NOI Online #3 Improve group T3 excellent subsequence

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Concise title

Request from $A$ selects subset$B$ makesBBNo two numbers in$B$ are bitwise and equal to$0$ . BBper subsetThe value of$B\; is$ the sum of all the numbers in the set plus one$\varphi$ value. Find all subsetsBBThe sum of$B$ 's value.

analysis

You can think because you have to know all the set $B\; is$ not easy, so considerBB with thesame valueThe number of$B.$Let$f_i$Indicates that all values ​​are $i$ 's subsetBBThe number of$B$ , we calculate$f_i$Multiply by $ph{i}_{i+1}$It is the value and sum of this plan, and adding up each value and sum is the answer. Consider how to calculate this $f$ , we can put$f\; is$ divided into two parts, and then one part is directly used$f_j$, The other part uses the set AAJust make up the number of$A.$Because the title isBBThe bitwise sum of all numbers in$B$ is equal to$0$ , so there will not be two numbers where one of the two numbers is$1$ situation. So we are dismantling$i$ can put$i$ is1 1The bit of$1$ is divided into two halves for calculation. To facilitate processing, we put$nu{m}_x$It represents $a_j$Median is xxthe number of$x$ ,$s$ meansiia number divided by$i$ ,$t$ is the remaining part, in order to ensure no double calculation, it is required that$s>=t$ , according to the principle of multiplication, there is a state transition equation$f_i+=f_s\times nu{m}_t$. But this has a problem: the upper limit of the sum of these numbers is too large, and the memory can't fit! Don’t worry, because the number we choose will not be both $In$ the case of$1$ , so the selected sum is naturally less than2 ⌊ log ⁡ 2 (max ⁡ {ai}) + 1 ⌋ 2^{\lfloor\log_2(\max\{a_i\})+1\rfloor}2⌊log2​(max{ ai​})+1⌋，以$2^{\lfloor {\log }_2\left(\max \right\{a_i\left\}\right)+1\rfloor }$ is the upper limit. But then there will be a situation:$a_i=0$ is missing! Actually$a_i$Equal to $It$ doesn’t matter whether it is$0$ or not. According to the principle of multiplication, the answer should be multiplied by$2^{nu{m}_0}$. As for the sieve method to find $phi$ (that is, when sieving prime numbers,$phi$ function) and find a subset of a set (to split$i$ ), everyone should know it?

Code

#include<bits/stdc++.h>
using namespace std;
const int NN=300000,P=1e9+7;
int num[NN],phi[NN],f[NN];
bool vis[NN];
int main()
{
    
    
	int n,maxx=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
    
    
		int x;
		scanf("%d",&x);
		num[x]++;
		maxx=max(maxx,x);
	}
	int m=0;
	while((1<<m)<=maxx)
		m++;
	phi[1]=1;
	for(int i=2;i<=(1<<m);i++)
		if(!vis[i])
			for(int j=i;j<=(1<<m);j+=i)
			{
    
    
				vis[j]=true;
				if(!phi[j])
					phi[j]=j;
				phi[j]=phi[j]/i*(i-1);
			}
	f[0]=1;
	for(int i=1;i<=(1<<m);i++)
		for(int s=i;;s=(s-1)&i)
		{
    
    
			int t=i^s;
			if(s<t)
				break;
			f[i]=(f[i]+1ll*f[t]*num[s])%P;
		}
	int ans=0;
	for(int i=0;i<=(1<<m);i++)
		ans=(ans+1ll*f[i]*phi[i+1])%P;
	for(int i=1;i<=num[0];i++)
		ans=ans*2ll%P;
	printf("%d",ans);
	return 0;
}