CF1148G

First set all digital prime factor indices to 1 without affecting the answer.

 

Considering the complement of the original image G, let f [i] indicate that f [i] of the given n numbers are multiples of i. So for each point, we can use the principle of tolerance and exclusion to calculate its degree (that is, the number of gcd = 1 in the original array. For example, the weight of point 1 is 6, then deg [1] = n- (f [2] + f [3] -f [6]).

 

If point F can be found, make deg [F]> 1, and let two points connected with no edges be G and H. After deleting the above three points, we will discuss in two cases.

 

A: There are at least k-3 points with degree> 0, and the smallest m is found by dichotomy, so that in the graph composed of the first m points, there are still k-3 points with degree> 0. Vigorously categorize the discussion, delete the part that only has the edge with the m-point, and add F, G, H to the answer if necessary (see the code for details).

 

B: There are no k-3 points with degree> 0, because 2k <= n, so there are at least k points with degree 0, and any k points can be selected (the weights are two or two gcd> 1).

 

If the point F is not found, the size of the connected blocks in the figure is less than 3, and a point in each connected block can be selected and added to the answer.

 

代码：
```cpp
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n,k,t;
int pcnt=0,pri[100000];
bool isp[1000000];
void getprime(void)
{
 memset(isp,1,sizeof(isp));
 for(int i=2;i<1000000;i++)
  if(isp[i]){
   pri[pcnt++]=i;
   for(int j=2;i*j<1000000;j++)
    isp[i*j]=0;
  }
 return;
}

struct num{
 int fac[10],cnt;
}arr[100000];
int f[10000000];
void add(int i,int pos,int S)
{
 if(pos==arr[i].cnt){
  if(S!=1) f[S]++;
  return;
 }
 add(i,pos+1,S);
 add(i,pos+1,S*arr[i].fac[pos]);
 return;
}
void del(int i,int pos,int S)
{
 if(pos==arr[i].cnt){
  if(S!=1) f[S]--;
  return;
 }
 del(i,pos+1,S);
 del(i,pos+1,S*arr[i].fac[pos]);
 return;
}
void read(void)
{
 scanf("%d%d",&n,&k);
 for(int i=0;i<n;i++){
  scanf("%d",&t);
  arr[i].cnt=0;
  for(int j=0;pri[j]*pri[j]<=t;j++)
   if(t%pri[j]==0){
    arr[i].fac[arr[i].cnt++]=pri[j];
    while(t%pri[j]==0)
     t/=pri[j];
   }
  if(t>1) arr[i].fac[arr[i].cnt++]=t;
  if(i<2*k) add(i,0,1);
 }
 n=2*k;
 return;
}

int deg[100000],F=-1,G=-1,H=-1;
int getdeg(int i,int pos,int S,int opt)
{
 if(pos==arr[i].cnt)
  return opt*f[S];
 return getdeg(i,pos+1,S,opt)+getdeg(i,pos+1,S*arr[i].fac[pos],-opt);
}
inline bool common(int a,int b)
{
 int i=0,j=0;
 while(i<arr[a].cnt&&j<arr[b].cnt){
  if(arr[a].fac[i]==arr[b].fac[j])
   return 1;
  if(arr[a].fac[i]>arr[b].fac[j])
   j++;
  else i++;
 }
 return 0;
}
void getFGH(int i)
{
 F=i;
 for(int j=0;j<n;j++)
  if(!common(i,j)){
   if(G==-1) G=j;
   else if(H==-1) H=j;
  }
 return;
}

void solve1(void)
{
 for(int i=0;i<n;i++)
  del(i,0,1);
 int cnt=0;
 for(int i=0;i<n&&cnt<k;i++)
  if(getdeg(i,0,1,-1)==cnt){
   cnt++;
   printf("%d ",i+1);
   add(i,0,1);
  }
 return;
}
int check(int m)
{
 int m0=m+1;
 for(int i=0;i<=m;i++)
  if(i!=F&&i!=G&&i!=H)
   add(i,0,1);
  else m0--;
 int x=0;
 for(int i=0;i<=m;i++){
  if(i==F||i==G||i==H)
   continue;
  deg[i]=m0-getdeg(i,0,1,-1);
  if(deg[i]>0)
   x++;
 }
 return x;
}

void solve2(void)
{
 for(int i=0;i<n;i++)
  if(i!=F&&i!=G&&i!=H)
   del(i,0,1);
 int l=k-4,r=n-2,m;
 while(l<r){
  m=(l+r+1)>>1;
  if(check(m)>=k-3)
   r=m-1;
  else l=m;
  for(int i=0;i<=m;i++)
   if(i!=F&&i!=G&&i!=H)
    del(i,0,1);
 }
 m=l+1;
 
 int p=check(m);
 if(p>=k){
  for(int i=0;i<m&&p>k;i++)
   if(i!=F&&i!=G&&i!=H&&deg[i]==1&&!common(i,m)){
    p--; deg[i]=-1;
   }
 }
 else{
  for(int i=0;i<m&&p>k-2;i++)
   if(i!=F&&i!=G&&i!=H&&deg[i]==1&&!common(i,m)){
    p--; deg[i]=-1;
   }
  printf("%d %d ",F+1,G+1);
  if(p==k-3)
   printf("%d ",H+1);
 }
 for(int i=0;i<=m;i++)
  if(i!=F&&i!=G&&i!=H&&deg[i]>0)
   printf("%d ",i+1);
 return;
}

int main(void)
{
 getprime();
 read();
 int w=0;
 for(int i=0;i<n;i++){
  deg[i]=getdeg(i,0,1,-1);
  if(deg[i]<n) w++;
  if(F==-1&&deg[i]<=n-2)
   getFGH(i);
 }
 if(F==-1){
  solve1();
  return 0;
 }
 for(int i=0;i<n;i++)
  del(i,0,1);
 if(check(n-1)<k-3){
  for(int i=0;i<n&&k>0;i++)
   if(deg[i]==0){
    printf("%d ",i+1);
    k--;
   }
 }
 else solve2();
 return 0;
}
```