結果

問題 No.3026 Range LCM (Online Version)
ユーザー KumaTachiRen
提出日時 2025-02-14 23:05:30
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
MLE  
実行時間 -
コード長 3,737 bytes
コンパイル時間 6,301 ms
コンパイル使用メモリ 336,620 KB
実行使用メモリ 814,708 KB
最終ジャッジ日時 2025-02-14 23:06:40
合計ジャッジ時間 68,648 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
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 * 10 MLE * 26
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vm>;
inline ostream &operator<<(ostream &os, const mint x)
{
return os << x.val();
};
inline istream &operator>>(istream &is, mint &x)
{
long long v;
is >> v;
x = v;
return is;
};
struct Fast
{
Fast()
{
std::cin.tie(nullptr);
ios::sync_with_stdio(false);
cout << setprecision(10);
}
} fast;
#define all(a) (a).begin(), (a).end()
#define contains(a, x) ((a).find(x) != (a).end())
#define rep(i, a, b) for (int i = (a); i < (int)(b); i++)
#define rrep(i, a, b) for (int i = (int)(b) - 1; i >= (a); i--)
#define YN(b) cout << ((b) ? "YES" : "NO") << "\n";
#define Yn(b) cout << ((b) ? "Yes" : "No") << "\n";
#define yn(b) cout << ((b) ? "yes" : "no") << "\n";
using ll = long long;
using vl = vector<ll>;
using vi = vector<int>;
using vvi = vector<vi>;
int opmax(int x, int y) { return max(x, y); }
int opor(int x, int y) { return x | y; }
int e() { return 0; }
void solve()
{
const int m = 200200;
const int sq = 500;
vi ps, lpf(m + 1);
rep(i, 0, m + 1) lpf[i] = i;
rep(i, 2, m + 1)
{
if (lpf[i] == i)
{
ps.push_back(i);
for (int j = i; j <= m; j += i)
if (lpf[j] > i) lpf[j] = i;
}
}
map<int, int> ps_idx;
rep(i, 0, ps.size()) ps_idx[ps[i]] = i;
const int B = 10;
int n;
cin >> n;
vi a(n);
rep(i, 0, n) cin >> a[i];
vi ps_small, ps_large;
for (auto p : ps)
{
if (p < sq)
ps_small.push_back(p);
else
ps_large.push_back(p);
}
int x0 = ps_idx[ps_large[0]];
vector<segtree<int, opmax, e>> seg_small(ps_small.size());
rep(i, 0, seg_small.size()) seg_small[i] = segtree<int, opmax, e>(n);
vector<segtree<int, opor, e>> seg_large((ps_large.size() + B - 1) / B);
rep(i, 0, seg_large.size()) seg_large[i] = segtree<int, opor, e>(n);
rep(i, 0, a.size())
{
int v = a[i];
while (v > 1)
{
int p = lpf[v];
int c = 0;
while (v % p == 0)
{
v /= p;
c++;
}
if (p < sq)
seg_small[ps_idx[p]].set(i, c);
else
{
int x = ps_idx[p] - x0;
seg_large[x / B].set(i, 1 << (x % B));
}
}
}
vvi pow(seg_small.size());
rep(i, 0, pow.size())
{
int p = ps_small[i];
int sz = seg_small[i].all_prod();
pow[i] = vi(sz + 1);
pow[i][0] = 1;
rep(j, 1, pow[i].size())
pow[i][j] = pow[i][j - 1] * p;
}
vvm prod(seg_large.size());
rep(i, 0, prod.size())
{
int sz = seg_large[i].all_prod();
int c = min(B, (int)ps_large.size() - i * B);
prod[i] = vm(sz + 1, 1);
rep(j, 0, prod[i].size()) rep(k, 0, c) if (((j >> k) & 1) == 1) prod[i][j] *= ps_large[i * B + k];
}
int q;
cin >> q;
const ll MOD = 998244353;
mint prv_ans = 1;
while (q-- > 0)
{
ll x, z;
cin >> x >> z;
x = (x * prv_ans).val(), z = (z * prv_ans).val();
ll y = x % n, w = z % n;
int l = (int)min(y, w), r = (int)max(y, w) + 1;
mint ans = 1;
rep(i, 0, seg_small.size())
ans *= pow[i][seg_small[i].prod(l, r)];
rep(i, 0, seg_large.size())
ans *= prod[i][seg_large[i].prod(l, r)];
cout << ans.val() << "\n";
prv_ans = ans.val();
}
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
solve();
}
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