​​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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#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 pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair 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;
}

 

EPFL - Fighting