P3226 [HNOI2012] 集合选数

神仙题。
发现不是 \(2\) 的倍数和不是 \(3\) 的倍数的数之间没有限制，可以分开考虑。
以 \(1\) 为例。
构造一个矩阵，左上角为 \(1\)，每个数右边是祂乘 \(3\),下面是祂乘 \(2\)，原问题为求在这个矩阵里选数，不能选相邻的数的方案数，用状压 \(dp\) 随便做，最后把所有左上角数字不同的矩阵的答案乘起来。要取模。

代码

#include<bits/stdc++.h>
using namespace std;
namespace IO{
    template<typename T>
    inline void read(T&x){
        x=0;char c=getchar();bool f=0;
        while(!isdigit(c)) c=='-'?f=1:0,c=getchar();
        while(isdigit(c)) x=x*10+c-'0',c=getchar();
        f?x=-x:0;
    }
    template<typename T>
    inline void write(T x){
        if(x==0){putchar('0');return ;}
        x<0?x=-x,putchar('-'):0;short st[50],top=0;
        while(x) st[++top]=x%10,x/=10;
        while(top) putchar(st[top--]+'0');
    }
    inline void read(char&c){c=getchar();while(isspace(c)) c=getchar();}
    inline void write(char c){putchar(c);}
    inline void read(string&s){s.clear();char c;read(c);while(!isspace(c)&&~c) s+=c,c=getchar();}
    inline void write(string s){for(int i=0,len=s.size();i<len;i++) putchar(s[i]);}
    template<typename T>inline void write(T*x){while(*x) putchar(*(x++));}
    template<typename T,typename...T2> inline void read(T&x,T2&...y){read(x),read(y...);}
    template<typename T,typename...T2> inline void write(const T x,const T2...y){write(x),putchar(' '),write(y...),sizeof...(y)==1?putchar('\n'):0;}
}using namespace IO;
template<signed mod>struct Modint{
    signed z;
    Modint(){z=0;}
    Modint(signed x){x%=mod;z=x<0?x+mod:x;}
    Modint(long long x){x%=mod;z=x<0?x+mod:x;}
    Modint(short x){x%=mod;z=x<0?x+mod:x;}
    Modint(char x){x%=mod;z=x<0?x+mod:x;}
    Modint(bool x){x%=mod;z=x<0?x+mod:x;}
    friend Modint operator+(Modint t,Modint t2){Modint ans;ans.z=(t.z+t2.z)%mod;return ans;}
    friend Modint operator*(Modint t,Modint t2){Modint ans;ans.z=1ll*t.z*t2.z%mod;return ans;}
    friend Modint operator-(Modint t,Modint t2){Modint ans;ans.z=(t.z-t2.z)%mod;return ans;}
    Modint operator<<(const signed t)const{Modint ans;ans.z=(z<<t)%mod;return ans;}
    Modint operator>>(const signed t)const{Modint ans;ans.z=(z>>t)%mod;return ans;}
    Modint&operator+=(const Modint t){z=(z+t.z)%mod;return *this;}
    Modint&operator*=(const Modint t){z=1ll*z*t.z%mod;return *this;}
    Modint&operator-=(const Modint t){z=(z-t.z)%mod;return *this;}
    Modint&operator<<=(const signed t){z=(z<<t)%mod;return *this;}
    Modint&operator>>=(const signed t){z=(z>>t)%mod;return *this;}
    friend Modint ksm(Modint a,signed b){
        Modint ans=1;
        while(b){if(b&1) ans=ans*a;a=a*a,b>>=1;}
        return ans;
    }
    friend void read(Modint&z){
        signed x=0;char c=getchar();bool f=0;
        while(!isdigit(c)) c=='-'?f=1:0,c=getchar();
        while(isdigit(c)) x=(x*10ll+c-'0')%mod,c=getchar();
        f?x=-x:0;
        z.z=x;
    }
    friend void write(Modint x){x.z<0?x.z+=mod:0;write(x.z);}
};
#define int long long
const int mod=1000000001,maxh=20,maxl=20,maxzt=400000,maxn=100010;
int n,jl=1,lt;
Modint<mod>f[maxh][maxzt],ans=1;
bool flag[maxn];
inline void js(int x){
	int s=x;
	f[0][0]=1;
	Modint<mod> ans=0;
	lt=0;
	for(int i=1;;i++){
		if(s>n) break;
		int ss=s,l=0;
		for(;ss<=n;l++) flag[ss]=1,ss*=2;
		for(int j=0;j<(1<<l);j++){
			f[i][j]=0;
			if(j&(j<<1)) continue;
			for(int k=0;k<(1<<lt);k++){
				if(j&k) continue;
				f[i][j]+=f[i-1][k];
			}
		}
		s*=3;
		if(s>n) for(int j=0;j<1<<l;j++) ans=ans+f[i][j];
		lt=l;
	}
	::ans=::ans*ans;
}
signed main(){
    read(n);
	for(int i=1;i<=n;i++){
		if(flag[i]) continue;
		js(i);
	}
    write(ans);
	return 0;
}