結果

問題 No.3250 最小公倍数
ユーザー Kude
提出日時 2025-08-29 21:56:05
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,171 ms / 2,000 ms
コード長 6,100 bytes
コンパイル時間 3,529 ms
コンパイル使用メモリ 322,128 KB
実行使用メモリ 90,172 KB
最終ジャッジ日時 2025-08-29 21:56:22
合計ジャッジ時間 15,032 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 21
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp:59:25: warning: ‘std::vector<std::pair<int, int> >& {anonymous}::factorize_lv(int)’ defined but not used [-Wunused-function]
   59 | vector<pair<int, int>>& factorize_lv(int x) {
      |                         ^~~~~~~~~~~~

ソースコード

diff # 

#include<bits/stdc++.h>
namespace {
#pragma GCC diagnostic ignored "-Wunused-function"
#include<atcoder/all>
#pragma GCC diagnostic warning "-Wunused-function"
using namespace std;
using namespace atcoder;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return true; } else return false; }
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
using mint = modint998244353;

pair<vector<int>, vector<int>> primes_lpf(const int n) {
  vector<int> primes; primes.reserve(n / 10);
  vector<int> lpf(n + 1);
  for (int i = 2; i <= n; i += 2) lpf[i] = 2;
  for (int i = 3; i <= n; i += 6) lpf[i] = 3;
  if (2 <= n) primes.push_back(2);
  if (3 <= n) primes.push_back(3);
  // 5 * x <= n, x <= floor(n / 5)
  const int n5 = n / 5;
  int x = 5;
  char add_next = 2;
  for (; x <= n5; x += add_next, add_next ^= 0x2 ^ 4) {
    int px = lpf[x];
    if (px == 0) {
      lpf[x] = px = x;
      primes.push_back(x);
    }
    for (int i = 2;; ++i) {
      int q = primes[i];
      int y = q * x;
      if (y > n) break;
      lpf[y] = q;
      if (q == px) break;
    }
  }
  for (; x <= n; x += add_next, add_next ^= 0x2 ^ 4) {
    if (lpf[x] == 0) {
      lpf[x] = x;
      primes.push_back(x);
    }
  }
  return {move(primes), move(lpf)};
}

constexpr int PSIZE = 1000000;
auto [primes, lpf] = primes_lpf(PSIZE);

vector<pair<int, int>>& factorize_lv(int x) {
  int ps[10], cs[10];
  int sz = 0;
  while (x != 1) {
    int p = lpf[x], c = 0;
    do {x /= p; c++;} while (x % p == 0);
    ps[sz] = p; cs[sz] = c; sz++;
  }
  static vector<pair<int, int>> fs;
  fs.clear();
  for (int i = 0; i < sz; i++) fs.emplace_back(ps[i], cs[i]);
  return fs;
}

struct HLD {
  const vector<vector<int>>& to;
  int root, n;
  vector<int> sz, parent, depth, idx, ridx, head, inv;

  HLD(const vector<vector<int>>& to, int root=0)
      : to(to), root(root), n(to.size()),
        sz(n), parent(n), depth(n), idx(n), ridx(n), head(n), inv(n) {
    init_tree_data(root, -1, 0);
    int nxt = 0;
    assign_idx(root, root, nxt);
  }
  void init_tree_data(int u, int p, int d) {
    parent[u] = p;
    depth[u] = d;
    int s = 1;
    for (int v: to[u]) if (v != p) {
      init_tree_data(v, u, d+1);
      s += sz[v];
    }
    sz[u] = s;
  }
  void assign_idx(int u, int h, int& nxt, int p=-1) {
    head[u] = h;
    idx[u] = nxt;
    inv[nxt] = u;
    nxt++;
    int heaviest = -1;
    int mxweight = 0;
    for (int v: to[u]) if (v != p) {
      if (sz[v] > mxweight) {
        heaviest = v;
        mxweight = sz[v];
      }
    }
    if (heaviest != -1) {
      assign_idx(heaviest, h, nxt, u);
      for (int v: to[u]) if (v != p && v != heaviest) {
        assign_idx(v, v, nxt, u);
      }
    }
    ridx[u] = nxt;
  }

  int lca(int u, int v) {
    while (head[u] != head[v]) {
      if (depth[head[u]] > depth[head[v]]) {
        u = parent[head[u]];
      } else {
        v = parent[head[v]];
      }
    }
    return depth[u] < depth[v] ? u : v;
  }
  // returns reference to tuple of (path fragments from x upto lca (excluding lca), those from y, lca)
  // storage of retval is reused to avoid creating short vectors on each query
  tuple<vector<pair<int, int>>, vector<pair<int, int>>, int> paths_res;
  auto& paths(int x, int y) {
    auto& [x_paths, y_paths, lca] = paths_res;
    x_paths.clear();
    y_paths.clear();
    while (head[x] != head[y]) {
      int hx = head[x], hy = head[y];
      if (depth[hx] > depth[hy]) {
        x_paths.emplace_back(x, hx); x = parent[hx];
      } else {
        y_paths.emplace_back(y, hy); y = parent[hy];
      }
    }
    if (depth[x] > depth[y]) {
      x_paths.emplace_back(x, inv[idx[y] + 1]); x = y;
    } else if (depth[x] < depth[y]) {
      y_paths.emplace_back(y, inv[idx[x] + 1]); y = x;
    }
    lca = x;
    return paths_res;
  }
  int dist(int u, int v) {
    int w = lca(u, v);
    return depth[u] + depth[v] - 2 * depth[w];
  }
  template <class F> int max_ancestor(int v, F f) {
    if (!f(v)) return -1;
    int hv = head[v];
    int p = parent[hv];
    while (p != -1 && f(p)) {
      v = p;
      hv = head[v];
      p = parent[hv];
    }
    int il = idx[hv] - 1, ir = idx[v];
    while (ir - il > 1) {
      int ic = (il + ir) / 2;
      (f(inv[ic]) ? ir : il) = ic;
    }
    return inv[ir];
  }
  int ascend(int v, int k) {
    assert(depth[v] >= k);
    int td = depth[v] - k;
    int hv = head[v];
    while (depth[hv] > td) {
      v = parent[hv];
      hv = head[v];
    }
    int rest = depth[v] - td;
    return inv[idx[v] - rest];
  }
  int move_to(int u, int v, int by) {
    int l = lca(u, v);
    int du = depth[u] - depth[l];
    int dv = depth[v] - depth[l];
    assert(by <= du + dv);
    if (by <= du) return ascend(u, by);
    else return ascend(v, du + dv - by);
  }
};


} int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  VI a(n);
  rep(i, n) cin >> a[i];
  VVI to(n);
  rep(_, n - 1) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    to[u].emplace_back(v);
    to[v].emplace_back(u);
  }
  HLD hld(to);
  static VI idx[1000010];
  rep(i, n) idx[a[i]].emplace_back(i);
  VI id;
  vector<mint> ans(n);
  rep(i, n) ans[i] = a[i];
  for (int p : primes) {
    mint inv = mint(p).inv();
    for (int pk = 1; !__builtin_mul_overflow(pk, p, &pk) && pk <= 1000000;) {
      id.clear();
      for (int v = pk; v <= 1000000; v += pk) {
        id.insert(id.end(), all(idx[v]));
      }
      ranges::sort(id, {}, [&](int u) { return hld.idx[u]; });
      int sz = id.size();
      rep(i, sz - 1) {
        int l = hld.lca(id[i], id[i+1]);
        ans[l] *= inv;
      }
    }
  }
  for (int u : hld.inv | views::reverse) if (u != 0) ans[hld.parent[u]] *= ans[u];
  for (mint x : ans) cout << x.val() << '\n';
}
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