HackerRank

https://vjudge.net/contest/279745#problem/G

每次将质数的倍数放进一个集合中，那么如果最后的集合数为n的话；

方案数： 2^n -2 ；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/
int a[maxn];
int fa[maxn];
int p[1000004];
int prime[1000004];
int tot;
bool vis[1000005];

void init() {
	for (int i = 2; i <= 1000000; i++) {
		if (!vis[i]) {
			prime[++tot] = i;
		}
		for (int j = 1; j <= tot &&(ll) i*(prime[j]) <= 1000000; j++) {
			vis[i*prime[j]] = 1;
			if (i%prime[j] == 0)break;
		}
	}
}
int findfa(int x) {
	if (x == fa[x])return x;
	else return fa[x] = findfa(fa[x]);
}
int qpow(int a, int b, int mod) {
	int ans = 1;
	while (b>0) {
		if (b & 1)ans = (ll)ans * a%mod;
		a =(ll) a * a%mod; b >>= 1;
	}
	return ans;
}

int main() {
	
	init();
	int T; cin >> T;
	while (T--) {
		int n; rdint(n);
		int cnt = 0;
		for (int i = 1; i <= n; i++) {
			rdint(a[i]);
			if (a[i] == 1) {
				cnt++; n--; i--;
			}
		}
		if (n) {
			sort(a + 1, a + 1 + n);
			n = unique(a + 1, a + 1 + n) - a - 1;
			ms(p);
			for (int i = 1; i <= n; i++) {
				p[a[i]] = i; fa[i] = i;
			}
			for (int i = 1; i <= tot; i++) {
				int pp = 0;
				for (int j = prime[i]; j <= 1000000; j += prime[i]) {
					if (p[j]) {
						if (!pp)pp = findfa(p[j]);
						else {
							int q = findfa(p[j]);
							fa[q] = pp;
						}
					}
				}
			}
			for (int i = 1; i <= n; i++) {
				findfa(i);
				if (i == fa[i])cnt++;
			}

		}
		cout << (qpow(2, cnt, mod) - 2 + mod) % mod << endl;
	}
	return 0;
}