CF803F

 题意要求gcd=1的子序列个数.

我们想到求gcd!=1的子序列个数,看到值域很小,我们考虑对值域下手,因为是gcd=1,我们联想到互质,于是我们可以使用容斥求个数,详细来说就是,含有1的因子的个数-含有2,3,5,7,11....一个质因子的个数+含有2*3,3*5....两个质因子的个数-......由于普通的容斥原理是2^n的,因为选的是质因数,所以我们用莫比乌斯函数实现.

#include 
#define int long long
#define IOS ios::sync_with_stdio(false), cin.tie(0) 
#define ll long long 
// #define double long double
#define ull unsigned long long 
#define PII pair 
#define PDI pair 
#define PDD pair 
#define debug(a) cout << #a << " = " << a << endl 
#define point(n) cout << fixed << setprecision(n)
#define all(x) (x).begin(), (x).end() 
#define mem(x, y) memset((x), (y), sizeof(x)) 
#define lbt(x) (x & (-x)) 
#define SZ(x) ((x).size()) 
#define inf 0x3f3f3f3f 
#define INF 0x3f3f3f3f3f3f3f3f
namespace nqio{const unsigned R = 4e5, W = 4e5; char *a, *b, i[R], o[W], *c = o, *d = o + W, h[40], *p = h, y; bool s; struct q{void r(char &x){x = a == b && (b = (a = i) + fread(i, 1, R, stdin), a == b) ? -1 : *a++;} void f(){fwrite(o, 1, c - o, stdout); c = o;} ~q(){f();}void w(char x){*c = x;if (++c == d) f();} q &operator >>(char &x){do r(x);while (x <= 32); return *this;} q &operator >>(char *x){do r(*x); while (*x <= 32); while (*x > 32) r(*++x); *x = 0; return *this;} template<typename t> q&operator>>(t &x){for (r(y),s = 0; !isdigit(y); r(y)) s |= y == 45;if (s) for (x = 0; isdigit(y); r(y)) x = x * 10 - (y ^ 48); else for (x = 0; isdigit(y); r(y)) x = x * 10 + (y ^ 48); return *this;} q &operator <<(char x){w(x);return *this;}q &operator<< (char *x){while (*x) w(*x++); return *this;}q &operator <<(const char *x){while (*x) w(*x++); return *this;}template<typename t> q &operator<< (t x) {if (!x) w(48); else if (x < 0) for (w(45); x; x /= 10) *p++ = 48 | -(x % 10); else for (; x; x /= 10) *p++ = 48 | x % 10; while (p != h) w(*--p);return *this;}}qio; }using nqio::qio;
using namespace std;
const int N = 1e6 + 10, MOD = 1e9 + 7;
int n, a[N], cc[N];
int mu[N], vis[N], primes[N], sum[N], cnt;
void get_mu(int n)
{
    cnt = 0, mu[1] = 1;
    for(int i = 2; i <= n; ++ i) {
        if(vis[i] == 0){
            primes[ ++ cnt] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= cnt && i * primes[j] <= n; ++ j) {
            vis[primes[j] * i] = 1;
            if(i % primes[j] == 0) break;
            mu[i * primes[j]] -= mu[i];
            //mi[i * primes[j]] = -mu[i];
        }
    }
    for(int i = 1; i <= n; ++ i)
        sum[i] += sum[i - 1] + mu[i];
}
int qmi(int a, int k, int p) {
	int res = 1;
	while (k) {
		if (k & 1) res = res * a % p;
		k >>= 1;
		a = a * a % p;
	}
	return res;
}
void solve() {
	qio >> n;
	get_mu(1e6);
	int mx = 0;
	for (int i = 1, x; i <= n; ++i) qio >> x, ++cc[x], mx = max(mx, x);
	int ans = 0;
	for (int i = 1; i <= mx; ++i) {
		int s = 0;
		for (int j = 1; j <= mx / i; ++j) s += cc[i * j];
		ans = ((ans + mu[i] * (qmi(2, s, MOD) - 1) % MOD) % MOD + MOD) % MOD;
	}
	qio << ans << "\n";
}
signed main()  {
	// IOS;
	int T = 1;
	// qio >> T;
	while (T--) solve();
}