牛客练习赛D数学家的谜题线段树+bitset优化

Posted 2021-12-12 hesorchen

题目

https://ac.nowcoder.com/acm/contest/11175/D

单点修改和区间查询，询问区间乘能被多少个素数整除。

解题思路

显然是维护区间质因子的数量，可以用带修莫队。

也可以用线段树+bitset优化。将质数映射到1-10000的位中。

代码

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5, M = 5e4 + 5;

vector<int> fac[N];

bool vis[N];
int rk[N];
void init()
{
    int ct = 0;
    for (int i = 2; i < N; i++)
    {
        if (!vis[i])
        {
            rk[i] = ++ct;
            for (int j = i; j < N; j += i)
            {
                vis[j] = 1;
                fac[j].emplace_back(i);
            }
        }
    }
    // for (int i = 1; i <= 11; i++)
    //     cout << i << ' ' << rk[i] << endl;
    // cout << ct << endl;
}

int a[M];

struct node
{
    int l, r;
    bitset<10000> bit;
} tr[M << 2];

void pushup(int k)
{
    tr[k].bit = tr[k * 2].bit | tr[k * 2 + 1].bit;
}

void build(int k, int l, int r)
{
    tr[k].l = l, tr[k].r = r;
    if (l == r)
    {
        tr[k].bit.reset();
        for (auto it : fac[a[l]])
            tr[k].bit.set(rk[it]);
        return;
    }
    int mid = l + r >> 1;
    build(k * 2, l, mid);
    build(k * 2 + 1, mid + 1, r);
    pushup(k);
}

void change(int k, int id, int x)
{
    if (tr[k].l == tr[k].r)
    {
        tr[k].bit.reset();
        for (auto it : fac[x])
            tr[k].bit.set(rk[it]);
        return;
    }
    int mid = tr[k].l + tr[k].r >> 1;
    if (id <= mid)
        change(k * 2, id, x);
    else
        change(k * 2 + 1, id, x);
    pushup(k);
}

bitset<10000> query(int k, int l, int r)
{
    if (tr[k].l == l && tr[k].r == r)
        return tr[k].bit;
    int mid = tr[k].l + tr[k].r >> 1;
    if (r <= mid)
        return query(k * 2, l, r);
    else if (l > mid)
        return query(k * 2 + 1, l, r);
    else
        return query(k * 2, l, mid) | query(k * 2 + 1, mid + 1, r);
}
void solve()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    build(1, 1, n);
    while (m--)
    {
        int op, l, r;
        scanf("%d %d %d", &op, &l, &r);
        if (op == 1)
            change(1, l, r);
        else
            cout << query(1, l, r).count() << endl;
    }
}

int main()
{
    init();
    solve();
    return 0;
}