正题

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题目大意

将$n$个数分为$2$个非空的集合，要求两个集合内数的乘积的$gcd$为$1$。求方案数

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解题思路

拥有相同质因子的数一定要在同一组，枚举质因数然后将拥有的数都用并查集合并然后用联通块数量来求即可。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=1e6+10,XJQ=1e9+7;
ll n,fa[N],T,cnt,vis[N];
bool v[N];
ll find(ll x)
{return (fa[x]==x)?x:(fa[x]=find(fa[x]));}
void uniquee(ll x,ll y){ll Fa=find(x),Fb=find(y);if(Fa==Fb)return;if(Fa<Fb)fa[Fb]=Fa;else fa[Fa]=Fb; 
}
int main()
{scanf("%lld",&T);while(T--){memset(vis,0,sizeof(vis));memset(v,0,sizeof(v));scanf("%lld",&n);cnt=0;int z=0;for(ll i=1;i<=n;i++){ll x;scanf("%lld",&x);if(x==1)cnt++;if(vis[x]||x==1)continue;vis[x]=++z;fa[z]=z;}vis[0]=0;for(int i=2;i<=1e6;i++){if(v[i])continue;int last=0;for(int j=i;j<=1e6;j+=i){v[j]=1;if(vis[j]){if(vis[last])uniquee(vis[last],vis[j]);last=j;}}}for(int i=1;i<=z;i++)if(find(i)==i)cnt++;long long k=1;for(ll i=1;i<=cnt;i++)k=k*2ll%XJQ;printf("%lld\n",(k-2ll+XJQ)%XJQ);}
}