#1026285 (C++17) No.2497 GCD of LCMs

• 問題: No.2497 GCD of LCMs
• ユーザー: srjywrdnprkt
• 提出日時: 2024-11-12 14:42:50
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 48 ms / 2,000 ms
• コード長: 2,472 bytes
• コンパイル時間: 2,621 ms
• コンパイル使用メモリ: 222,128 KB
• 最終ジャッジ日時: 2025-02-25 03:59:23
• ジャッジサーバーID （参考情報）: judge4 / judge5

ファイルパターン	結果
sample	AC * 3
other	AC * 14

ソースコード

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#include <bits/stdc++.h>
#include <atcoder/modint>
using namespace std;
using namespace atcoder;
using ll = long long;
using mint = modint998244353;
map<ll, ll> prime;
void prime_factor(ll n){
    prime.clear();
    ll m = n;
    if (n % 2 == 0){
        while(n % 2 == 0){
            prime[2]++;
            n /= 2;
        }
    }
    for (ll i = 3; i*i <= m; i+=2){
        if (n % i == 0){
            while(n % i == 0){
                prime[i]++;
                n /= i;
            }
        }
    }
    if (n != 1){
        prime[n]++;
    }
}
template<typename T> using pq = priority_queue<T, vector<T>, greater<T>>;
vector<ll> dijkstra(vector<vector<ll>> &E, vector<ll> &B){
    ll N = E.size();
    ll from, alt, d;
    pq<pair<ll, ll>> que;
    vector<ll> dist(N, 1e18);
    vector<bool> vst(N);
    dist[0] = B[0];
    que.push({B[0], 0});
    while(!que.empty()){
        tie(d, from) = que.top();
        que.pop();
        if (vst[from]) continue;
        vst[from] = 1;
        for (auto to : E[from]){
            alt = max(d, B[to]);
            if (alt < dist[to]){
                dist[to] = alt;
                que.push({dist[to], to});
            }
        }
    }
    return dist;
}
int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    /*
       Aの素因数pごとに考える。
       素因数pの指数eを考えると、LCMのGCD->MAXのMIN
       1->xに行く時のeの最大値の最小値でdist(x)を定める。
       p^dist(x)の積
    */
    ll N, M, u, v;
    cin >> N >> M;
    //素因数の集合
    vector<ll> v2, A(N);
    vector<mint> ans(N, 1);
    for (int i=0; i<N; i++){
        cin >> A[i];
        prime_factor(A[i]);
        for (auto [p, e] : prime) v2.push_back(p);
    }
    sort(v2.begin(), v2.end());
    v2.erase(unique(v2.begin(), v2.end()), v2.end());
    vector<vector<ll>> E(N);
    for (int i=0; i<M; i++){
        cin >> u >> v; u--; v--;
        E[u].push_back(v);
        E[v].push_back(u);
    }
    for (auto p : v2){
        vector<ll> B(N);
        for (int i=0; i<N; i++){
            ll cnt=0, X;
            X = A[i];
            while(X % p == 0){
                X /= p;
                cnt++;
            }
            B[i] = cnt;
        }
        vector<ll> dist = dijkstra(E, B);
        for (int i=0; i<N; i++){
            ans[i] *= mint(p).pow(dist[i]);
        }
    }
    for (int i=0; i<N; i++) cout << ans[i].val() << endl;
    return 0;
}
 
 
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