Edu rounds:

• Problems are meant for educational, so the technique is pretty much can be found in textbooks or similar problems in the past.
• The standings/scoring/penalty will be a little bit different from normal rounds.
• The difficulty of Edu round usually have the same range as normal Div. 2.

Including minuki's points biggest difference is that there is a 12 hour open hacking phase.

During this time, you/someone else can attempt to create new, valid test cases that adhere to the problem's logic and constraints, but they can only target and test the code of one other specific participant at a time.

If a submitted solution fails on one of these newly created test cases (resulting in Time Limit Exceeded or incorrect output), the creator of the successful hack earns points (i think its +100 or something). (UPD: INCORRECT DOESNT HAPPEN)

After the 12 hour hacking phase, these newly generated test cases are incorporated into the final system testing to identify any other solutions that fail under these conditions.

why rating updation is delayed?

I'm assuming they are all cheaters, since they probably don't have to ability to solve C in 3 minutes or solve E in 9 minutes (unless your like a LGM, but they aren't, so...)

The person in spot #1 literally registered right before the contest... unbelievable

I have been on Codeforces few weeks ago. Every time I looked at the scoreboard, I felt like I was intellectually disabled...

I mean the problems don't feel that hard to me. I'm currently upsolving and was able to do solve A-D in ~25 minutes (then get stuck on E). I don't mean it in the way that there's no cheaters, but the first 4 questions aren't that hard to decide.

Maybe you can turn on Warning All mode.And I write down the bugs I have made and classify them,including bugs on problems understanding and on details.Maybe this will work

dp digit, although I couldn't do it on time. We can iterate from lower to higher bit, and for each bit in the even position, we try to iterate the number of 1 which will be add in that position. Notice that the number of 1 won't be increase, so we need to store the upper bound of this number each time we iterate.

which is Digit DP ? D? if so whats the intution

its observation and i'm sure most of them used gpt as it's very classical one for GPT.

Permutation cycle and its length is already very old idea. I'll point out just 1 problem for example that use the same idea and have almost the same people solved. (This was Div. 3 from 5 months ago)

I learned various ideas associated to permutations thanks to a great blog written by nor, although they deleted all of their blogs on codeforces a couple of months ago. Luckily, I've found their original blogs hosted on this website:

one of the best introductions to permutations

So, in essence: yes, problem C was relatively straightforward if you knew the theory + saw an observation or two, although those observations would be easier to find if you knew the theory beforehand...

it is 2.5e5*2.5e5*26 sir, how can't u TLE?

you don't need odd_remain, you can get it using suffix sum...

bro got the right mindset is already a bless, after contest I do see people instantly solve D by dp since it's classical... except me because I'm still suck at dp badly.

me too,thought they asked to find no of pairs of [L,R] or else i would have got some time for D.

me too :(

what are u even trying to do here? and why are you taking MOD ?

you were 99% correct except the floor part should have taken ceil and use this formula

want_sum=x-rem_sum
how_many=ceil(want_sum/tot)+1
if how_many<=k:
   ans+=k-how_many+1

Decide , which characters you want to place at even positions (or odd positions) , then the sum of $$$c[x]$$$ corresponding to them must be equal to number of even positions. So just find those ways using knapsack.

Every letter must be all placed in either odd or even place.Then we can DP,the state is the count of letters placed in odd and even places.Let n be the total count of letters,this is O(n^2)

But if we know the count of odd places,the count of even places can also be known.So the time can be O(26n)

The transfer is to place all same letters in either odd or even places,this can be calculated by Combination Number in O(1) if we have preprocessed.

My code

guys just wait until editorial comes out

Get sum of all ci's. now get k which is ceil(sum/2), form subseqeunces of array(remove all the zeros for better understanding) which give the sum as k. for this collections of characters the number of permutation would be=(number of elements in the subseqeunce)!*(number of non zero ci's — number of elements in the subseqeunce)!/(product of factorial of all ci's). do the same for each subseqeunce you get and add them up and mod whatever value they gave. you would need to use dp to get these subseqeunces

Every letter must be all placed in either odd or even place.Then we can DP,the state is the count of letters placed in odd and even places.Let n be the total count of letters,this is O(n^2)

But if we know the count of odd places,the count of even places can also be known.So the time can be O(26n)

The transfer is to place all same letters in either odd or even places,this can be calculated by Combination Number in O(1) if we have preprocessed.

My code

You need to calculate how many permutaions you can to do when you fixed %2 for ever character. And then you can calculate how may combinations of %2 for ever character have (DP). Check my code, if you don`t understand me.

For any valid distrbution of char (ex: 'a' in odd postion, 'b' in even position, ...), the number of sequence are the same.

• Parity Constraint:

  • Each letter must be placed entirely in even or odd indices.

• Total Positions:

  • Let n = sum(c[i]) be the total string length. assume 0 indexed string.
  • Define E = ceil(n/2) (even positions) and O = floor(n/2) (odd positions).

• Subset Sum Condition:

  • We must choose a subset of letter counts summing to E (remaining letters go to O).
  • Use dynamic programming (DP) to count valid selections.

• Once an assignment is fixed:

  • The even positions can be filled in: $$$\frac{E!}{\prod_{i \in \text{even}} c[i]!}$$$
  • The odd positions can be filled in: $$$\frac{O!}{\prod_{i \in \text{odd}} c[i]!}$$$
  • Combined, the total permutations for any valid assignment are: $$$\frac{E! \times O!}{\prod_{i=0}^{25} c[i]!}$$$

• Since this value is independent of specific assignment:

  • we multiply it by the DP count X (number of valid ways to distribute letters).

• Final Answer $$$= X \cdot \frac{E! \times O!}{\prod_{i=0}^{25} c[i]!}\mod998244353$$$

⠀

#include <bits/stdc++.h>
using namespace std;
 
#define Author_HarMit ios_base::sync_with_stdio(0); cin.tie(0);
#define ll long long
 
#define fo(i, n) for (ll i = 0; i < n; i++)
#define rfo(i, n) for (ll i = n - 1; i >= 0; i--)
 
#define in(n) ll n; cin >> n;
#define inn(n, k) ll n, k; cin >> n >> k;
#define vin(a, n) vector<ll> a(n); for (ll i = 0; i < n; i++) cin >> a[i];
#define strin(s) string s; cin >> s;
 
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
 
const ll inf = 1e18;
const ll mod = 998244353;
 
// In the end, when it's all over, all that matters is what you have done.
 
const ll N = 5e5 + 55;
vector<ll> inv(N);
vector<ll> fact(26*N);
 
ll binpow(ll a, ll b, ll m) {
    a %= m;
    ll res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}
 
void pre(){
    inv[1] = 1;
    for(ll i=2; i<N; i++)
        inv[i] = (inv[i-1] * (binpow(i,mod-2,mod)))%mod;
 
    fact[0] = 1;
    for(ll i=1; i<(26*N); i++)
        fact[i] = (fact[i-1]*i)%mod;
}
 
ll calc(ll id, ll rem, vector<ll>& a, vector<vector<ll>>& dp){
    if(id >= a.size())
        return (rem == 0);
 
    if(dp[id][rem] != -1)
        return dp[id][rem];
 
    ll ways = calc(id+1, rem, a, dp);
 
    if(a[id] <= rem)
        ways = ways + calc(id+1, rem-a[id], a, dp);
 
    return dp[id][rem] = ways;
}
 
void useless() {
    
    vin(occ,26);
    ll n = 26;
 
    ll len = 0;
 
    vector<ll> a;
 
    fo(i,n){
        if(occ[i])
            a.push_back(occ[i]);
        len += occ[i];
    }
 
    ll eve = (len+1)/2;
    ll odd = (len)/2;
 
    vector<vector<ll>> dp(30, vector<ll> (eve+10, -1));
 
    ll ways = calc(0,eve,a,dp);
 
    ll ansh = (((fact[eve] * fact[odd])%mod)* ways)%mod;
 
    ll chhed = 1;
    for(auto x : a)
        chhed = (chhed * inv[x])%mod;
 
    cout << (ansh * chhed)%mod << '\n';
}
 
signed main() {
 
    Author_HarMit;
 
    pre();
   
    ll tt = 1;
    cin >> tt;
 
    // cout << fixed << setprecision(6);
    
    while (tt--) {
        useless();
    }
}

Every letter must be all placed in either odd or even place.Then we can DP,the state is the count of letters placed in odd and even places.Let n be the total count of letters,this is O(n^2)

But if we know the count of odd places,the count of even places can also be known.So the time can be O(26n)

The transfer is to place all same letters in either odd or even places,this can be calculated by Combination Number in O(1) if we have preprocessed.

My code

For me, solution for C was obvious right after reading test cases.

Why you are think that many cheaters now?

Well,I thought my rank was 172 on the board but I got rank 82.Does this mean some cheaters were discovered

313805985 My AC recursive D

Recursion tends to be a bit slower compared to iterative methods.

But yeah, the time limit in this problem is a bit strict. My initial solution did not precompute the inverse factorial, so it had an extra log n complexity every time nCr was calculated, which caused a TLE. It passed after I precomputed the inverse factorial.

He's not using AI, There was a blog about a telegram group with around 800 members.

I was checking the group messages, and his codes are very similar to the codes that were sent in that group.