typename T> void read (T &x){char c = getchar ();x = 0;int f = 1;while (c < '0' || c > '9') f = (c == '-' ? -1 : 1),c = getchar ();while (c >= '0' && c <= '9') x = x * 10 + c - '0',c = getchar ();x *= f;}
template <typename T,typename ... Args> void read (T &x,Args& ... args){read (x),read (args...);}
template <typename T> void write (T x){if (x < 0) x = -x,putchar ('-');if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void Mx (T &a,T b){a = max (a,b);}
template <typename T> void Mi (T &a,T b){a = min (a,b);}

bool mark[MAXN];
int n,a[25][25],lim[25];

void init (int x){
	a[1][1] = x;
	for (Int i = 1;i <= 15;++ i){
		if (i > 1) a[i][1] = a[i - 1][1] * 3;
		if (a[i][1] > n) break;
		for (Int j = 2;j <= 20;++ j){
			a[i][j] = a[i][j - 1] * 2;
			if (a[i][j] > n){
				lim[i] = j - 1;
				break;
			}
		}
		for (Int j = 1;j <= lim[i];++ j) mark[a[i][j]] = 1;
	}
} 

bool chk[1 << 21];
int dp[2][1 << 22];

int WorkDP (){
	int endl = 0;
	for (Int i = 0;i < (1 << lim[1]);++ i) dp[1][i] = chk[i];
	for (Int i = 2;i <= 15;++ i){
		if (a[i][1] > n){
			endl = i - 1;
			break;
		}
		for (Int S = 0;S < (1 << lim[i]);++ S) if (chk[S]){
			dp[i & 1][S] = 0;
			for (Int S1 = 0;S1 < (1 << lim[i - 1]);++ S1)
				if ((S & S1) == 0) dp[i & 1][S] += dp[i - 1 & 1][S1],dp[i & 1][S] %= mod;
		}
		else dp[i & 1][S] = 0;
	}
	int ans = 0;
	for (Int S = 0;S < (1 << lim[endl]);++ S) ans += dp[endl & 1][S],ans %= mod;
	return ans;
}

signed main(){
	read (n);int res = 1;
	for (Int i = 0;i < (1 << 20);++ i) chk[i] = ((i & (i >> 1)) == 0);
	for (Int i = 1;i <= n;++ i) if (!mark[i]) init (i),res = 1ll * res * WorkDP () % mod;
	write (res),putchar ('\n');
	return 0;
}