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-4typedef 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
