中国(北方)大学生程序设计训练赛(第二周)(Problem G: Connected Components-并查集)

给一张图，点数n<=100000，每个点有一个权$1\le X_i\le1000000$，两点不互质则连边，求连通块个数。

对于每个权值把它的素因子也纳入图，然后每个权值与其素因子连边。
注意特判1

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <functional>
#include <cstdlib>
#include <queue>
#include <stack>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define MAXN (1000000+10) 
typedef long long ll;
int p[MAXN],tot;
bool b[MAXN]={0};
int f[MAXN];
void make_prime(int n)
{
    tot=0;f[1]=0;
    Fork(i,2,n)
    {
        if (!b[i]) p[++tot]=i,f[i]=1;
        For(j,tot)
        {
            if (i*p[j]>n) break;
            b[i*p[j]]=1;f[i*p[j]]=i;
            if (i%p[j]==0) break;  
        }
    }
}

class bingchaji
{
public:
    int father[MAXN],n,cnt;
    void mem(int _n)
    {
        n=cnt=_n;
        For(i,n) father[i]=i;
    }
    int getfather(int x) 
    {
        if (father[x]==x) return x;

        return father[x]=getfather(father[x]);
    }
    void unite(int x,int y)
    {
        // cout<<x<<" "<<y<<endl;
        x=getfather(x);
        y=getfather(y);
        if (x^y) {
            --cnt;
            father[x]=y;
        }
    }
    bool same(int x,int y)
    {
        return getfather(x)==getfather(y);
    }
}S;
int a[MAXN];
int main() {
    // freopen("G.in","r",stdin);
    make_prime(1000000);
    // For(i,100) cout<<i<<' '<<f[i]<<endl;
    int T=read();
    For(kcase,T) {
        int n=read();
        S.mem(1000000);
        For(i,n) a[i]=read();
        int ans=0;
        For(i,n) if (a[i]==1) ++ans;

        sort(a+1,a+1+n);
        n=unique(a+1,a+1+n)-a-1;
        //For(i,n) cout<<a[i]<<' ';cout<<endl;
        For(i,n) if (a[i]>1) {
            int t=a[i]; 
            while(f[t]!=1) {
                S.unite(a[i],t/f[t]);
                t=f[t];
            }
            S.unite(a[i],t);
        }
        MEM(b)
        For(i,n) if (a[i]>1)
        {
            int t=S.getfather(a[i]);
            if (!b[t]) b[t]=1,++ans;
        }
        printf("Case %d: %d\n",kcase,ans);
    }

    return 0;
}