P6298 齿轮

发表于 | 阅读数

给$n$个数，求取$k$个数的$gcd=m$的方案数，求出所有$m$。

$g(x)=\sum [x|a[i]]$

$gcd=x$如果选择$C_{g(x)}^k$。会重复到$f(xk)$

倒过来枚举，调和级数$O(m\log m)$

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
ll qpow(ll a, ll x, ll mo)
{
    ll res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        x >>= 1;
    }
    return res;
}
int fac[N], facinv[N];
void prepare()
{
    fac[0] = 1;
    facinv[0] = 1;
    for (int i = 1; i < N; i++)
        fac[i] = 1ll * fac[i - 1] * i % mod;
    facinv[N - 1] = qpow(fac[N - 1], mod - 2, mod);
    for (int i = N - 2; i >= 1; i--)
        facinv[i] = 1ll * facinv[i + 1] * (i + 1) % mod;
}
ll C(ll i, ll n)
{
    if (n <= 0)
        return 0;
    if (i > n)
        return 0;
    if (i == 0)
        return 1;
    return 1ll * fac[n] * facinv[i] % mod * facinv[n - i] % mod;
}
int v[N], a[N];
ll f[N];
int n, m, k;
int main()
{
    prepare();
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]), v[a[i]]++;
    for (int i = m; i >= 1; i--)
    {
        int num = 0;
        for (int j = i; j <= m; j += i)
            num += v[j];

        int p = 0;
        for (int j = i; j <= m; j += i)
            p = (p + f[j]) % mod;
        f[i] = (0ll + C(k, num) - p + mod + mod) % mod;
    }
    for (int i = 1; i <= m; i++)
        printf("%lld ", f[i]);
}