Codeforces Round #850 Editorial

№	Пользователь	Рейтинг
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

№	Пользователь	Вклад
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	159
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	Dominater069	153
9	nor	153

1786A2 - Чередующаяся колода (сложная версия)

Problem author: KAN

Explanation

Code by KAN

NT = int(input())
 
for T in range(NT):
	n = int(input())
	answer = [0, 0, 0, 0]
	first_card = 1
	for it in range(1, 20000):
		who = 0 if it % 4 == 1 or it % 4 == 0 else 1
		cnt = it
		if n < cnt:
			cnt = n
		cnt_white = (cnt + first_card % 2) // 2
		cnt_black = cnt - cnt_white
		answer[who * 2 + 0] += cnt_white
		answer[who * 2 + 1] += cnt_black
		first_card += cnt
		n -= cnt
		if n == 0:
			break
	assert(n == 0)
	print(*answer)

1786B - Производство тортов

Problem author: KAN

Explanation

Code by KAN

#include <bits/stdc++.h>

using namespace std;

const int inf = 1000000000;

int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    int n, w, h;
    cin >> n >> w >> h;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<int> b(n);
    for (int i = 0; i < n; i++) {
      cin >> b[i];
    }
    int minshift = -inf;
    int maxshift = inf;
    for (int i = 0; i < n; i++) {
      minshift = max(minshift, (b[i] + h) - (a[i] + w));
      maxshift = min(maxshift, (b[i] - h) - (a[i] - w));
    }
    if (minshift <= maxshift) {
      cout << "YES" << '\n';
    } else {
      cout << "NO" << '\n';
    }
  }
  return 0;
}

1785A - Монстры (простая версия)

Problem author: tourist

Explanation

Code by tourist

#include <bits/stdc++.h>

using namespace std;

int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    vector<int> b(n);
    b[0] = 1;
    for (int i = 1; i < n; i++) {
      b[i] = min(b[i - 1] + 1, a[i]);
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      ans += a[i] - b[i];
    }
    cout << ans << '\n';
  }
  return 0;
}

1785B - Обмен буквами

Problem author: tourist

Explanation

Code by PavelKunyavskiy

private fun IntArray.countOf(value: Int) = count { it == value }

private fun solve() {
    val s = "win"
    fun IntArray.bad() = (0 until 3).singleOrNull { c -> count { it == c } > 1 }
    val data = List(readInt()) { readLn().map { s.indexOf(it) }.toIntArray() }
    val ans = mutableListOf<String>()
    fun exchange(c1: Int, c2: Int, i1: Int, i2: Int) {
        ans.add("$$${i1+1} $$${s[c1]} $$${i2+1} $$${s[c2]}")
        val index1 = data[i1].indexOf(c1)
        val index2 = data[i2].indexOf(c2)
        data[i1][index1] = c2
        data[i2][index2] = c1
    }
    val todo = List(3) { List(3) { mutableListOf<Int> () } }
    for (i in data.indices) {
        val bad = data[i].bad() ?: continue
        for (j in 0 until 3) {
            if (data[i].countOf(j) == 0) {
                todo[bad][j].add(i)
            }
        }
    }

    for (i in 0 until 3) {
        for (j in 0 until 3) {
            if (i != j) {
                while (todo[i][j].isNotEmpty() && todo[j][i].isNotEmpty()) {
                    exchange(i, j, todo[i][j].removeLast(), todo[j][i].removeLast())
                }
            }
        }
    }
    while (todo[0][1].isNotEmpty() && todo[1][2].isNotEmpty() && todo[2][0].isNotEmpty()) {
        val a = todo[0][1].removeLast()
        val b = todo[1][2].removeLast()
        val c = todo[2][0].removeLast()
        exchange(0, 1, a, b)
        exchange(0, 2, b, c)
    }
    while (todo[1][0].isNotEmpty() && todo[2][1].isNotEmpty() && todo[0][2].isNotEmpty()) {
        val a = todo[1][0].removeLast()
        val b = todo[2][1].removeLast()
        val c = todo[0][2].removeLast()
        exchange(1, 0, a, c)
        exchange(2, 1, b, c)
    }
    println(ans.size)
    println(ans.joinToString("\n"))
}

fun main() {
    repeat(readInt()) {
        solve()
    }
}

private fun readLn() = readLine()!!
private fun readInt() = readLn().toInt()

1785C - Монстры (сложная версия)

Problem author: tourist

Explanation

Code by tourist

#include <bits/stdc++.h>

using namespace std;

int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<int> mn(2 * n - 1);
    vector<int> add(2 * n - 1, 0);
    vector<int> pos(2 * n - 1);
    auto Pull = [&](int x, int z) {
      mn[x] = min(mn[x + 1], mn[z]) + add[x];
      pos[x] = (mn[x + 1] <= mn[z] ? pos[x + 1] : pos[z]);
    };
    function<void(int, int, int, int, int, int)> Modify = [&](int x, int l, int r, int ll, int rr, int v) {
      if (ll <= l && r <= rr) {
        mn[x] += v;
        add[x] += v;
        return;
      }
      int y = (l + r) >> 1;
      int z = x + 2 * (y - l + 1);
      if (ll <= y) {
        Modify(x + 1, l, y, ll, rr, v);
      }
      if (rr > y) {
        Modify(z, y + 1, r, ll, rr, v);
      }
      Pull(x, z);
    };
    function<void(int, int, int)> Build = [&](int x, int l, int r) {
      if (l == r) {
        mn[x] = l;
        pos[x] = l;
        return;
      }
      int y = (l + r) >> 1;
      int z = x + 2 * (y - l + 1);
      Build(x + 1, l, y);
      Build(z, y + 1, r);
      Pull(x, z);
    };
    Build(0, 1, n);
    long long s = 0;
    long long m = 0;
    for (int i = 0; i < n; i++) {
      s += a[i];
      m += 1;
      Modify(0, 1, n, a[i], n, -1);
      if (mn[0] < 0) {
        s -= pos[0];
        m -= 1;
        Modify(0, 1, n, pos[0], n, +1);
      }
      cout << s - m * (m + 1) / 2 << " \n"[i == n - 1];
    }
  }
  return 0;
}

1785D - Деревянная ложка

Problem author: tourist

Explanation

Code by tourist

#include <bits/stdc++.h>

using namespace std;

template <typename T>
T inverse(T a, T m) {
  T u = 0, v = 1;
  while (a != 0) {
    T t = m / a;
    m -= t * a; swap(a, m);
    u -= t * v; swap(u, v);
  }
  assert(m == 1);
  return u;
}

template <typename T>
class Modular {
 public:
  using Type = typename decay<decltype(T::value)>::type;

  constexpr Modular() : value() {}
  template <typename U>
  Modular(const U& x) {
    value = normalize(x);
  }

  template <typename U>
  static Type normalize(const U& x) {
    Type v;
    if (-mod() <= x && x < mod()) v = static_cast<Type>(x);
    else v = static_cast<Type>(x % mod());
    if (v < 0) v += mod();
    return v;
  }

  const Type& operator()() const { return value; }
  template <typename U>
  explicit operator U() const { return static_cast<U>(value); }
  constexpr static Type mod() { return T::value; }

  Modular& operator+=(const Modular& other) { if ((value += other.value) >= mod()) value -= mod(); return *this; }
  Modular& operator/=(const Modular& other) { return *this *= Modular(inverse(other.value, mod())); }

  template <typename U = T>
  typename enable_if<is_same<typename Modular<U>::Type, int>::value, Modular>::type& operator*=(const Modular& rhs) {
    value = normalize(static_cast<int64_t>(value) * static_cast<int64_t>(rhs.value));
    return *this;
  }

 private:
  Type value;
};

template <typename T> Modular<T> operator*(const Modular<T>& lhs, const Modular<T>& rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T, typename U> Modular<T> operator*(const Modular<T>& lhs, U rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T, typename U> Modular<T> operator*(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T, typename U> Modular<T> operator/(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) /= rhs; }

constexpr int md = 998244353;
using Mint = Modular<std::integral_constant<decay<decltype(md)>::type, md>>;

vector<Mint> fact(1, 1);
vector<Mint> inv_fact(1, 1);

Mint C(int n, int k) {
  if (k < 0 || k > n) {
    return 0;
  }
  while ((int) fact.size() < n + 1) {
    fact.push_back(fact.back() * (int) fact.size());
    inv_fact.push_back(1 / fact.back());
  }
  return fact[n] * inv_fact[k] * inv_fact[n - k];
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  C(1 << n, 0);
  vector<Mint> dp(1 << n);
  dp[0] = (1 << n) * fact[1 << (n - 1)];
  for (int rd = n - 2; rd >= 0; rd--) {
    vector<Mint> new_dp(1 << n);
    Mint sum = 0;
    for (int i = 0; i < (1 << n); i++) {
      new_dp[i] = sum * C((1 << n) - 1 - i - (1 << rd), (1 << rd) - 1) * fact[1 << rd];
      sum += dp[i];
    }
    swap(dp, new_dp);
  }
  Mint sum = 0;
  for (int i = 0; i < (1 << n); i++) {
    cout << sum() << '\n';
    sum += dp[i];
  }
  return 0;
}

1785E - Бесконечная игра

Problem author: tourist

Explanation

Code by tourist

#include <bits/stdc++.h>

using namespace std;

template <typename T>
class Modular {
 public:
  using Type = typename decay<decltype(T::value)>::type;

  constexpr Modular() : value() {}
  template <typename U>
  Modular(const U& x) {
    value = normalize(x);
  }

  template <typename U>
  static Type normalize(const U& x) {
    Type v;
    if (-mod() <= x && x < mod()) v = static_cast<Type>(x);
    else v = static_cast<Type>(x % mod());
    if (v < 0) v += mod();
    return v;
  }

  const Type& operator()() const { return value; }
  template <typename U>
  explicit operator U() const { return static_cast<U>(value); }
  constexpr static Type mod() { return T::value; }

  Modular& operator+=(const Modular& other) { if ((value += other.value) >= mod()) value -= mod(); return *this; }
  template <typename U> Modular& operator+=(const U& other) { return *this += Modular(other); }

  template <typename U>
  friend bool operator==(const Modular<U>& lhs, const Modular<U>& rhs);

 private:
  Type value;
};

template <typename T> bool operator==(const Modular<T>& lhs, const Modular<T>& rhs) { return lhs.value == rhs.value; }
template <typename T, typename U> bool operator==(const Modular<T>& lhs, U rhs) { return lhs == Modular<T>(rhs); }

constexpr int md = 998244353;
using Mint = Modular<std::integral_constant<decay<decltype(md)>::type, md>>;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int len = (int) s.size();
  vector<Mint> ans(3);
  for (int use = 1; use < (1 << 4); use++) {
    int pw = 1 << 8;
    int init = 0;
    for (int i = 0; i < 4; i++) {
      init += i << (2 * i);
    }
    vector<vector<int>> go_state(pw, vector<int>(2));
    vector<vector<int>> go_diff(pw, vector<int>(2));
    for (int state = 0; state < pw; state++) {
      for (int put = 0; put < 2; put++) {
        int new_diff = 0;
        int new_state = 0;
        for (int i = 0; i < 4; i++) {
          int to = (state >> (2 * i)) & 3;
          if (put == 0) {
            if (to & 1) {
              if (use & (1 << i)) {
                new_diff += 1;
              }
              to = 0;
            } else {
              to |= 1;
            }
          } else {
            if (to & 2) {
              if (use & (1 << i)) {
                new_diff -= 1;
              }
              to = 0;
            } else {
              to |= 2;
            }
          }
          new_state += to << (2 * i);
        }
        go_state[state][put] = new_state;
        go_diff[state][put] = new_diff;
      }
    }
    int limit = (len + 1) / 2 * 2 * __builtin_popcount(use);
    vector<vector<Mint>> dp(2 * limit + 1, vector<Mint>(pw));
    dp[limit][init] = 1;
    for (char c : s) {
      vector<vector<Mint>> new_dp(2 * limit + 1, vector<Mint>(pw));
      for (int sum = 0; sum < (int) dp.size(); sum++) {
        for (int state = 0; state < pw; state++) {
          if (dp[sum][state] == 0) {
            continue;
          }
          for (int put = 0; put < 2; put++) {
            if ((put == 0 && c == 'b') || (put == 1 && c == 'a')) {
              continue;
            }
            int new_sum = sum + go_diff[state][put];
            int new_state = go_state[state][put];
            new_dp[new_sum][new_state] += dp[sum][state];
          }
        }
      }
      swap(dp, new_dp);
    }
    for (int sum = 0; sum < (int) dp.size(); sum++) {
      for (int state = 0; state < pw; state++) {
        if (dp[sum][state] == 0) {
          continue;
        }
        vector<int> to(4);
        for (int i = 0; i < 4; i++) {
          to[i] = (state >> (2 * i)) & 3;
        }
        vector<int> seq(1, 0);
        while (true) {
          int nxt = to[seq.back()];
          bool done = false;
          for (int i = 0; i < (int) seq.size(); i++) {
            if (seq[i] == nxt) {
              done = true;
              seq.erase(seq.begin(), seq.begin() + i);
              break;
            }
          }
          if (done) {
            break;
          }
          seq.push_back(nxt);
        }
        int real_use = 0;
        for (int x : seq) {
          real_use |= (1 << x);
        }
        if (use == real_use) {
          ans[sum > limit ? 0 : (sum < limit ? 2 : 1)] += dp[sum][state];
        }
      }
    }
  }
  for (int i = 0; i < 3; i++) {
    cout << ans[i]() << '\n';
  }
  return 0;
}

1785F - Минимумы или медианы

Problem author: tourist

Explanation

Code by tourist, O(n^2)

#include <bits/stdc++.h>

using namespace std;

template <typename T>
T inverse(T a, T m) {
  T u = 0, v = 1;
  while (a != 0) {
    T t = m / a;
    m -= t * a; swap(a, m);
    u -= t * v; swap(u, v);
  }
  assert(m == 1);
  return u;
}

template <typename T>
class Modular {
 public:
  using Type = typename decay<decltype(T::value)>::type;

  constexpr Modular() : value() {}
  template <typename U>
  Modular(const U& x) {
    value = normalize(x);
  }

  template <typename U>
  static Type normalize(const U& x) {
    Type v;
    if (-mod() <= x && x < mod()) v = static_cast<Type>(x);
    else v = static_cast<Type>(x % mod());
    if (v < 0) v += mod();
    return v;
  }

  const Type& operator()() const { return value; }
  template <typename U>
  explicit operator U() const { return static_cast<U>(value); }
  constexpr static Type mod() { return T::value; }

  Modular& operator+=(const Modular& other) { if ((value += other.value) >= mod()) value -= mod(); return *this; }
  Modular& operator/=(const Modular& other) { return *this *= Modular(inverse(other.value, mod())); }

  template <typename U = T>
  typename enable_if<is_same<typename Modular<U>::Type, int>::value, Modular>::type& operator*=(const Modular& rhs) {
    value = normalize(static_cast<int64_t>(value) * static_cast<int64_t>(rhs.value));
    return *this;
  }

 private:
  Type value;
};

template <typename T, typename U> Modular<T> operator*(const Modular<T>& lhs, U rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T, typename U> Modular<T> operator/(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) /= rhs; }

constexpr int md = 998244353;
using Mint = Modular<std::integral_constant<decay<decltype(md)>::type, md>>;

vector<Mint> fact(1, 1);
vector<Mint> inv_fact(1, 1);

Mint C(int n, int k) {
  if (k < 0 || k > n) {
    return 0;
  }
  while ((int) fact.size() < n + 1) {
    fact.push_back(fact.back() * (int) fact.size());
    inv_fact.push_back(1 / fact.back());
  }
  return fact[n] * inv_fact[k] * inv_fact[n - k];
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  Mint ans = 1;
  for (int p = 1; p <= k; p++) {
    for (int q = 0; q <= max(0, k - 1 - p); q++) {
      if (2 * (k - p - q) + 1 <= n - p) {
        ans += (q == 0 ? 1 : C((k - 1 - p) - (q - 1), q));
      }
    }
  }
  for (int q = 1; q <= k - 1; q++) {
    int pos = 2 * (k - q) + 2;
    int shifts = max(1, pos - n);
    int left = k - shifts;
    ans += C(left - (q - 1), q);
  }
  cout << ans() << '\n';
  return 0;
}

Code by tourist, O(n)

#include <bits/stdc++.h>

using namespace std;

template <typename T>
T inverse(T a, T m) {
  T u = 0, v = 1;
  while (a != 0) {
    T t = m / a;
    m -= t * a; swap(a, m);
    u -= t * v; swap(u, v);
  }
  assert(m == 1);
  return u;
}

template <typename T>
class Modular {
 public:
  using Type = typename decay<decltype(T::value)>::type;

  constexpr Modular() : value() {}
  template <typename U>
  Modular(const U& x) {
    value = normalize(x);
  }

  template <typename U>
  static Type normalize(const U& x) {
    Type v;
    if (-mod() <= x && x < mod()) v = static_cast<Type>(x);
    else v = static_cast<Type>(x % mod());
    if (v < 0) v += mod();
    return v;
  }

  const Type& operator()() const { return value; }
  template <typename U>
  explicit operator U() const { return static_cast<U>(value); }
  constexpr static Type mod() { return T::value; }

  Modular& operator+=(const Modular& other) { if ((value += other.value) >= mod()) value -= mod(); return *this; }
  Modular& operator/=(const Modular& other) { return *this *= Modular(inverse(other.value, mod())); }

  template <typename U = T>
  typename enable_if<is_same<typename Modular<U>::Type, int>::value, Modular>::type& operator*=(const Modular& rhs) {
    value = normalize(static_cast<int64_t>(value) * static_cast<int64_t>(rhs.value));
    return *this;
  }

 private:
  Type value;
};

template <typename T, typename U> Modular<T> operator*(const Modular<T>& lhs, U rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T, typename U> Modular<T> operator/(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) /= rhs; }

constexpr int md = 998244353;
using Mint = Modular<std::integral_constant<decay<decltype(md)>::type, md>>;

vector<Mint> fact(1, 1);
vector<Mint> inv_fact(1, 1);

Mint C(int n, int k) {
  if (k < 0 || k > n) {
    return 0;
  }
  while ((int) fact.size() < n + 1) {
    fact.push_back(fact.back() * (int) fact.size());
    inv_fact.push_back(1 / fact.back());
  }
  return fact[n] * inv_fact[k] * inv_fact[n - k];
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  Mint ans = 1;
  for (int q = 0; q <= k - 1; q++) {
    int bound = k - q - max(1, 2 * (k - q) - n + 1);
    ans += C(bound + 1, q + 1);
  }
  for (int q = 1; q <= k - 1; q++) {
    int pos = 2 * (k - q) + 2;
    int shifts = max(1, pos - n);
    int left = k - shifts;
    ans += C(left - (q - 1), q);
  }
  cout << ans() << '\n';
  return 0;
}

Комментарии (8)

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no_oneplb

21 месяц назад, # |

+11

Thanks for the editorial,waiting from a long time.

→ Ответить

Satrpx

Finally an editorial! After a long week and a half.

AshutoshChoudhary

11 месяцев назад, # |

Another approach for B

If we are shifting to left it means we are moving in negative direction If we are shifting to right it means we are moving in positive direction

We can try all possible values of shifting by binary search, and compare if we want to shift in positive and negative direction and reduce the search space on each try

https://codeforces.net/contest/1786/submission/238301378

priyamrai

5 месяцев назад, # ^ |

← Rev. 2 →

Why is it obvious that the ith cake should be below the ith dispenser ??It maybe that a cake's last part is below start of some dispenser and another cake is below the same dispenser ?@AshutoshChoudhary

Sunnetbayev

8 месяцев назад, # |

hi tourist

kir_tan22

6 месяцев назад, # |

help with B please? what goes wrong here?

appreciated. 262313860

xie_lzh

2 месяца назад, # |

Did anyone notice that the binomal in the the solution of F is wrong ?

2 месяца назад, # ^ |

Sorry I'm wrong

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