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题目大意:给定一个长度为n的正整数序列.将 {1,2,...,n} 划分成两个非空集合S,T.使得 \(gcd(∏_{i∈S}a_i,∏_{i∈T}a_i)\) = 1.求划分方案数，对 \(10^9+7\)取模.

显然的性质是:有相同因数的数字们必须在一个集合里面.

所以我们用并查集来维护.最后统计分成了几个集合.答案就是\(2^{cnt} - 2\)

线性筛+并查集即可.

代码如下:

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#include <bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;

const int N = 1e6 + 66, mod = 1e9 + 7;

int n, cnt;
int v[N], mp[N], pri[N], vis[N], las[N];
vector <int> p[N];

inline void pres_dou()
{
    int i, j;
    for (i = 2; i < 1e6; ++ i)
    {
        if (! v[i]) pri[++ cnt] = i, mp[i] = i;
        for (j = 1; j <= cnt && pri[j] * i < 1e6; ++ j)
        {
            v[i * pri[j]] = 1;
            mp[i * pri[j]] = pri[j];
            if (i % pri[j] == 0) break;
        }
    }
    return;
}

inline void dfs(int x)
{
    vis[x] = 1;
    for (int i = 0; i < (int)p[x].size(); ++ i) if (! vis[p[x][i]]) dfs(p[x][i]);
    return;
}

inline void yhm_func()
{
    int i, x; n = read();
    for (i = 1; i <= cnt; ++ i) las[pri[i]] = 0;
    for (i = 1; i <= n; ++ i)
    {
        vis[i] = 0, p[i].clear();
        x = read();
        while (x > 1)
        {
            int tmp = mp[x];
            while (x % tmp == 0) x /= tmp;
            if (las[tmp]) p[i].push_back(las[tmp]), p[las[tmp]].push_back(i);
            las[tmp] = i;
        }
    }
    int res = 1;
    for (i = 1; i <= n; ++ i) if (! vis[i]) dfs(i), res = res * 2 % mod;
    return (void)(put((res - 2 + mod) % mod));
}

signed main()
{	
    pres_dou();
    int T = read();
    while (T --) yhm_func();

    return 0;
}

后记:

巧妙的思维题