Traded respect for 2000$

I second thia

That's why you shouldn't sell your cf account

Are you implying 300iq sold the access to his codeforces account to someone else? That is a strong claim, do you have any proof of this claim?

Thou hast shattered the streak, disrupting the continuum. Thy transgression shall not find solace in forgiveness.

I don't have any real proof, and I'm not sure if I believe it either. But it's sort of an urban legend that has been going around for a while.

But there are some pieces of evidence pointing towards that:

• Large drop in rating and performance. I think if I'd stop for 2 years, my performance in contests would still be higher than what 300iq has now. (I had a long pause in 2022 and had little change afterwards). And he was a LGM, I'm only low red.
• At around the same time, marked changes in coding style: using <bits/stdc++.h> instead of individual headers, no longer using #ifdef iq, changing indent size to 4 spaces instead of 2.

the most credible online course

$100 in cryptocurrency to enroll in this course

Didn't know he meant this when he said that he is leaving CP for cryptofinance.

The problem from CEOI 2020, Roads, is in fact an easier version of Exercise 2.13 from "Computational Geometry: Algorithms and applications". Instead of connecting disjoint segments, you're asked to connect disjoint triangles.

First they took our rating with their "plag check", now they're taking our money! Stop that! Grey Lives Matter!

I think the interesting part is that both happened at the same time.

I love codeforces as well, but I don't think this blog is worth posting...

despite the common conception that all chn sounding names are the same person, maybe they j have the same code style and templates cuz they go to the same school and were taught the same style?

I'm not aware of many articles on techniques to reduce the number of FFT iterations and specific applications of the transposition principle, not limited to this theme. Are there any prominent references within the CodeForces community?

I do not fifth thia

A bonus problem: I just remembered that I liked 104848I - 1\%-Euclidean a lot. So consider yourself lucky (or unlucky) with a 13th geometry problem.

You are right. But the organization "Anshi Dafu Middle" has many ZhuJianfeng. You can request to ban them.

There are rumors that Ildar has returned to Russia. This information is also confirmed by a very important person for him in the past (I remembered communicating with this person recently).

I do not want to spread this and I am not responsible for this information.

Ildar has returned to Russia

that must be terrible for his mental health

I fifth thia

Yes, I noticed. Before yesterday's bug, had been at 1460 so I got a net +6. You're back to pupil too. GL!

Registered 5 years ago, Contribution : 55, Divisible by 5.

Awesome!!

Or they should just start holding them on codeforces. It's a different format and a different kind of problems but they're still generally good quality so I suspect the low attendance is due to the disgusting platform and them not being advertised in the right place (eg CF main page).

Basically, the idea is: we want to apply one of the usual solutions to the maximum sum on subarray problem with an additional constraint.

The additional constraint is the following one: basically, we get $$$k$$$ penalty if the subarray contains at least $$$1$$$ element, additional $$$k$$$ penalty for a subarray containing at least $$$m+1$$$ elements, additional $$$k$$$ for a subarray with at least $$$2m+1$$$ elements, and so on.

Suppose we iterate on the left border of the subarray we're interested in. We can reduce some elements by $$$k$$$ to maintain the penalty. So, the first element which we decrease by $$$k$$$ will be the $$$1$$$-st after the left border; the second element will be the $$$(m+1)$$$-th after the left border, and so on.

Obviously, iterating on the left border is too slow, if we want to model these decreases and search for the best right border. However, we can instead iterate on the remainder of the left border modulo $$$m$$$, because this leads to similar decreases. If our left border $$$l$$$ has remainder modulo $$$m$$$ equal to $$$x$$$, then we need to decrease the $$$x$$$-th element of the array, the element $$$(x+m)$$$, the element $$$(x+2m)$$$, and so on.

And now we need to find the maximum sum on subarray with the remainder of $$$l$$$ modulo $$$m$$$ equal to $$$x$$$. We can modify our prefix sum approach to consider only such subarrays of follows: while we keep track of the minimum value of $$$p_l$$$, we consider only values of $$$l$$$ having $$$l \bmod m = x$$$.

So, this is how we get a solution in $$$O(nm)$$$.

You are back specialist :( :)

I third thia

yandex cup qual? semifinal? or final? (Because if not final,I can participate.)

I'm waiting the JOISC 2024 solution page for 4 weeks.. now I might collapse and will make blunder on all contest

I wish you all the best of luck.

Hmm, why did you decide to post it now?

Correct. People have just given up on the platform since Topcoder has been putting negative effort into competitive programming.

Any updates on this?

OTZ

There is a huge difference between a messy organization and health risks.

Even if only 1-2% of all participants will get sick, it's not good. These kids aren't supposed to climb high mountains (2500+ m) to get ready for a school competition in programming, it's just totally wrong.

Topcoder has 2 divisions, first time participants would be assigned to Div2. There are only 3 problems per round, so it's hard to fully compare to the Codeforces divisions.

welcome to crypto!

300iq will win IOI of scam

I really liked TimeisMoney problems, although I don't see it as a Geometry problem :) If you like it, then I want to recommend 2014-2015 ACM-ICPC, Asia Tokyo Regional Contest J. Exhibition.

2015-2016 NEERC Northern K. Kingdom trip is one of my favorite Geometry problem. I usually do not prefer geometry problems, but I like them if it has a nice observation and is not too painful to code.

I want to introduce my problem from ptzcamp. Can you solve it in $$$O(n \log W)$$$ time?

cf: 2182

lichess rapid: 2227

where's the interview

yes!

"your boos mean nothing to me, I have seen what makes you cheer" — 300iq, probably

I hope we will be honest enough to refrain who have participated in Yandex Cup. Otherwise, it will adversely affect the rating distribution. Because you know, the problem authored by "tourist" will surely hold a huge score.

Thanks, also:

"Because you won't become GM before me hehe" — GlowCheese 2024.

They will probably appear closer to contest start, i have seen it many times

Fun problems, kudos to the authors!

I liked the blog, and I say as an expert that it helps me a lot, especially because I tend to get stuck in problems.

But most of the time it's not a random guess (only when I'm out of ideas), but something that comes naturally by imagining the problem and making examples on paper, something related to intuition.

And there's a skill that helps me a lot: knowing how to create good test cases. You can observe many things by simulating a good test case (in addition to avoiding WAs due to edge cases haha).

Codeforces: 1712

Chess.com:

Blitz: 787

Rapid: 892

I usually play blitz more, but I guess this is abnormal for you guys.

CF peak: 2437 Chess.com rapid peak: 907

My code passed your case, but for this case:

Input:
5 3
1 4 3
4 5 4
4 3 4
Output:
0

my code will give 4.

I'm not sure whether this is common.

I hacked a lot people when I try to use their codes to check my wrong code. I'm so helpless. Who could help me?

fangahawk akcube orz!

I'm talking about the overall quality and usefulness of a blog. I feel like you shouldn't post things that most of the people on Codeforces agree on just for + contribution.

Ignore this. I misunderstood the problem.

For me, 100$ is a decent amount of money. We need our money back at least.

cool

You get +10 mins added to your total submission time for each wrong verdict you receive.

Only final. Sorry for misunderstanding

no

My solution discussion stream

Waiting for Happy Hackers Programming Challenge II without patience!!!

Chess.com: 2050 (Omar_Hafez_Chess) Codeforces: 1410

Unfortinatly I didn't give competitive programming so much time Although I really enjoyed it

But it becomes so boring for me to learn new topics in Competitive Programming and Reading tutorials which is a real pain for me.. But in chess, I don't need to read long boring tutorials or watch badly prepared YouTube videos to understand a problem. In chess, I just need to imagine the board. I didn't read any book about chess or watch tutorial in chess or get a coach in it.

According to this paper, the problem is NP-hard, so if n and k are large you're out of luck. However, you can do a bit better ($$$\mathcal O\left(n 2^k\right)$$$) with a simple DP: for each $$$0 \le i \le n$$$, $$$0 \le a \le 2^k$$$, let $$$dp_{i,a}$$$ be the minimum positive number of the first $$$i$$$ elements you need to get an XOR of 0. $$$dp_{0,a} = \infty$$$, $$$dp_{i,a_i} = 1$$$, and otherwise $$$dp_{i,a} = \min \left ( dp_{i-1,a},1+dp_{i-1,a \oplus a_i} \right ) $$$. The answer is $$$dp_{n,0}$$$.

I second this thia!!1!

Get a life

There is a mistake in the last line, it should be best_i + sum(i+1,i+len) — k (it is written correctly in the sample code).

As for the reasoning, Suppose you want to find score of array of length len starting at index, then let rem=len%m. The your score would be sum(i,i+rem-1) -k + max(i+rem).

where max(i+rem) is the maximum score you can obtain if you are starting at index (i+rem) and your sub-array's length is multiple of m (i.e. m, 2m, 3m...)

Now, you can notice that max[i] = sum(i,i+m-1) + max[i+m]. Your base case would be max[n-m]= MAX(0, sum(n-m,n-1)], and max[i]=0 for all i greater than n-m and less than n-1.

After your max array is calculated you can go through all the indices and for each rem = 1,2,3....m, find out the maximum score.

CF peak: 2102
chess.com peak: 2092

Hey, can someone help me with test case 4?

Consider $$$n=4$$$. What I did is just take $$$x_{1}, x_{2}$$$ contigous numbers from 2 columns, and shift them to get the desired result. So, $$$x_{1}=1,2$$$, and correspondingly, $$$x_{2}=3,2$$$. This is just one case. We can take numbers from 3 columns, and get $$${x_{1},x_{2},x_{3}}$$$={$$$1,1,2$$$}, $$${x_{1},x_{2},x_{3}}$$$={$$$1,2,1$$$}, and $$${x_{1},x_{2},x_{3}}$$$={$$$2,1,1$$$}. Seeing the number of cases to be small, I just bruteforced over the number of integers $$$p$$$, such that I choose $$$p$$$ columns, and take $$$x_{i}$$$ many contigous values from column $$$i$$$.

#include <bits/stdc++.h>
using namespace std;
#define int int64_t

int32_t main(){
        ios::sync_with_stdio(0), cin.tie(0);

        #ifndef ONLINE_JUDGE
        freopen("inp.txt","r",stdin);
        freopen("out.txt","w",stdout);
        #endif

        int _test;
        cin>>_test;
        while(_test--){
                int n, m, ans=0;
                cin>>n>>m;
                vector<vector<int>> a(n,vector<int>(m)), d(n+1,vector<int>(m));
                for(int i=0; i<n; ++i){
                        for(int j=0; j<m; ++j){
                                cin>>a[i][j];
                        }
                }
                for(int t=1; t<=n; ++t){  //  length
                        for(int j=0; j<m; ++j){  //  column
                                for(int i=0; i<n; ++i){  //  starting point
                                        int sum=0;
                                        for(int z=0; z<t; ++z){  
                                                sum+=a[(i+z)%n][j];
                                        }
                                        d[t][j]=max(d[t][j],sum);
                                }
                        }
                }
                // bruteforce over combinations of sum n
                // taking 1 column
                for(int j=0; j<m; ++j){
                        ans=max(ans,d[n][j]);
                }
                // taking 2 columns
                for(int i=0; i<m; ++i){
                        for(int j=i+1; j<m; ++j){
                                if(n==2){
                                        ans=max(ans,d[1][i]+d[1][j]);
                                }else if(n==3){
                                        ans=max(ans,d[1][i]+d[2][j]);
                                        ans=max(ans,d[2][i]+d[1][j]);
                                }else if(n==4){
                                        ans=max(ans,d[1][i]+d[3][j]);
                                        ans=max(ans,d[3][i]+d[1][j]);
                                        ans=max(ans,d[2][i]+d[2][j]);
                                }
                        }
                }
                // taking 3 columns
                for(int i=0; i<m; ++i){
                        for(int j=i+1; j<m; ++j){
                                for(int k=j+1; k<m; ++k){
                                        if(n==3){
                                                ans=max(ans,d[1][i]+d[1][j]+d[1][k]);
                                        }else if(n==4){
                                                ans=max(ans,d[1][i]+d[1][j]+d[2][k]);
                                                ans=max(ans,d[1][i]+d[2][j]+d[1][k]);
                                                ans=max(ans,d[2][i]+d[1][j]+d[1][k]);
                                        }
                                }
                        }
                }
                if(m>=n){
                        vector<int> lst;
                        for(int i=0; i<m; ++i){
                                lst.push_back(d[1][i]);
                        }
                        sort(lst.rbegin(),lst.rend());
                        int np=0;
                        for(int i=0; i<n; ++i){
                                np+=lst[i];
                        }
                        ans=max(ans,np);
                }
                cout<<ans<<'\n';

        }
        return 0;
}

3/4 people did it with a simple $$$O(q x\log^2{n})$$$ (x is unordered map) cheese.

The two solutions I had:

1. $$$O(q\log^2{n})$$$ using a segment tree, with a treap stored in each node. Although this is theoretically better than the second solution, it would probably run slower.
2. $$$O(q\sqrt{n \log{n}})$$$ using square root decomposition and maintaining a fenwick tree for each block. For the $$$b$$$-th block, the $$$l$$$-th element of its fenwick tree stores the sum of all lengths of the elements in the $$$b$$$-th block which are the first of their kind after indice $$$l$$$. This has a very good constant factor and runs in just 600 ms (code).

Looks like really high constant factor. There are a lot of calls to inv. Changing power from recursive to iterative gives AC. Code Link. Also $$$log^2(n)$$$ is intended to fail F2. It is possible to modify the parallel binary search solution to also solve F2 in $$$knlog^2(n)$$$ by maintaining some deltas, but this will not be fast enough.

Sorry, not sure what the issue is, updated the link. Also for your implementation, I'm not sure but I believe it should be possible to store the inverse values for all the Segment tree nodes when building it. You would need to call inv only at the leaf nodes. Might also be possible to just use a double hash with a smaller modular field $$$\approx 10^6$$$ too. But all of these would increase the constant factor too, so not sure if it will AC even after removing the $$$log$$$.

Q. Why doesn't anybody talk to circles ?

Ans- Because it's pointless

Some statistics from today:

Div1: Registered — 46 Opened at least 1 problem — 38

Div2: Registered — 59 Opened at least 1 problem — 46

Total: Registered — 105 Opened at least 1 problem — 84

Codeforces Round 940 (Div. 2) and CodeCraft-23: Registered — 26432 Submitted at least 1 problem — 13722

Can we make a conclusion that Topcoder SRMs are basically dead? What's your best guess when Topcoder will stop holding them altogether given such a low attendance?

I never even got to specialist , but hopefully any day now :)

check this material: https://cp-algorithms.com/string/suffix-array.html#comparing-two-substrings-of-a-string

why are you asking? what's your concern?

No... There are still 3 tests left for the team to be chosen (COI and 2 Final TSTs).

The average altutide of a country is clearly not the same as the average altitude of where people live.

What is the meaning of "thia"?

wow!! OP

Keyser_soze_09 orz. Oh Wow, so much rating.

I'm glad someone got that reference :D

MikeMirzayanov listened to users complaining about not enough CF rounds, so he invented rounds that write themselves

It's tough to say because I'm not sure what rating I'd achieve on Codeforces if I put in the same amount of time as I did with chess. I'd probably lean towards saying that chess seemed easier for me to improve in, considering I had a higher rating after 3 years of playing chess, which is the same amount of time I've spent on Codeforces.

2030 in chess.com

CF peak: 1428, Chess: 2260 Bullet (Lichess).

Chess is a waste of time, CF is fun and rewarding, leading to valuable skills, job opportunities, and a better future.

: I'M TRYING TO RETIRE FROM CHESS BUT I'M ADDICTED :

It took long time

300iq(in base 4) move

$$$a$$$ size is n

you have a[i+1] inside a loop

if $$$i$$$ is $$$n-1$$$ you will try to take value out of $$$a$$$ bounds which is undefined behaviour

:O tourist problem

why base case dp[0]=1 ? i couldn't undestand that

Auto comment: topic has been updated by Bhanuvenkat (previous revision, new revision, compare).

before understanding the order of practice or anything else you need to learn how to be consistent. You have only solved 2 or 3 problems in months. I don't think that is enough to even meet the requirement of the practice which people tend to say.

Fabi was so close :(

CF: 2132 Lichess: 1821

did the same!

what you need to do is to divide this big number i.e. 720 with 3!.

It is because there are repeated groups in counting. and the number of repetitions for each group is 3!.

I had the same problem too. AC code:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m,h,t;
    cin>>n>>m;
    multiset <int> s;
    for(int i=0;i<n;i++){
        cin>>h;
        s.insert(h);
    }
    while(m--){
        cin>>t;
        multiset<int>::iterator it=s.upper_bound(t);
        if(it==s.begin()){
            cout<<"-1\n";
            continue;
        }
        it--;
        cout<<*it<<"\n";
        s.erase(it);
    }
}

TLE code:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m,h,t;
    cin>>n>>m;
    vector <int> v;
    for(int i=0;i<n;i++){
        cin>>h;
        v.push_back(h);
    }
    sort(v.begin(),v.end());
    while(m--){
        cin>>t;
        auto it=upper_bound(v.begin(),v.end(),t);
        if(it==v.begin()){
            cout<<"-1\n";
            continue;
        }
        it--;
        cout<<*it<<"\n";
        v.erase(it);
    }
}

As organizing IOI takes years, I don't believe the local organizers will change it by themselves. IOI 2023 was organized for 4 years and even with this amount of time spent for organization, there were some blunders $$$-$$$ in example the departure day was a horrible mess, we almost missed our flight back to Bulgaria. If a change of city is to be made, then the global IOI organizers should make such a call, with the thought in mind that there could be potential problems with the restart of IOI preparations.

CF peak — 1671

Chess.com peak — 2200 (rapid)