UOJ188. 【UR13】Sanrd

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看到因子，还是多想想Min_25筛吧。

解释下

  • 如果$p_k$是次大因子，那么只要选出$[p_k,n/p_j^e]$中的质数.
  • 如果$p_k$不是是次大因子，那么次大因子就是$\sum_{i=1}^{n/p_k^e}[min(i)>p_k] f(i)$.
  • 注意边界$p_j> n^2$，还是恒没有贡献。
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;
struct Min_25
{
    ll n, Lim;
    ll g[N], p[N];
    int pos1[N], pos2[N], cnt;
    int pcnt;
    ll pr[N], npr[N];
    void Prime_init(int X)
    {
        for (int i = 2; i <= X; i++)
        {
            if (!npr[i])
            {
                pr[++pcnt] = i;
            }
            for (int j = 1; j <= pcnt && pr[j] * i <= X; j++)
            {
                npr[i * pr[j]] = 1;
                if (i % pr[j] == 0)
                    break;
            }
        }
    }
    int Getcnt(ll x)
    {
        return (x <= Lim) ? pos1[x] : pos2[n / x];
    }
    ll S(ll x, int y)
    {
        if (pr[y] >= x || x <= 2)
            return 0;
        ll ans = 0;
        for (int i = y + 1; i <= pcnt && pr[i] * pr[i] <= x; i++)
        {
            ll pe = pr[i];
            for (int e = 1; pe * pr[i] <= x; e++, pe = pe * pr[i])
            {
                int k = Getcnt(x / pe);
                ans += (S(x / pe, i));

                //cout << pr[i] << " " << e << ' ' << g[k] - (i - 1) << endl;
                ans += pr[i] * (g[k] - (i - 1));
            }
        }
        return ans;
    }

    ll get(ll x)
    {
        n = x;
        Lim = sqrt(n);
        Prime_init(Lim);

        for (ll i = 1, j; i <= n; i = j + 1)
        {
            j = n / (n / i);
            ll k = n / i;
            p[++cnt] = k;
            g[cnt] = k - 1;
            if (k <= Lim)
                pos1[k] = cnt;
            else
                pos2[n / k] = cnt;
        }

        for (int i = 1; i <= pcnt; i++)
        {
            for (int j = 1; j <= cnt && 1ll * pr[i] * pr[i] <= p[j]; j++)
            {
                int k = Getcnt(p[j] / pr[i]);
                g[j] -= g[k] - (i - 1);
            }
        }

        return S(n, 0);
    }
} T1, T2;
ll l, r;
int main()
{
    // cin >> r;
    // cout << T1.get(r) << endl;
    cin >> l >> r;
    cout << T2.get(r) - T1.get(l - 1) << endl;
}