結果

問題 No.2589 Prepare Integers
コンテスト
ユーザー KumaTachiRen
提出日時 2023-12-06 20:21:12
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 3,848 bytes
コンパイル時間 6,467 ms
コンパイル使用メモリ 255,352 KB
最終ジャッジ日時 2025-02-18 08:43:47
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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 * 3 WA * 1
other AC * 5 WA * 19
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 # 

#include <bits/stdc++.h>
#include <atcoder/all>

using namespace std;
using namespace atcoder;

struct Fast {
  Fast() {
    std::cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << setprecision(10);
  }
} fast;

#define all(a) (a).begin(), (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)--)
using vi = vector<int>;
using ll = long long;
using vl = vector<ll>;

ll safe_mod(ll a, ll b) {
  ll c = a % b;
  return c < 0 ? c + b : c;
}

ll ext_gcd(ll a, ll b, ll& x, ll& y) {
  for (ll u = y = 1, v = x = 0; a;) {
    ll q = b / a;
    swap(x -= q * u, u);
    swap(y -= q * v, v);
    swap(b -= q * a, a);
  }
  return b;
}

const ll A_MAX = 1e9;
int lg;
ll k;
vl powk, basis, cnt, msd;
bool is_prime;

ll op(ll a, ll b) {
  if (k == 2) return a ^ b;
  ll ret = 0, pow = 1;
  while (a | b) {
    ret += (a + b) % k * pow;
    a /= k;
    b /= k;
    pow *= k;
  }
  return ret;
}

ll op_pow(ll a, ll n) {
  if (k == 2) return n & 1 ? a : 0;
  if (n == 0) return 0;
  n = safe_mod(n, k);
  ll ret = 0, pow = 1;
  while (a) {
    ret += safe_mod(a % k * n, k) * pow;
    a /= k;
    pow *= k;
  }
  return ret;
}

void query1(ll x) {
  if (k == 2) {
    rrep(i, 0, lg) {
      if (x >> i) {
        if (basis[i] == 0) {
          basis[i] = x;
          msd[i] = 1;
          break;
        }
        x ^= basis[i];
      }
    }
  } else if (is_prime) {
    rrep(i, 0, lg) {
      if (x / powk[i]) {
        if (basis[i] == 0) {
          ll s, t;
          ext_gcd(x / powk[i], k, s, t);
          basis[i] = op_pow(x, s);
          msd[i] = 1;
          break;
        }
        x = op(x, op_pow(basis[i], -(x / powk[i])));
      }
    }
  } else {
    vl a{x};
    rrep(i, 0, lg) {
      ll g = msd[i];
      vl v(a.size()), prod(a.size() + 1);
      rep(j, 0, a.size()) {
        ll u = a[j] / powk[i];
        ext_gcd(g, u, prod[j], v[j]);
        g = g * prod[j] + u * v[j];
        prod[j] = safe_mod(prod[j], k);
        v[j] = safe_mod(v[j], k);
      }
      rrep(j, 1, a.size()) prod[j - 1] = safe_mod(prod[j - 1] * prod[j], k);
      ll t = op_pow(basis[i], prod[0]);
      prod[a.size()] = 1;
      rep(j, 0, a.size()) t = op(t, op_pow(a[j], v[j] * prod[j + 1] % k));

      rep(j, 0, a.size()) a[j] = op(a[j], op_pow(t, -a[j] / powk[i] / g));
      a.push_back(op(basis[i], op_pow(t, -msd[i] / g)));

      if (g != k) {
        basis[i] = t;
        msd[i] = g;
      }
    }
  }
  rep(i, 0, lg) cnt[i + 1] = cnt[i] * (k / msd[i]);
}

void query2(ll x) {
  if (x > cnt[lg]) {
    cout << -1 << "\n";
    return;
  }
  x--;
  ll ans = 0;
  rrep(i, 0, lg) {
    if (basis[i] == 0) continue;
    ll u = msd[i];
    ll v = ans / powk[i];
    ans = op(ans, op_pow(basis[i], x / cnt[i] - v / u));
    x %= cnt[i];
  }
  cout << ans << "\n";
}

void query3(ll x) {
  x++;
  ll ans = 0, y = 0;
  rrep(i, 0, lg) {
    if (basis[i] == 0) {
      ll u = x / powk[i] % k, v = y / powk[i] % k;
      if (u == v) continue;
      if (u > v) ans += cnt[i];
      break;
    } else {
      ll d = y / powk[i] % k;
      ll u = msd[i];
      y = op(y, op_pow(basis[i], -(d / u)));
      d %= u;
      ll v = x / powk[i] % k - d;
      ans += (v + u - 1) / u * cnt[i];
      if (v % u) break;
      y = op(y, op_pow(basis[i], v / u));
    }
  }
  cout << ans << "\n";
}

int main() {
  int q;
  cin >> k >> q;

  is_prime = true;
  for (ll d = 2; d * d <= k; d += (d & 1) + 1)
    if (!(k % d)) {
      is_prime = false;
      break;
    }

  powk = vl();
  for (ll p = 1; p <= A_MAX; p *= k) powk.push_back(p);
  lg = (int)powk.size();
  basis = vl(lg, 0);
  msd = vl(lg, k);
  cnt = vl(lg + 1);
  cnt[0] = 1;

  while (q--) {
    int t;
    ll x;
    cin >> t >> x;
    if (t == 1) query1(x);
    if (t == 2) query2(x);
    if (t == 3) query3(x);
  }
}
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