Editorial of CodeTON Round 9 (Div. 1 + Div. 2)

#	User	Rating
1	tourist	3993
2	jiangly	3743
3	orzdevinwang	3707
4	Radewoosh	3627
5	jqdai0815	3620
6	Benq	3564
7	Kevin114514	3443
8	ksun48	3434
9	Rewinding	3397
10	Um_nik	3396

#	User	Contrib.
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	159
6	-is-this-fft-	158
7	awoo	156
8	TheScrasse	154
9	Dominater069	153
9	nor	153

Thanks for participating in the contest . We hope you liked the problems. We would love to hear your feedback in the comments.

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2039A - Shohag Loves Mod

Author: YouKn0wWho

Tutorial

Code

#include<bits/stdc++.h>
using namespace std;

const int N = 3e5 + 9;
using ll = long long;

void solve() {
  int n; cin >> n;
  for (int i = 1; i <= n; i++) {
    cout << 2 * i - 1 << ' ';
  }
  cout << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

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2039B - Shohag Loves Strings

Author: YouKn0wWho

Tutorial

Code

#include<bits/stdc++.h>
using namespace std;

const int N = 3e5 + 9;
using ll = long long;

void solve() {
  string s; cin >> s;
  int n = s.size();
  for (int i = 0; i + 1 < n; i++) {
    if (s[i] == s[i + 1]) {
      cout << s.substr(i, 2) << '\n';
      return;
    }
  }
  for (int i = 0; i + 2 < n; i++) {
    if (s[i] != s[i + 1] and s[i] != s[i + 2] and s[i + 1] != s[i + 2]) {
      cout << s.substr(i, 3) << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

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2039C1 - Shohag Loves XOR (Easy Version)

Author: YouKn0wWho

Tutorial

Code

#include<bits/stdc++.h>
using namespace std;

using ll = long long;

void solve() {
  int x; ll m; cin >> x >> m;

  int ans = 0;
  for (int y = 1; y <= min(2LL * x, m); y++) {
    if (x != y and ((x % (x ^ y)) == 0 or (y % (x ^ y) == 0))) {
      ++ans;
    }
  }
  cout << ans << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

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2039C2 - Shohag Loves XOR (Hard Version)

Author: YouKn0wWho

Tutorial

Code

#include<bits/stdc++.h>
using namespace std;

using ll = long long;

void solve() {
  int x; ll m; cin >> x >> m;

  // divisible by x
  ll p = m - m % x;
  ll ans = p / x - (x < p);
  if ((x ^ p) >= 1 and (x ^ p) <= m) ++ans;
  p += x;
  if ((x ^ p) >= 1 and (x ^ p) <= m) ++ans;

  // divisibly by y
  for (int y = 1; y <= min(1LL * x, m); y++) {
    ll cur = x ^ y;
    if (cur % y == 0) {
      ++ans;
    }
  }

  // divisible by both
  if (x <= m) {
    --ans;
  }

  cout << ans << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

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2039D - Shohag Loves GCD

Author: YouKn0wWho

Tutorial

Code (Solution 1)

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5 + 9;
int p[N];
void solve() {
  int n, m; cin >> n >> m;
  vector<int> s(m + 1);
  for (int i = 1; i <= m; i++) {
    cin >> s[i];
  }
  if (m < __lg(n) + 1) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 1; i <= n; i++) {
    cout << s[m - p[i]] << ' ';
  }
  cout << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 2; i < N; i++) {
    if (p[i]) continue;
    for (int j = i; j < N; j += i) {
      int x = j;
      while (x % i == 0) x /= i, ++p[j];
    }
  }
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

Code (Solution 2)

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5 + 9;
using ll = long long;

vector<int> d[N];
void solve() {
  int n, m; cin >> n >> m;
  vector<int> s(m + 1);
  for (int i = 1; i <= m; i++) {
    cin >> s[i];
  }
  vector<int> a(n + 1, -1);
  for (int i = 1; i <= n; i++) {
    set<int> banned;
    for (int j: d[i]) {
      banned.insert(a[j]);
    }
    for (int k = m; k >= 1; k--) {
        if (banned.find(s[k]) == banned.end()) {
            a[i] = s[k];
            break;
        }
    }
    if (a[i] == -1) {
        cout << -1 << '\n';
        return;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << a[i] << ' ';
  }
  cout << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < N; i++) {
    for (int j = i + i; j < N; j += i) {
        d[j].push_back(i);
    }
  }
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

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2039E - Shohag Loves Inversions

Author: YouKn0wWho

Tutorial

Code

#include<bits/stdc++.h>
using namespace std;

const int N = 1e6 + 9, mod = 998244353;
using ll = long long;

int dp[N]; // count of arrays that we can get if the current number of inversions is > max element of the array
void solve() {
  int n; cin >> n;
  int sum = 0;
  for (int i = n; i >= 1; i--) {
    dp[i] = (1LL * i * sum % mod + 1) % mod;
    sum = (sum + dp[i]) % mod;
  }
  int ans = n - 1; // arrays having 0 and 1 inversions
  for (int k = 3; k <= n; k++) {
    int ways = (1LL * (k - 1) * (k - 2) / 2 - 1 + mod) % mod; // count of arrays achievable such that > 1 inversion count was inserted for the first time
    ans += 1LL * ways * dp[k] % mod;
    ans %= mod;
  }
  cout << ans << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

Code (by wyrqwq, without initial casework)

#include <bits/stdc++.h>

#include <chrono>
std::mt19937 eng(std::chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return std::uniform_int_distribution<int>(l, r)(eng); }

namespace FastIO {
//	char buf[1 << 21], *p1 = buf, *p2 = buf;
//	#define getchar() (p1 == p2 && (p1 = buf, p2 = (p1 + fread(buf, 1, 1 << 21, stdin))) == p1 ? EOF : *p1++)
	template <typename T> inline T read() { T x = 0, w = 0; char ch = getchar(); while (ch < '0' || ch > '9') w |= (ch == '-'), ch = getchar(); while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar(); return w ? -x : x; }
	template <typename T> inline void write(T x) { if (!x) return; write<T>(x / 10), putchar((x % 10) ^ '0'); }
	template <typename T> inline void print(T x) { if (x > 0) write<T>(x); else if (x < 0) putchar('-'), write<T>(-x); else putchar('0'); }
	template <typename T> inline void print(T x, char en) { print<T>(x), putchar(en); }
//	inline char rChar() { char ch = getchar(); while (!isalpha(ch)) ch = getchar(); return ch; }
}; using namespace FastIO;

using i32 = int32_t;
using u32 = uint32_t;
using u64 = uint64_t;
template <uint32_t MOD> struct mint {
	static constexpr u32 get_r() {
		u32 ret = MOD;
		for (i32 i = 0; i < 4; ++i) ret *= 2 - MOD * ret;
		return ret;
	}
	static constexpr u32 r = get_r();
	static constexpr u32 n2 = -u64(MOD) % MOD;
	static_assert(r * MOD == 1, "invalid, r * MOD != 1");
	static_assert(MOD < (1 << 30), "invalid, MOD >= 2 ^ 30");
	static_assert((MOD & 1) == 1, "invalid, MOD % 2 == 0");
	u32 a;
	constexpr mint() : a(0) {}
	constexpr mint(const int64_t &b) : a(reduce(u64(b % MOD + MOD) * n2)){};
	static constexpr u32 reduce(const u64 &b) { return (b + u64(u32(b) * u32(-r)) * MOD) >> 32; }
	 constexpr mint &operator += (const mint &b) { if (i32(a += b.a - 2 * MOD) < 0) a += 2 * MOD; return *this; }
	constexpr mint &operator -= (const mint &b) { if (i32(a -= b.a) < 0) a += 2 * MOD; return *this; }
	constexpr mint &operator *= (const mint &b) { a = reduce(u64(a) * b.a); return *this; }
	constexpr mint &operator /= (const mint &b) { *this *= b.inverse(); return *this; }
	constexpr mint operator + (const mint &b) const { return mint(*this) += b; }
	constexpr mint operator - (const mint &b) const { return mint(*this) -= b; }
	constexpr mint operator * (const mint &b) const { return mint(*this) *= b; }
	constexpr mint operator / (const mint &b) const { return mint(*this) /= b; }
	constexpr bool operator == (const mint &b) const { return (a >= MOD ? a - MOD : a) == (b.a >= MOD ? b.a - MOD : b.a); }
	constexpr bool operator != (const mint &b) const { return (a >= MOD ? a - MOD : a) != (b.a >= MOD ? b.a - MOD : b.a); }
	constexpr mint operator-() const { return mint() - mint(*this); }
	constexpr mint pow(u64 n) const { mint ret(1), mul(*this); while (n > 0) { if (n & 1) ret *= mul; mul *= mul, n >>= 1; } return ret; }
	constexpr mint inverse() const { return pow(MOD - 2); }
	friend std::ostream &operator<< (std::ostream &os, const mint &b) { return os << b.get(); }
	friend std::istream &operator>> (std::istream &is, mint &b) { int64_t t; is >> t; b = mint<MOD>(t); return (is); }
	constexpr u32 get() const { u32 ret = reduce(a); return ret >= MOD ? ret - MOD : ret; }
	static constexpr u32 get_MOD() { return MOD; }
    explicit operator u32() const { return get(); }
}; using modint = mint<998244353>;

// Let's write some brute first
// dp[i][j] := current length is i, current number of inversions is j (not inserted)
// dp[i][j] -> dp[>= i + 1][[j + 1, j + i]]
// this is true for j >= 1, so let's do something when j = 0
// we can generate [0, (0 ... ), 1, 0] -> dp[>= 3][1]
// this is still kinda annoying because 1 > 1 does not hold, we process it till j >= 2
// [0, 0, ..., 0, 1, 0] -> [0, 0, ..., 0, 1, 0, 1, ..., 1]
// after that we insert an 1 before some numbers of 0 and we get dp[i][1] -> dp[>= i + 1][[j + 1, j + i - 1]]
// the answer is sum dp[i][j] for all 1 <= i <= n, j >= 1, plus 1 ([0, 0, 0 ... 1])
// actually we care nothing 'bout, j so let's say f[i] = sum dp[i][j]
// (f[i] * i - 1) -> f[i + 1], f[i + 2], ..., f[n]

#define MAXN 1000001
modint f[MAXN];
void solve() {
	int n = read<int>(); modint ans = 1, pre = 2;
	f[3] = 1;
	for (int i = 4; i <= n; ++i) 
		f[i] = pre + modint(1), pre += f[i] * modint(i) - modint(1);
	for (int i = 3; i <= n; ++i) ans += f[i];
	// f[3] : [0, 1, 0]
	// f[4] : [0, 0, 1, 0] (+1), [0, 1, 1, 0], [1, 0, 1, 0] (dp[3][1] * 2)
	print<int>(ans.get(), '\n');
}

int main() { int T = read<int>(); while (T--) solve(); return 0; }

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2039F1 - Shohag Loves Counting (Easy Version)

Author: YouKn0wWho

Tutorial

Code (by LipArcanjo)

#include<bits/stdc++.h>
using namespace std;

const int N = 2e5 + 9, mod = 998244353;
using ll = long long;

int add(int a, int b){
	a += b;
	if(a > mod) a -= mod;
	if(a < 0) a += mod;
	return a;
}

// dp[i][j] = number of arrays where starting element is i and gcd of the array is j
int dp[N], cur[N], uni[N];
int sum[N];
vector<int> d[N];
void solve() {
  int m; cin >> m;
  for (int i = 1; i <= m; i++) {
    dp[i] = cur[i] = 0;
	uni[i] = 0;
	sum[i] = 0;
  }
  int ans = 0;
  ans = 0;
  for (int i = m; i >= 1; i--) {
    for (int j: d[i]) {
      cur[j] = 0;
    }
	int sz = d[i].size();
	for(int idj = sz-1; idj >= 0; idj--){
		int j = d[i][idj];
		uni[j] = add(sum[j],sum[j]);
		for(int idk = idj+1; idk < sz; idk++){
			int k = d[i][idk];
			if(k%j) continue;
			uni[j] = add(uni[j],-uni[k]);
		}
		cur[j] = add(uni[j], - add(dp[j],dp[j]));
	}

    cur[i] += 1;

    for (int j : d[i]) {
	  dp[j] = add(dp[j],cur[j]);
	  for(auto k : d[j]){
	  	sum[k] = add(sum[k],cur[j]);
	  }
	  ans = add(ans,cur[j]);
    }
	
  }
  cout << ans << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < N; i++) {
    for (int j = i; j < N; j += i) {
      d[j].push_back(i);
    }
  }
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

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2039F2 - Shohag Loves Counting (Hard Version)

Author: YouKn0wWho

Tutorial

Code (unoptimized)

#include<bits/stdc++.h>
using namespace std;
 
const int N = 1e6 + 9, mod = 998244353;
 
inline void add(int &x, int y) {
  x = x + y >= mod ? x + y - mod : x + y;
}
int mob[N];
void mobius() {
  mob[1] = 1;
  for (int i = 2; i < N; i++){
    mob[i]--;
    for (int j = i + i; j < N; j += i) {
      mob[j] -= mob[i];
    }
  }
  for (int i = 1; i < N; i++) {
    mob[i] = (mob[i] % mod + mod) % mod;
  }
}
vector<int> divs[N];
int dp[N];
int f[N];
int tmp[N], ans[N];
void solve() {
  for (int i = 1; i < N; i++) {
    for (int d: divs[i]) {
      tmp[d] = (mod - f[d]) % mod;
      for (int c: divs[d]) {
        add(tmp[d], dp[c]);
      }
      tmp[d] = (2 * tmp[d] + 1) % mod;
    }

    // apply mobius inversion formula
    for (int d: divs[i]) {
      for (int c: divs[d]) {
        add(dp[d], 1LL * mob[c] * tmp[d / c] % mod);
      }
      add(f[d], tmp[d]);
    }

    ans[i] = ans[i - 1];
    add(ans[i], f[i]);
  }
}
 
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < N; i++) {
    for (int j = i; j < N; j += i) {
      divs[j].push_back(i);
    }
  }
  mobius();
  solve();
  int t = 1;
  cin >> t;
  while (t--) {
    int m; cin >> m;
    cout << ans[m] << '\n';
  }
  return 0;
}

Code (optimized)

#include<bits/stdc++.h>
using namespace std;
 
const int N = 1e6 + 9, mod = 998244353;
 
inline void add(int &x, int y) {
  x = x + y >= mod ? x + y - mod : x + y;
}
int spf[N];
void sieve() {
  vector<int> p;
  for(int i = 2; i < N; i++) {
    if (spf[i] == 0) spf[i] = i, p.push_back(i);
    int sz = p.size();
    for (int j = 0; j < sz && i * p[j] < N && p[j] <= spf[i]; j++) {
      spf[i * p[j]] = p[j];
    }
  }
}
int mob[N];
void mobius() {
  mob[1] = 1;
  for (int i = 2; i < N; i++){
    mob[i]--;
    for (int j = i + i; j < N; j += i) {
      mob[j] -= mob[i];
    }
  }
  for (int i = 1; i < N; i++) {
    mob[i] = (mob[i] % mod + mod) % mod;
  }
}
int c[N];
vector<int> divs[N];
void gen_divs(int n) { // not sorted
  int id = 1, x = n;
  divs[n][0] = 1;
  while (n > 1) {
    int k = spf[n];
    int cur = 1, sz = id;
    while (n % k == 0) {
      cur *= k;
      n /= k;
      for (int i = 0; i < sz; i++) {
        divs[x][id++] = divs[x][i] * cur;
      }
    }
  }
}

void prec() {
  sieve();
  // generate divisors without using push_back as its really slow on Codeforces
  for (int i = 1; i < N; i++) {
    for (int j = i; j < N; j += i) {
      c[j]++;
    }
    divs[i].resize(c[i]);
    gen_divs(i);
  }
  mobius();
}
int dp[N];
int f[N];
int tmp[N], ans[N];
void solve() {
  for (int i = 1; i < N; i++) {
    for (int d: divs[i]) {
      tmp[d] = (mod - f[d]) % mod;
      for (int c: divs[d]) {
        add(tmp[d], dp[c]);
      }
      tmp[d] = (2 * tmp[d] + 1) % mod;
    }

    // apply mobius inversion formula
    for (int d: divs[i]) {
      for (int c: divs[d]) {
        add(dp[d], 1LL * mob[c] * tmp[d / c] % mod);
      }
      add(f[d], tmp[d]);
    }

    ans[i] = ans[i - 1];
    add(ans[i], f[i]);
  }
}
 
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  prec();
  solve();
  int t = 1;
  cin >> t;
  while (t--) {
    int m; cin >> m;
    cout << ans[m] << '\n';
  }
  return 0;
}

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2039G - Shohag Loves Pebae

Author: YouKn0wWho

Tutorial

Code

#include<bits/stdc++.h>
using namespace std;
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};

const int N = 1e6 + 9, T = 1e7 + 9, RT = 33333, mod = 998244353; 
using ll = long long;

int power(int n, long long k) {
  int ans = 1 % mod;
  while (k) {
    if (k & 1) ans = (long long) ans * n % mod;
    n = (long long) n * n % mod;
    k >>= 1;
  }
  return ans;
}

int SQRT(int n) {
  int x = sqrt(n);
  while (x * x < n) ++x;
  while (x * x > n) --x;
  return x;
}

int spf[T], id[T], DIAMETER, mu[T];
vector<int> primes; // 1 indexed
int prefix_prime_count[T], prefix_sum_mu[T];
void init() {
  mu[1] = 1;
  for(int i = 2; i < T; i++) {
    if (spf[i] == 0) spf[i] = i, mu[i] = i <= DIAMETER ? 0 : -1, primes.push_back(i);
    int sz = primes.size();
    for (int j = 0; j < sz && i * primes[j] < T && primes[j] <= spf[i]; j++) {
      spf[i * primes[j]] = primes[j];
      if (i % primes[j] == 0) mu[i * primes[j]] = 0;
      else mu[i * primes[j]] = mu[i] * (primes[j] <= DIAMETER ? 0 : -1);
    }
  }
  primes.insert(primes.begin(), 0);
  for (int i = 1; i < primes.size(); i++) {
    id[primes[i]] = i;
  }
  for (int i = 2; i < T; i++) {
    prefix_prime_count[i] = prefix_prime_count[i - 1] + (spf[i] == i);
  }
  for (int i = 1; i < T; i++) prefix_sum_mu[i] = prefix_sum_mu[i - 1] + mu[i];
}
int cnt[N]; // count of nodes having each diameter
int m;
namespace GoodNumbers { // numbers which aren't divisible by the first k primes
  gp_hash_table<int, int, custom_hash> mp[RT << 1];
  int count_num(int n, int k) { // n is a floor value, returns good numbers <= n
    if (k == 0 or n == 0) return n;
    if (primes[k] >= n) return 1;
    if (n < T and 1LL * primes[k] * primes[k] > n) {
      return 1 + prefix_prime_count[n] - k;
    }
    if (mp[k].find(n) != mp[k].end()) return mp[k][n];
    int ans;
    if (1LL * primes[k] * primes[k] > n) {
      int x = upper_bound(primes.begin(), primes.begin() + k, (int)SQRT(n)) - primes.begin() - 1;
      ans = count_num(n, x) - (k - x);
    }
    else ans = count_num(n, k - 1) - count_num(n / primes[k], k - 1);
    mp[k][n] = ans;
    return ans;
  }
};

vector<pair<int, int>> v;
namespace Dirichlet {
  // good number = numbers that aren't divisible by any prime <= DIAMETER
  // we will run dirichlet imagining there exists no prime <= DIAMETER
  gp_hash_table<int, int, custom_hash> mp;
  int p_c(int n) {
    return n < 1 ? 0 : 1;
  }
  int p_g(int n) {
    return GoodNumbers::count_num(n, v.back().first);
  }
  int solve (int x) { // sum of mob[i] over 1 <= i <= x and i is a good number
    if (x < T) return prefix_sum_mu[x];
    if (mp.find(x) != mp.end()) return mp[x];
    int ans = 0;
    for (int i = 2, last; i <= x; i = last + 1) {
      last = x / (x / i);
      ans += solve(x / i) * (p_g(last) - p_g(i - 1));
    }
    ans = p_c(x) - ans;
    return mp[x] = ans;
  }
};

int count_primes(int n) {
  if (n < T) return prefix_prime_count[n];
  int x = SQRT(n);
  int k = upper_bound(primes.begin(), primes.end(), x) - primes.begin() - 1;
  return GoodNumbers::count_num(n, k) + k - 1;
}


// diameter > 2 * sqrt(m)
void solve_large() {
  // only primes are good, so count total ways
  // and subtract where gcd is prime (means all nodes have a fixed prime)
  int total_ways = 1;
  int primes_under_m = count_primes(m);
  for (auto [k, c]: v) {
    if (m <= primes[k]) break;
    total_ways = 1LL * total_ways * power((primes_under_m - k + 1) % mod, c) % mod; // 1 or a prime > k
  }
  int bad_ways = (max(0, primes_under_m - v.back().first)) % mod;
  int ans = (total_ways - bad_ways + mod) % mod;
  cout << ans << '\n';
}

// diameter <= 2 * sqrt(m)
void solve_small() {
  int ans = 0;
  for (int l = 1, r; l <= m; l = r + 1) {
    int x = m / l;
    r = m / x;
    int cur = ((Dirichlet::solve(r) - Dirichlet::solve(l - 1)) % mod + mod) % mod;
    if (cur) {
      int mul = 1;
      for (auto [k, c]: v) {
        if (x <= primes[k]) break;
        mul = 1LL * mul * power(GoodNumbers::count_num(x, k) % mod, c) % mod;
      }
      ans += 1LL * cur * mul % mod;
      ans %= mod;
    }
  }
  cout << ans << '\n';
}

vector<int> g[N];
int dp[N], up[N];
void dfs(int u, int p = 0) {
  dp[u] = 0;
  if (p) g[u].erase(find(g[u].begin(), g[u].end(), p));
  for (auto v: g[u]) {
    if (v ^ p) {
      dfs(v, u);
      dp[u] = max(dp[u], dp[v] + 1);
    }
  }
}
int pref[N], suf[N];
void dfs2(int u) {
  int sz = g[u].size();
  for (int i = 0; i < sz; i++) {
    int v = g[u][i];
    pref[i] = dp[v] + 1;
    if (i) pref[i] = max(pref[i], pref[i - 1]);
  }
  for (int i = sz - 1; i >= 0; i--) {
    int v = g[u][i];
    suf[i] = dp[v] + 1;
    if (i + 1 < sz) suf[i] = max(suf[i], suf[i + 1]);
  }
  for (int i = 0; i < sz; i++) {
    int v = g[u][i];
    int cur = up[u];
    if (i) cur = max(cur, pref[i - 1]);
    if (i + 1 < sz) cur = max(cur, suf[i + 1]);
    up[v] = cur + 1;
  }
  for (auto v: g[u]) {
    dfs2(v);
  }
}
int mx_d[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n; cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int u, v; cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  dfs2(1);
  for (int u = 1; u <= n; u++) {
    vector<int> vec;
    if (u != 1) vec.push_back(up[u]);
    for (auto v: g[u]) {
      vec.push_back(dp[v] + 1);
    }
    sort(vec.rbegin(), vec.rend());
    mx_d[u] = vec[0];
    if (vec.size() > 1) {
      mx_d[u] += vec[1];
    }
    mx_d[u] += 1;
  }
  for (int i = 1; i <= n; i++) {
    cnt[mx_d[i]]++;
    DIAMETER = max(DIAMETER, mx_d[i]);
  }

  init();

  int last_prime = 0;
  for (int i = 2; i <= DIAMETER; i++) {
    if (spf[i] == i) last_prime = i;
    if (cnt[i]) {
      int k = id[last_prime];
      if (!v.empty() and v.back().first == k) {
        v.back().second += cnt[i];
      } else {
        v.push_back({k, cnt[i]});
      }
    }
  }

  if (DIAMETER > 2 * SQRT(m)) solve_large();
  else solve_small();
  return 0;
}

Rate the Problem

2039H1 - Cool Swap Walk (Easy Version)

Author: wuhudsm

Tutorial

We can observe that this kind of path is imporatnt — when we are in $$$(x, x)$$$, we only perform one of the following two kind of moves:

Move 1

$$$ (x, x) \rightarrow (x, x+1) \rightarrow (x+1, x+1) $$$

This move transforms $$$[\ldots,a_x, a_{x+1},\ldots]$$$ into $$$[\ldots,a_{x+1}, a_{x},\ldots]$$$.

Move 2

$$$ (x, x) \rightarrow (x, x+1) \rightarrow (x, x+2) \rightarrow (x+1, x+2) \rightarrow (x+2, x+2) $$$

This move transforms $$$[\ldots,a_x, a_{x+1},a_{x+2},\ldots]$$$ into $$$[\ldots,a_{x+2}, a_{x+1},a_{x},\ldots]$$$.

Summary of the path:

Note the arrays before and after the path as $$$a$$$ and $$$a'$$$, respectively. We can see $$$a'_n=a_1$$$, and $$$[a'_1,\ldots,a'_{n-1}]$$$ can be obtained from $$$[a_2,\ldots,a_{n}]$$$ through the following transformation:

Swap any two adjacent numbers of $$$[a_2,\ldots,a_{n}]$$$, but each number can be swapped at most once.

This inspires us to use Odd-Even Sort algorithm.

Steps to Achieve the Sorted Array:

Step $$$1$$$: Initialize $$$a_1 = mn$$$: If $$$a_1 \neq mn$$$, where $$$mn$$$ is the minimum of the array, use the following path:

$$$ (1, 1) \rightarrow (1, p_1) \rightarrow (p_1, p_1) \rightarrow (p_1, n) \rightarrow (n, n) $$$

This sequence ensures that $$$a_1 = mn$$$.

Then, repeat steps $$$2$$$ and $$$3$$$ until the array is sorted.

Step $$$2$$$: Perform Odd-Even Sorting: Perform an Odd-Even Sort (a round of comparison) using the key path above on the subarray $$$a_2, \dots, a_n$$$.

Step $$$3$$$: Maintain the orderliness of $$$[a_{2}, \dots ,a_{n}]$$$ while repeatedly making $$$a_1 = mn$$$: After step $$$2$$$, we want $$$mn$$$ back to the head of the array. To achieve this, perform the following operations:

$$$ (1, 1) \rightarrow (1, n) \rightarrow (n, n) $$$

This sequence transforms the array as follows:

$$$ [a_1, a_2, \dots, a_n(a_n=mn)] \rightarrow [a'_1, a'_2, \dots, a'_n]=[a_n, a_{n-1}, a_1, a_2, \dots, a_{n-2}] $$$

When this is performed after an odd-even sort, it ensures that:

$$$mn$$$ is back to the head of the array.
The subarray $$$a_1, \dots, a_{n-1}$$$ has been cyclically shifted.

Handling Continuous Cyclic Shifts in Odd-Even Sort:

Even Length ($$$n-1$$$ is even): Cyclic shifting does not affect the odd-even sort. You can continue applying the sort as usual.
Odd Length ($$$n-1$$$ is odd): A small modification is needed. Specifically, First compare $$$(a_3,a_4),(a_5,a_6),\ldots$$$ instead of $$$(a_2,a_3),(a_4,a_5),\ldots$$$ This adjustment ensures that the odd-even sort operates correctly despite the continuous cyclic shifts.

Overall, we obtained a sorted array using $$$2n$$$ walks.

Code

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <bitset>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int N=2010;
int T,n,mn,tot;
int a[N];
vector<int> X[N],Y[N];

void path1(int num) //(1,1)->(1,2)->(2,2)->(2,3)->(3,3)->...
{
	for(int i=1;i<=n;i++)
	{
		X[num].push_back(i),Y[num].push_back(i);
		if(i!=n)
		{
			X[num].push_back(i),Y[num].push_back(i+1);
			swap(a[i],a[i+1]);
		}
	}
}

void path2(int num) //(1,1)->(1,n)->(n,n)
{
	for(int i=1;i<=n;i++)
	{
		X[num].push_back(1),Y[num].push_back(i);
		swap(a[1],a[i]);
	}
	for(int i=2;i<=n;i++)
	{
		X[num].push_back(i),Y[num].push_back(n);
		swap(a[i],a[n]);
	}
}

void walk1(int j)
{
	X[tot].push_back(j-1),Y[tot].push_back(j);
	X[tot].push_back(j-1),Y[tot].push_back(j+1);
    X[tot].push_back(j),Y[tot].push_back(j+1);
	X[tot].push_back(j+1),Y[tot].push_back(j+1);
    swap(a[j-1],a[j+1]);
}

void walk2(int j)
{
	X[tot].push_back(j-1),Y[tot].push_back(j);
	X[tot].push_back(j),Y[tot].push_back(j);
	X[tot].push_back(j),Y[tot].push_back(j+1);
	X[tot].push_back(j+1),Y[tot].push_back(j+1);
	swap(a[j-1],a[j]);
	swap(a[j],a[j+1]);
}

int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		mn=n;tot=0;
		for(int i=1;i<=n;i++)   mn=min(mn,a[i]);
		for(int i=1;i<=3*n;i++) X[i].clear(),Y[i].clear();
		int p1;
		for(int i=1;i<=n;i++) if(a[i]==mn) p1=i;
		if(p1!=1)
		{
		    tot++;
		    for(int i=1;i<=p1;i++) X[tot].push_back(1),Y[tot].push_back(i),swap(a[1],a[i]);
		    for(int i=2;i<=p1;i++) X[tot].push_back(i),Y[tot].push_back(p1),swap(a[i],a[p1]);
		    for(int i=p1+1;i<=n;i++) X[tot].push_back(p1),Y[tot].push_back(i),swap(a[p1],a[i]);
		    for(int i=p1+1;i<=n;i++) X[tot].push_back(i),Y[tot].push_back(n),swap(a[i],a[n]);
		}
		for(int i=2;i<=n;i++)
		{
			tot++;
			X[tot].push_back(1),Y[tot].push_back(1);
			if(n&1)
			{
				if(i&1)
				{
					for(int j=2;j<=n;j+=2)
					{
						if(j+1==i) walk2(j);
						else if(a[j]>a[j+1]) walk1(j);
						else walk2(j);
					}
				}
				else
				{
					for(int j=2;j<=n;j+=2)
					{
						if(a[j]>a[j+1]) walk1(j);
						else walk2(j);
					}
				}
			}
			else
			{
				if(i&1)
				{
					for(int j=2;j<=n;j+=2)
					{
						if(j==i-1)
						{
							X[tot].push_back(j-1),Y[tot].push_back(j);
							X[tot].push_back(j),Y[tot].push_back(j);
							swap(a[j-1],a[j]);
							j--;
						}
						else if(a[j]>a[j+1]) walk1(j);
						else walk2(j);
					}
				}
				else
				{
					for(int j=2;j<=n;j+=2)
					{
						if(j==i)
						{
							X[tot].push_back(j-1),Y[tot].push_back(j);
							X[tot].push_back(j),Y[tot].push_back(j);
							swap(a[j-1],a[j]);
							j--;
						}
						else if(a[j]>a[j+1]) walk1(j);
						else walk2(j);
					}
				}
			}
			path2(++tot);
		}
		printf("%d\n",tot);
		for(int i=1;i<=tot;i++)
		{
			for(int j=1;j<2*n-1;j++)
			{
			    if(X[i][j]==X[i][j-1]) printf("R");
			    else printf("D");
			}
			printf("\n");
		}

	}

	return 0;
}

Rate the Problem

2039H2 - Cool Swap Walk (Hard Version)

Author: wuhudsm

Tutorial

Code

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <bitset>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int N=2010;
int T,n,tot;
int a[N];
vector<int> X[N],Y[N];

void path1(int num)  //(1,1)->(1,2)->(2,2)->(2,3)->(3,3)->...
{
	for(int i=1;i<=n;i++)
	{
		X[num].push_back(i),Y[num].push_back(i);
		if(i!=n)
		{
			X[num].push_back(i),Y[num].push_back(i+1);
			swap(a[i],a[i+1]);
		}
	}
}

void path2(int num) //(1,1)->(1,n)->(n,n)
{
	for(int i=1;i<=n;i++)
	{
		X[num].push_back(1),Y[num].push_back(i);
		swap(a[1],a[i]);
	}
	for(int i=2;i<=n;i++)
	{
		X[num].push_back(i),Y[num].push_back(n);
		swap(a[i],a[n]);
	}
}

void path3(int num,vector<int> p) //swap(1,p[0]),(2,p[1]),... note p[0]!=1
{
	for(int i=1;i<=p[0];i++)
	{
		X[num].push_back(1),Y[num].push_back(i);
		swap(a[1],a[i]);
	}
	for(int i=1;i<p.size();i++)
	{
		for(int j=p[i-1];j<=p[i];j++)
		{
			X[num].push_back(i+1),Y[num].push_back(j);
		    swap(a[i+1],a[j]);
		}
	}
	int x=p.size(),y=p.back();
	while(x!=n)
	{
	    x++;
	    X[num].push_back(x),Y[num].push_back(y);
		swap(a[x],a[y]);
	}
	while(y!=n)
	{
	    y++;
	    X[num].push_back(x),Y[num].push_back(y);
		swap(a[x],a[y]);
	}
}

void walk1(int j)
{
	X[tot].push_back(j-1),Y[tot].push_back(j);
	X[tot].push_back(j-1),Y[tot].push_back(j+1);
    X[tot].push_back(j),Y[tot].push_back(j+1);
	X[tot].push_back(j+1),Y[tot].push_back(j+1);
    swap(a[j-1],a[j+1]);
}

void walk2(int j)
{
	X[tot].push_back(j-1),Y[tot].push_back(j);
	X[tot].push_back(j),Y[tot].push_back(j);
	X[tot].push_back(j),Y[tot].push_back(j+1);
	X[tot].push_back(j+1),Y[tot].push_back(j+1);
	swap(a[j-1],a[j]);
	swap(a[j],a[j+1]);
}

void walk3(int j)
{
	X[tot].push_back(j-1),Y[tot].push_back(j);
	X[tot].push_back(j),Y[tot].push_back(j);
	swap(a[j-1],a[j]);
}

void init()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	tot=0;
	for(int i=1;i<=3*n;i++) X[i].clear(),Y[i].clear();
	vector<pair<int,int> > pr;
	for(int i=1;i<=n;i++) pr.push_back(make_pair(a[i],i));
	sort(pr.begin(),pr.end());
	for(int i=1;i<=n;i++) a[pr[i-1].second]=i;
}

void step1()
{
	int p1,pn;
	vector<int> p;
	for(int i=1;i<=n;i++) if(a[i]==1) p1=i;
	if(p1!=1)
	{
        p.push_back(p1);
        path3(++tot,p);
	}
	if(n==2) return ;
	tot++;
	X[tot].push_back(1),Y[tot].push_back(1);
	for(int j=2;j<=n;j+=2)
	{
	    if(j+1>n) walk3(j);
	    else if(a[j]==n) walk1(j);
		else walk2(j);
	}
	p1=n;
	for(int i=1;i<=n;i++) if(a[i]==n) pn=i;
    p.clear();
    p.push_back(pn);p.push_back(p1);
    path3(++tot,p);
    p.clear();
    for(int i=1;i<=n;i++) if(a[i]<=(n+1)/2) p.push_back(i);
    path3(++tot,p);
}

void step2()
{
	int head;
	if(n&1)
	{
	    for(int t=1;t<=2;t++)
		{
            head=n/2+2;
            for(int i=1;i<=n/2+(t==1);i++)
            {
                tot++;
                X[tot].push_back(1),Y[tot].push_back(1);
                for(int j=2;j<=n;j++)
                {
                    if(!(head<=j&&j<=head+n/2-1)) walk3(j);
                    else if(j==head&&(head&1)) walk3(j);
                    else
                    {
                        if(!(head<=j+1&&j+1<=head+n/2-1)) walk3(j);
                        else if(a[j]>a[j+1]) walk1(j),j++;
                        else walk2(j),j++;
                    }
                }
                head--;
            }
		}
	}
	else
	{
		for(int t=1;t<=2;t++)
		{
            head=n/2+1;
            for(int i=1;i<=n/2;i++)
            {
                tot++;
                X[tot].push_back(1),Y[tot].push_back(1);
                for(int j=2;j<=n;j++)
                {
                    if(!(head<=j&&j<=head+n/2-1)) walk3(j);
                    else if(j==head&&(head&1)) walk3(j);
                    else
                    {
                        if(!(head<=j+1&&j+1<=head+n/2-1)) walk3(j);
                        else if(a[j]>a[j+1]) walk1(j),j++;
                        else walk2(j),j++;
                    }
                }
                head--;
            }
		}
	}
}

void output()
{
	printf("%d\n",tot);
	for(int i=1;i<=tot;i++)
	{
		for(int j=1;j<2*n-1;j++)
		{
		    if(X[i][j]==X[i][j-1]) printf("R");
		    else printf("D");
		}
		printf("\n");
	}
}

int main()
{
	scanf("%d",&T);
	while(T--)
	{
		init();
		step1();
		step2();
		output();
	}

	return 0;
}

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Comments (35)

Write comment?

Tsar1453

5 hours ago, # |

Even after solving A,B my rating went from 999 to 994 why?

→ Reply

HarshitKumarGupta

5 hours ago, # ^ |

Many people solved C1 and higher difficulty problems, so your rank dropped to 6000+ out of ~9100+ which gets calculated as a slightly lower performance than your last rating.

You can check more on rating calculation at https://mirror.codeforces.com/blog/entry/20762 or try out some CodeForces Rating Predictor for a better view on the rank vs rating.

Oh! Okay I guess I have to grind more to reach pupil

Mhn_Neek

+71

This round was a peace of shit

shiinamashiro_

+21

can't be said better

Pa_sha

-8

Why? Looks like really good round for me

eternal_happiness

4 hours ago, # ^ |

+22

probably related to the fact that E was oeis-able

methanol

Yeah that's sad. I think it is a good problem otherwise

Asuna_Yuuki

+12

PaciukZvichainyi

← Rev. 2 →

Personally, I didn't like this contest because half of the problems involve gcd, half of the problems involve combinatorics, and half of the problems involve gcd + combinatorics.

FellingRain

Alireza_OLY

can't agree more Got 1599 for this ...

amirhanordobaev

Bitmask + XOR = Pain

tafsiruzzaman

Shohag Loves Bitmasks :')

Leonardo_Paes

+29

Shohag loves OEIS (specially A079752)

Gromp16

bro changed all the bad downvotes to average, disgusting

vudinhlongg

4 hours ago, # |

_LonggVuz. bitmasks almost killed me, the problem D saved my rating ._.

Benq

← Rev. 3 →

F2: Cool problem, but why set the time limit so tight? The model solution (and most of the contestants' solutions) run in over 3s, and I couldn't get my solution to run much faster than 5s during the contest. :(

Pyqe

3 hours ago, # ^ |

I agree. It's a cool problem but the time limit is too tight

Dominater069

I liked B, C1, C2, E. Thanks for the round. H also looks interesting

the_last_smilodon

Anyone else solved C2 using binary search ? 😂 😂

risingStark

I did. I found the highest y <= m, such that x^y is a multiple of x.

Russian_Chessist

What's the approximate rating of D?

gaspar.kristijan

https://clist.by/problems/

Here you can see approximate ratings of all problems

Stalin67

Probably 1800 to 1900.

Cuellius

+31

The problem E does not request even ability to think. We need only slow solution that can calculate the number of possible arrays without modulo for $$$n < 14$$$ (which is easy to implement) and Wolfram Mathematica function FindSequenceFunction, which attempts to find a simple function that yields the sequence, when given successive integer arguments. Using the slow solution, we get the following sequence $$$(0, 1, 2, 5, 19, 102, 682, 5321, 47071, 464570, 5057058, 60166913, 776595027)$$$ of the answers. Now, we can use FindSequenceFunction to obtain the solution:

FindSequenceFunction[{0, 1, 2, 5, 19, 102, 682, 5321, 47071, 464570, 
  5057058, 60166913, 776595027}]

It gives the answer:

DifferenceRoot[Function[{y, n}, 
    {1 + (2 + n)*y[n] + (-3 - n)*y[1 + n] + 
          y[2 + n] == 0, y[1] == 0, y[2] == 1, y[3] == 2}]]

It is the recurrence relation, which can be simply coded: 292998575.

YouKn0wWho

3 hours ago, # |

+33

Some learnings as an Author: Judging by the votes and comments it seems like people mostly liked the problems individually but didn't like that most problems were mathy and didn't have enough variations. So I will try to keep that in mind and will add more variations to the contest in the future.

And I am also not happy that during the contest we found out that some people found the second difference/derivative of the sequence on OEIS. I searched on OEIS before but couldn't find it (I saw that it's not there but I didn't notice that there are some deltas matching for another sequence that were written on the OEIS page, I guess I have to be better regarding finding sequences on OEIS). Otherwise, we would have definitely modified the problem. Sorry for this issue.

Hopefully next time I will try to bring a more diverse round. Thanks for participating in the round!

Why Shohag loves so much number theory, although problems are really good.

dsarules

Bro how does D have more solves than C2

Aziz.Ammourah

Hello every one I have been trying to learn more about mobius function and it’s application but I couldn’t find any resources Can some one help me in this matter ?

arvindf232

The time limit of F2 is very bad. The solution which runs in $$$\sum_i \sigma^2(i)$$$ should have passed easily. It should not be a close call. The optimisations into $$$\sum_i \sigma(i) \times p(i)$$$ do not add any new ideas, and the constraints suggest there is a solution faster than this.

The inclusion of both C1 and C2 may make sense in terms of round balance (which frankly, is still pretty bad), but I don't see it being a good idea to test the same skill-set twice.

real.emerald

There's a cute solution to D.

Let cnt(x) be the number of divisors of x, not including 1 and x.

Then, set answer[i] to be equal to S[m - cnt(i)].

That's it :)

And if the index is ever equal to zero or negative (assuming indexing starts at 1), then output -1.

The reasoning stems from the same thoughts as the editorial, though I am too tired to write out a formal proof.

2 hours ago, # |

My solution of how managed to solve C2 but also at the same time failed miserably at it. I am able to skip an observation, but with paid a serious price in terms of implementation.

We need to check for $$$x| x \ xor \ y$$$ or $$$y | x\ xor\ y$$$. The second case is only possible if $$$y <= 2x$$$. We focus on the first case, rewrite $$$x xor y$$$ as $$$x + y - 2 (x\ and\ y)$$$. Let $$$k$$$ be the smallest integer such that $$$2^k > x$$$. Then, the value of $$$y\ mod\ 2^k$$$ is enough to determine $$$x\ and\ y$$$. For each $$$0 \leq c < 2^k$$$, we consider $$$y = a2^k + c$$$, this becomes a certain divisbility condition imposed on $$$a$$$, which can be solved with a certain CRT. This solution is now $$$O(x \log x)$$$.

To optimise, It can be seen (by looking at the implementation), that the required extended modular inverse to calculate is same for every $$$c$$$, so the $$$log x$$$ time to calculate inverse is common. This gives a solution of $$$O(x)$$$.

Being able to come up with a highly technical alternative solution, but unable to actually follow through with it in implementation, turns out to costed me greatly in this round. Submission: 292984478.

110 minutes ago, # |

I might have a little less mathy and simpler explanation for C2.

Let’s say x XOR y is a multiple of x.

So,
x XOR y = c * x (where c is any integer >= 0).

If c = 0:
x XOR y = 0 implies x = y (which is a valid case).
If c = 1:
x XOR y = x implies y = 0 (which is an invalid case since y > 0).
If c = 2:
x XOR y = 2 * x implies y = (2 * x) XOR x.

In this case, y > x, because the leftmost bit in 2 * x is not set in x. Therefore, the XOR operation introduces a set bit that makes y > x.

Thus, x XOR y is a multiple of x if and only if y > x.

Now, consider y = (c * x) XOR x for some constant c.

From this, we can deduce:
y <= c * x + x and y >= c * x - x.
So, for any multiple of x, the value of y can differ from c * x by at most x.

To solve this:

Choose all multiples of x that are <= m.
For the last multiple, check to the left and right of it to ensure the corresponding range of y is valid within m.

Personally, I used binary search to find the largest y such that y = (c * x) XOR x <= m. Then, I checked the next two multiples of x beyond this point to see if their corresponding range of y also falls within m.

Dividing the Range of `y`

Since 1 <= y <= m, we can split the range of y into two parts:

For 1 <= y <= x:
Here, we can simply loop over all values of x in O(n) and check if x XOR y is divisible by x or y.
For x < y <= m:
From the earlier reasoning, x XOR y is divisible by x. But not divisible by y, as shown in the editorial in Case 2.

This approach ensures correctness and efficiency with a O(n) solution.

Ankit_singh_sisodya

35 minutes ago, # |

xor is like subtraction without borrowing.(problem c2 editorial correction)YouKn0wWho

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Dividing the Range of y

Dividing the Range of `y`