Dear Codeforces Team,

Thank you for notifying me about the code similarity issue. I'd like to respectfully explain the situation regarding my submissions for problems.

Contest 2143 D1

1. Problem Constraints Lead to Natural Solutions:

• This is an easy version with n ≤ 300, which naturally suggests using dynamic programming with O(n^3) complexity

• The state definition dp[l][r] is a common approach for subsequence counting problems with ordering constraints

1. Standard Coding Patterns:

• The modulo arithmetic handling (+= cur; if (>= MOD) -= MOD) is a well-known optimization technique
• The dp/ndp swap pattern is standard in iterative DP implementations to save memory
• The triple nested loop structure is typical for subsequence DP problems

1. My Problem-Solving Process:

• I approached this by considering how to track the "color constraints" through the largest red and blue elements seen so far
• The state (l, r) represents the maximum values in the two color groups, which is a logical way to ensure the coloring condition

Contest 2143 D2

1. Evidence from the author

• So now we only have O(n^2) "state updates", but each takes linear time, keeping us at O(n^3). However, the transitions are now simple enough to apply a data structure on them, such as a fenwick tree. This sentence come from the editorial, the author also said that the solution should be fenwick tree, so it is Straightforward to it.

1. Beyond compare with other similar codes

• lukagvritishvili/339140741 vs Terrorb1ade/339121844. The result of beyond compare is below:

  

  they are totally different.

• -tyler_durden/339140213 vs Terrorb1ade/339121844:

  

  a lot of details are different.

Reduction to absurdity

Assuming that I plagiarize code from other places, let's say, chatgpt. More exactly, contest 2147 G, which is a tough one, but a lot of constants solved it, this is the ac code generated by chatgpt:

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

using int64 = long long;
const int MAXA = 1000000;
const int64 MOD = 998244353;

// ----- SPF (最小素因数) を線形ふるいで作る -----
vector<int> spf(MAXA + 1);
vector<int> primes_list;

void build_spf() {
    spf[1] = 1;
    for (int i = 2; i <= MAXA; ++i) {
        if (spf[i] == 0) {
            spf[i] = i;
            primes_list.push_back(i);
        }
        for (int p : primes_list) {
            long long ip = 1LL * i * p;
            if (p > spf[i] || ip > MAXA) break;
            spf[ip] = p;
        }
    }
}

vector<pair<int,int>> factorize_int(int n) {
    vector<pair<int,int>> res;
    while (n > 1) {
        int p = spf[n];
        int c = 0;
        while (n % p == 0) { n /= p; ++c; }
        res.emplace_back(p, c);
    }
    return res;
}

int64 mod_pow(int64 a, long long e) {
    int64 r = 1 % MOD;
    a %= MOD;
    while (e > 0) {
        if (e & 1) r = (__int128)r * a % MOD;
        a = (__int128)a * a % MOD;
        e >>= 1;
    }
    return r;
}
int64 mod_inv(int64 a) { return mod_pow(a, MOD - 2); }

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    build_spf();

    int T;
    if (!(cin >> T)) return 0;
    while (T--) {
        int x, y, z;
        cin >> x >> y >> z;

        // m = x*y*z (mod MOD)
        int64 m_mod = x % MOD;
        m_mod = ( (__int128)m_mod * (y % MOD) ) % MOD;
        m_mod = ( (__int128)m_mod * (z % MOD) ) % MOD;

        // m の異なる素因数集合 {p}
        // ついでに "q divides m?" 判定用に set かソート済みベクタ
        vector<pair<int,int>> fx = factorize_int(x);
        vector<pair<int,int>> fy = factorize_int(y);
        vector<pair<int,int>> fz = factorize_int(z);

        // p -> dummy exponent（合算しても区別なく「存在」判定に使う）
        unordered_map<int,int> pExp; pExp.reserve(16);
        auto add_pf = [&](const vector<pair<int,int>>& pf){
            for (auto [p,c] : pf) pExp[p] += c;
        };
        add_pf(fx); add_pf(fy); add_pf(fz);

        // {p} をベクタに
        vector<int> pList;
        pList.reserve(pExp.size());
        for (auto &kv : pExp) pList.push_back(kv.first);
        sort(pList.begin(), pList.end());

        // E_q = sum_{p|m} v_q(p-1)
        unordered_map<int,int> Eq; Eq.reserve(32);
        for (int p : pList) {
            int t = p - 1;
            if (t <= 1) continue;
            auto pf = factorize_int(t);
            for (auto [q, e] : pf) {
                Eq[q] += e;
            }
        }

        // ans = (xyz)^(-1) * Π_{q not| m} ((q^{E_q}+q-1)/q)
        int64 ans = mod_inv(m_mod);

        auto divides_m = [&](int q)->bool{
            return binary_search(pList.begin(), pList.end(), q);
        };

        for (auto &kv : Eq) {
            int q = kv.first;
            int e = kv.second;
            if (e == 0) continue;
            if (divides_m(q)) {
                // q | m のとき寄与は 1（= 何もしない）
                continue;
            }
            int64 q_mod = q % MOD;
            int64 term_num = (mod_pow(q_mod, e) + (q_mod + MOD - 1)) % MOD; // q^e + q - 1
            int64 term = (__int128)term_num * mod_inv(q_mod) % MOD;         // (q^e + q - 1)/q
            ans = (__int128)ans * term % MOD;
        }

        cout << ans << '\n';
    }
    return 0;
}

the code come from physics0523 shared in entry 146629.

Plz don't focus on what I gained, look at what I lost, I lost 252 rating in global round 29, think about it, why did not I use chatgpt to generate the ac code?

Conclusion

I understand that code similarity detection is important for maintaining contest integrity, and I appreciate your efforts in this regard.

Thank you for your understanding and consideration.