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OMG a Chinese cheater ;-;

It's over for codeforces

Codeforces should really put a report feature because 90% of blogs now are just people calling out cheaters

obfuscation of your code isn't even allowed, is it? how come there's still no punishment for those blatant rulebreakers even after numerous posts?

i asked chatgpt to deobfuscate luuia's code, guess what it gave me:https://chatgpt.com/share/680e5c4a-077c-8006-b5bc-be68aac51e25

huhuhuhuhuhuh...

As you can see, I was the first to pass every solution. Zhoujiarun and I are from the same school. Therefore, during his Codeforces competition to fulfill his commitment, he requested the code from me and had an AI modify the code style.

He will soon be banned. Wait and see.

I asked before.

We need a reporting feature, and we also hope to give some long-time users certain powers to help maintain our community environment as volunteers.

a dramatic end for a shit-poster RIP luuia

He is even boasting around like this is an honor or something, i guess that's just another 12 yrs old kid who wants attention.

I spent about five minutes trying to decode https://mirror.codeforces.com/contest/2086/submission/313800858 and got this:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 998244353;
ll ksm(ll a,ll q,int md) {
    ll res = 1;
    a %= md;
    while(q > 0) {
        if(q & 1) res = res * a % md;
        a = a * a % md;
        q >>= 1;
    }
    return res;
}
const int maxn = 500000;
vector<ll> fac(maxn + 1),inv(maxn + 1);
void init() {
    fac[0] = 1;
    for (int i = 1;i <= maxn;i++){
        fac[i] = fac[i-1] * i % mod;
    }
    inv[maxn] = ksm(fac[maxn],mod-2,mod);
    for (int i = maxn;i >= 1;i--){
        inv[i-1] = inv[i] * i % mod;
    }
}
int main(){
    init();
    int T;cin >> T;
    while(T--){
        vector<int> a(26);
        int cnt = 0;
        for (int i = 0;i < 26;i++){
            cin >> a[i];
            cnt += a[i];
        }
        int nl = (cnt + 1) / 2;
        int nr = cnt / 2;
        vector<ll> dp(nl + 1,0);
        dp[0] = 1;
        for (int i = 0;i < 26;i++){
            if(a[i] == 0) continue;
            for (int j = nl;j >= a[i]; j--){
                dp[j] = (dp[j] + dp[j - a[i]]) % mod;
            }
        }
        ll ansl = dp[nl];
        ll ansr = fac[nl] * fac[nr] % mod;
        ll ltot = 1;
        for (int i = 0;i < 26;i++){
            ltot = ltot * fac[a[i]] % mod;
        }
        ansr = ansr * ksm(ltot,mod-2, mod) % mod;
        ll ans = ansr * ansl % mod;
        cout << ans << "\n";
    }
    return 0;
}

In some places the variable names I changed may not be suitable for their original purpose, but at least they can be viewed.

just wait and see quietly.

.

What is "xian"? Is it "咸" but may not right