Codeforces Round #835 (Div. 4)

4 года назад, скрыть # ^ |

+11

You have exposed yourself to danger by using unordered_set. During the hacking stage in div3/div4/edu codeforces contests, such solutions are specifically targeted by hackers. See the https://codeforces.me/blog/entry/62393 blog for more details. Even if your solution wasn't hacked directly, all successful hacks become a part of the system test and still ruin your day.

Python users are also in danger for exactly the same reason, see the https://codeforces.me/blog/entry/101817 blog. This made solutions like 182003822 vulnerable.

→ Ответить

»

Nakano_Miku

4 года назад, скрыть # ^ |

0

So do you mean no one should use hash maps on cf because there will always be someone who can hack your solution and make the hash operations O(N) instead of O(1)?

→ Ответить

»

4 года назад, скрыть # ^ |

0

You can use hash maps wherever they meet your needs. But you have to ensure that there is little to no possibility of constructing custom input data by abusing the hashing function, provided that an attacker knows it.

Personally, I would not rely on out-of-the-box usage of unordered_map even at Div.1/Div.2 contests. Hash map hacks did happen there as well.

→ Ответить

»

prxshu

4 года назад, скрыть # |

0

I'd really appreciate if someone could help me with the logic for F.

My approach

→ Ответить

»

puffinmuffin

4 года назад, скрыть # ^ |

0

The answer can also be infinity : when the sum of a[i] over min(days,n) >= C

→ Ответить

»

Matveyko

4 года назад, скрыть # ^ |

0

I think answer is Infinity if sum of min(n, d) greatest A[i] >= C since we still can do different tasks

→ Ответить

»

rayated

4 года назад, скрыть # |

0

Can anyone help me with my submission for 1760G - SlavicG's Favorite Problem : 182028374.

→ Ответить

»

variety-jones

4 года назад, скрыть # ^ |

0

Take a look at Ticket 16547 from CF Stress for a counter example.

→ Ответить

»

rayated

4 года назад, скрыть # ^ |

0

Thank you.

→ Ответить

»

Batman_love_dp

4 года назад, скрыть # |

0

can someone share the idea and intuition behind G?

→ Ответить

»

xiaossr

4 года назад, скрыть # ^ |

0

One important property of XOR is that a^a = 0. So if you do two BFS, one rooted at A and one rooted at B, you can find nodes in the A BFS and B BFS where the XOR value is equal, so you can teleport from one path to another and reach B with XOR of 0. If this exists (there are some edge cases to consider), then it is possible.

→ Ответить

»

Batman_love_dp

4 года назад, скрыть # ^ |

0

oh thanks alot. sorry i have another question is applying dfs with the same concept you mentioned should fail? because i tried to do this using two dfs but it failled in test2 or its just an implementation issue?

→ Ответить

»

chromate00

4 года назад, скрыть # ^ |

0

I think you forgot this case: You should not consider the vertices in A's DFS/BFS where B coincides with the (unique) simple path starting from A.

→ Ответить

»

4 года назад, скрыть # ^ |

0

Try this testcase (It's possible to teleport from node 1 to node 2. And then go '2 -> 3 -> 4', so the answer is YES):

→ Ответить

»

lazy_loop

4 года назад, скрыть # |

0

solved all problems in div4 for the 2nd time. feels amazing.

→ Ответить

»

sperm

4 года назад, скрыть # |

0

What's the 53rd case of second test case in G?

→ Ответить

»

rdjava9

4 года назад, скрыть # ^ |

+1

It's about the fact that we can teleport from a too. That means if there exist any (prefix xor starting from b) == 0 then also answer is True.

→ Ответить

»

Dghemy

4 года назад, скрыть # |

← Rev. 2 →

0

can one explain why in D

10 10 8 10 10 4

output NO ?

→ Ответить

»

lazy_loop

4 года назад, скрыть # ^ |

0

because it has more than 1 valley. one at value 8 and another at value 4.

→ Ответить

»

cleanser

4 года назад, скрыть # |

0

Did any1 else mess up F for the infinity case?

→ Ответить

»

lazy_loop

4 года назад, скрыть # ^ |

0

yup

→ Ответить

»

cleanser

4 года назад, скрыть # ^ |

0

Lmao. I just checked if biggest element was enough instead of the biggest d-elements, or just all if d is big.

→ Ответить

»

Imran435

4 года назад, скрыть # |

0

My solution (as following) gives wrong answer 2653rd numbers differ - expected: '4', found: '3' for 1760E — Binary Inversions.

Can anyone help me with testcases that gives wrong answer on test 2.

public void solve() {
	int n = in.nextInt();
	int[] a = new int[n];
	int tcount_0 = 0;
	for (int i = 0; i < n; i++) {
		a[i] = in.nextInt();
		if (a[i] == 0)
			tcount_0++;
	}
	long inversions = 0, count_0 = 0;
	for (int i = 0; i < n; i++) {
		if (a[i] == 0)
			count_0++;
		else
			inversions += tcount_0 - count_0;
	}
	long ans = Long.MIN_VALUE;
	long count_1 = count_0 = 0;
	for (int i = 0; i < n; i++) {
		if (a[i] == 0) {
			count_0++;
			ans = Math.max(ans, tcount_0 - count_0 - count_1);
		} else {
			ans = Math.max(ans, count_1 - tcount_0 + count_0);
			count_1++;
		}
	}
	out.println(inversions + ans);
}

→ Ответить

»

B_Consistent

4 года назад, скрыть # ^ |

+5

check once with input "10". I think you need to initialize with ans=0.

→ Ответить

»

unnatkumar

4 года назад, скрыть # |

0

My rating is 1400 and my standings show that this contest is rated for me. So, I am a little bit confused that will my rating will change or not ...

→ Ответить

»

abcsumits

4 года назад, скрыть # ^ |

0

:)

→ Ответить

»

fintai

4 года назад, скрыть # |

← Rev. 2 →

+2

t seemed like mimemamomu warmuping

→ Ответить

»

Kartik.

4 года назад, скрыть # |

0

How to solve F?

→ Ответить

»

smilinginpain

4 года назад, скрыть # ^ |

0

Binary Search on Answer(value of k)

→ Ответить

»

pathak_12

4 года назад, скрыть # |

0

Can anyone help in problem G? I'm getting MLE. My Solution

Approach: Starting bfs from 'a' and calculating zor as i move and checking if zor becomes equal to any edge weight that is adjacent to 'b'(so that i will teleport from here).

→ Ответить

»

ZrjaK

4 года назад, скрыть # ^ |

0

There are too many statements during bfs. You can save the weight of not only the adjacent edge but from any point to "b", so that the amount of statement will be small.

→ Ответить

»

usehttps

4 года назад, скрыть # |

+7

Apologies if this isn't the correct place to raise suspicions, I'm new here and couldn't figure out how to report users.

There is something weird I noticed about this user — https://codeforces.me/submissions/9999999999/contest/1760

You can see him making multiple submissions specifically designed to be hacked if there is a single testcase (on the first and trivial problem). He does it by hardcoding wrong input when there is only 1 test case like so:

if(t==1) cout<<"500949585450"<<endl;

(example submission https://codeforces.me/contest/1760/submission/182047190 )

I'm not sure what the purpose is. Maybe it's a multi-account of the hacker — https://codeforces.me/profile/Adel_Mahmoud and he's just farming points? Can you think of any other explanation?

→ Ответить

»

Away_in_the_heavens

4 года назад, скрыть # |

-17

gimme downvotes pls

→ Ответить

»

spearOfGod

4 года назад, скрыть # |

+30

So, I promise that as soon as I become pupil, I will try to give my perspective of how I make my solutions, I don't think that my code is worth but maybe my perspective can give you help in the future.

A

This one is the easiest, but I think that always that you read that you must give a number that is not the minimum or maximum of a list, you must think in make a vector, sort and print the position that you are interested. Like for example the second minimum would be a[1] or the second maximum a[n-2].

#include <bits/stdc++.h>

using namespace std;
/*

  


*/
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);  
}

int lcm(int a,int b)
{
    return (a*b)/gcd(a,b);
}

void solve(){  
    vector<int> a(3);
    for(auto &x : a){cin >> x;}
    sort(a.begin(),a.end());
    cout << a[1] << '\n';
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    //cout.precision(10);
    int t = 0;
    cin >> t;
    for(int p = 0; p < t;p++){
        solve();
    }
}

B

The important thought that you must find is that if the maximum letter is 'a' you only need one and 'z' you need 26 and with only make in your mind the solution of two case you began to see what you need to do, I was a little nervous when I was writing the code and was thinking what is better if sort the string and see the last letter or used the function max_element, when you have dudes but you know that the code that you are in the middle of writing will work too, is best to commit to what you are writing in the moment that change to another solution in my experience. This work good in first or second problem for me because the solutions that you think probably will be right because they try to make easy for beginners like me, this is my way to be faster.

Another useful type for the problems that want you to make a letter a number is remember that 'a'-'a' is 0 or a character minus the same character will return 0.

#include <bits/stdc++.h>

using namespace std;
/*

  


*/
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);  
}

int lcm(int a,int b)
{
    return (a*b)/gcd(a,b);
}

void solve(){  
    int n;
    cin >> n;
    string s;
    cin >> s;
    sort(s.begin(),s.end());
    cout << s[n-1] - 'a' + 1 << '\n';
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    //cout.precision(10);
    int t = 0;
    cin >> t;
    for(int p = 0; p < t;p++){
        solve();
    }
}

C

Here strike again the idea that we see in the problem A, in this problem you only care about the maximum and the second maximum so you make a vector, sort and use a[n-1] when the value x is not the maximum and a[n-2] if x is equal to the maximum, but I used a different method of only stored the two maximum numbers in a vector because I think a better version before beginning to write.

#include <bits/stdc++.h>

using namespace std;
/*

  


*/
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);  
}

int lcm(int a,int b)
{
    return (a*b)/gcd(a,b);
}

void solve(){  
    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(2);
    for(auto &x : a){cin >> x; 
        if(b[1] < x){
            b[0] = b[1];
            b[1] = x;
        }
        else if(b[1] == x){
            b[0] = b[1];
        }
        else if(b[0] < x){
            b[0] = x;
        }
        
    }
    for(auto x : a){
        if(x != b[1]){
            cout << x - b[1] << " ";
        }
        else{
            cout << x - b[0] << " ";
        }
    }
    cout << '\n';
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    //cout.precision(10);
    int t = 0;
    cin >> t;
    for(int p = 0; p < t;p++){
        solve();
    }
}

D

Is curious because I have come across this kind of problems many times that they want you to search there is a "Valley" or a "Peak" in the vector I still don't have a good way of approach this kind, but they always give you the conditions that must meet so the idea is always searching in a for if the condition is meet and see what are the corners case.

In this case you want to focus in every point because you can make L=R, the segments with same numbers but for example if you have a segment equal to {3,3,3,3} you don't want to waste your time with {3},{3,3},{3,3,3} because they will contradict the condition a[l−1]>a[l] or a[r] < a[r+1] other that this is good to keep in mind that l=0 or r = n-1 are free passes.

#include <bits/stdc++.h>

using namespace std;
/*

  


*/
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);  
}

int lcm(int a,int b)
{
    return (a*b)/gcd(a,b);
}

void solve(){  
    int n;
    cin >> n;
    int i = 1;
    int l = 0;
    vector<int> a(n);
    cin >> a[0];
    int h = a[0];
    int ans = 0;
    int cnt = 1;
    for(i;i < n;i++){
        int x;
        cin >> a[i];
        x = a[i];
        if(x != h){
            // cout << l << " " << i <<  " ";
            if(l == 0){
                if(a[i-1] < a[i]){ans++;}
            }
            else if(a[l-1] > a[l]){
                if(a[i-1] < a[i]){ans++;}

            }
            // cout << ans << '\n';
            l = i;
            h = x;
        }
    }
    // cout << ans << " ";
    if(n == 1){
        cout << "YES\n";
        return;
    }

    if(a[n-1] == a[n-2]){
        if(l == 0){ans++;}
        else if(a[l-1] > a[l]){ans++;}
    }
    else{
        if(a[n-2] >  a[n-1]){ans++;}
    }
    // cout << ans << " ";
    ans == 1 ? cout << "YES\n" : cout << "NO\n";
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    //cout.precision(10);
    int t = 0;
    cin >> t;
    for(int p = 0; p < t;p++){
        solve();
    }
}

E

this one is just in my reach of knowledge so when I find problems that seem difficult at the beginning, I always try to see what type of input I can solve and corner case to corner case I will build my solution.

The first one is what happen if n == 1? in this case no matter why I do the answer will be 0
what happen if there are only ones? then the best is to make the last number a 0 and the answer will be n-1
if there are only zeros? make the first number a 1 and the answer will be n-1

Now it's apparent that the best is to make the last one a zero or the first one converted to one and see how much we gain or lost for each operation and after seeing if it is good to make a change, go for all the vector and every see how much it will give to the answer. Another tip is to use comments when you feel that you are confused about what you have done and what not.

If you see my code is very inefficient but show that divide the problem in easy problems help you to see a solution.

#include <bits/stdc++.h>

using namespace std;
/*

  


*/
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);  
}

int lcm(int a,int b)
{
    return (a*b)/gcd(a,b);
}

void solve(){  
    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> pos;
    long long tots = 0;
    int i = 1;
    int fz = -1;
    int posz = 0;

    for(auto &x : a){
        cin >> x;
        if(x == 1){pos.push_back(i);}
        if(x == 0 && fz == -1){fz = i;posz =  i-1;}
        i++;
    }
    int n1 = pos.size();
    int cntz = n-n1;
    if(n == 1){cout << "0\n";return;}
    if(n1 == 0 || fz == -1 || cntz == 0){
        cout << n-1 << '\n';return;
    }
    int val1 = cntz-1-posz;// first z to 1
    int val2 = (n1-1)-(n-pos.back());//last 1 to z
    // cout << val1 << " " << val2 << '\n';
    if(val1 > val2){
        a[posz] = 1;
        cntz--;
    }
    else if(val2 > 0){
        a[pos.back()-1] = 0;
        cntz++;
    }
    for(int i = 0;i < n;i++){
        if(a[i] == 0){cntz--;}
        else{
            tots += cntz;
        }
    }
    cout << tots << '\n'; 
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    //cout.precision(10);
    int t = 0;
    cin >> t;
    for(int p = 0; p < t;p++){
        solve();
    }
}

I hope that was helpful this in some way, sorry if was not as good as you expect, when I became a better, I will make up for these bad tutorials.

→ Ответить

»

Everule

4 года назад, скрыть # ^ |

← Rev. 5 →

+25

It's not a bad tutorial. For some formatting tips I would recommend enclosing variables in dollar signs to get the mathematical look of $$$x$$$. You can use a_{x}for $$$a_x$$$ and a^{x} for $$$a^{x}$$$. This is called latex and its quite commonly used in math. You should also use links for your codes instead of pasting them.

For the solution themselves, you can use some advanced implementation tricks to make the code really simple and bug proof.

For problem D for example, we see 2 phases. decreasing, then increasing.

So we can define a phase function like


vector<int> a(n);
...
auto walk_while_decreasing = [&](int start){
    while(start + 1 < n && a[start] >= a[start + 1]) ++start;
    return start;
    //Similarly for increasing
};
int reach = walk_while_increasing(walk_while_decreasing(0));
if(reach == n - 1) //the array consists of these 2 phases
else // it doesn't

You could also get the exact values of $$$l$$$ and $$$r$$$ using the intermediate values of this calculation.

For problem E you need to only consider the change of inversion. This is the same as number of pairs $$$(i, j)$$$ such that $$$i \lt j$$$ and $$$a_i = 1$$$ and $$$a_j = 0$$$. If we consider what happens when we change a $$$0$$$ at index $$$x$$$ to a $$$1$$$ for example, we get more inversions from the $$$0$$$s to the right of $$$x$$$, and less inversions from the $$$1$$$s to the left of $$$x$$$. Therefore the best $$$0$$$ to flip would be on the left. But this also allows you to calculate the change in inversions from changing any index $$$x$$$.

→ Ответить

»

4 года назад, скрыть # |

0

Am I the only weirdo who used segment tree to solve C?

→ Ответить

»

4 года назад, скрыть # ^ |

0

How did you do it?

→ Ответить

»

4 года назад, скрыть # ^ |

0

How do we easily calculate the maximum of all array elements excluding the i-th element? It's possible to just do two range queries in a segment tree (the array part before the i-th element and the array part after the i-th element) and use the largest of these two results. In submission 181925712 I have only written the "main" function. Everything else is a copy of the segment tree implementation from the atcoder library.

→ Ответить

»

4 года назад, скрыть # ^ |

0

Nice idea lol

→ Ответить

»

4 года назад, скрыть # ^ |

0

Well, I just wanted to save time instead of inventing some original solution for this particular problem (and then proving that it correctly handles all corner cases). I'm not a particularly fast problem solver and taking care of A+B+C took me 13 minutes. Without segment tree this time could be even worse.

Your solution uses sort and you have beaten me by solving A+B+C in only 5 minutes. But now imagine not having sort function in the standard library. In this case you would need to find some decent bug free $$$O(N log N)$$$ sort implementation and copy/paste it as a part of your solution. Implementing your own sort from scratch during the contest would be a fool's errand and waste of time. Segment tree is pretty much the same and the only difference is that it is typically not provided in standard libraries of programming languages, so I need to copy/paste the atcoder's implementation.

My $$$O(N log N)$$$ segment tree solution implemented in D language can be even tweaked to remove all loops and branches 182184534 (except for the main loop iterating over testcases) and this reduces cyclomatic complexity. For comparison, a faster Wanderkind's $$$O(N)$$$ solution 182072909 also implemented in D language is more complicated. But both are fast enough and coding speed/convenience is IMHO more important during a contest.

→ Ответить

»

4 года назад, скрыть # ^ |

0

I agree with you on that. But most programming languages already have sort function. I personally read the problem quickly and the sample explanations and coded it. I know that I can't solve a lot of problems so I depend on speed instead. The problem doesn't really need proving but if you're looking for special cases and a full proven solution then yes Segment Tree was much better

→ Ответить

»

Wanderkind

4 года назад, скрыть # ^ |

0

Is there a reason you didn't use the sort function in std.algorithm?

I didn't because I am new to this language and I am not yet familiar with handling arrays.

→ Ответить

»

4 года назад, скрыть # ^ |

0

Because segment tree just happened to be my first idea after reading the problem statement. But I used sort for solving problems A and F.

→ Ответить

»

dalex

4 года назад, скрыть # ^ |

0

I imagine not having sort and finding first and second maximums

→ Ответить

»

4 года назад, скрыть # ^ |

0

This was exactly the point of my comment. Both sort and segment tree are not necessary for solving this problem. They both actually degrade time complexity. But they both are also easy to use and save time (tourist used sort 181903982 and the editorial suggests sort too). Sort is included in the standard library, while segment tree has to be copy/pasted.

→ Ответить

»

chrisTris

4 года назад, скрыть # |

← Rev. 2 →

0

For E, missed test case 7, just changed int to long after the contest, it got passed.

→ Ответить

»

cout_manish

4 года назад, скрыть # ^ |

← Rev. 2 →

+2

Rule 1 : No matter what, always use long int

Rule 2 : Always follow Rule 1.

→ Ответить

»

Klixel

4 года назад, скрыть # ^ |

+1

Or long long

→ Ответить

»

4 года назад, скрыть # ^ |

0

Rule 3: sometimes memory limit might be tight and you'll need to use int

→ Ответить

»

cout_manish

4 года назад, скрыть # ^ |

+1

Yet to encounter such problems. Thanks in advance, will remember this.

→ Ответить

»

AaronHe

4 года назад, скрыть # |

+22

My solutions (video):

A. Medium Number

Solution

B. Atilla's Favorite Problem

Solution

C. Advantage

Solution

D. Challenging Valleys

Solution

E. Binary Inversions

Solution

F. Quests

Solution

G. SlavicG's Favorite Problem

Solution

→ Ответить

»

qsmcgogo

4 года назад, скрыть # |

+2

problem G： Your task is to go from vertex a to vertex b, but you are allowed to enter node b if and only if after traveling to it, the value of x will become 0.

Why use "enter" instead of "arrive at" explaining the statement? I misunderstood when solving this problem and wasted 1 hour!

→ Ответить

»

Reduan9701

4 года назад, скрыть # |

0

What does ios::sync_with_stdio(0); cin.tie(0); these 2 lines mean?

→ Ответить

»

Wasi_Ur

4 года назад, скрыть # ^ |

← Rev. 4 →

+3

ios_base::sync_with_stdio(0); this code is used to make synchronization false. Because synchronization is initially true. And synchronization allows multiple objects to use a common resource (for example, Methods). cin.tie(0); tie(0) is a method which simply flushes before std::cin accepts an input. This is useful for interactive console programs which require the console to be updated constantly but also slows down the program for large input/output. Source: https://en.cppreference.com/w/cpp/io/ios_base/sync_with_stdio https://stackoverflow.com/questions/31162367/significance-of-ios-basesync-with-stdiofalse-cin-tienull

→ Ответить

»

Kamranali

4 года назад, скрыть # |

+1

is there any source apart from codeforces so that people from India and China communicate?

→ Ответить

»

steve004

4 года назад, скрыть # |

+8

Can someone give a test case for which my code for D is failing?
https://codeforces.me/contest/1760/submission/182044338

→ Ответить

»

4 года назад, скрыть # ^ |

+15

Check this out:

1
6
2 2 3 3 2 2

The answer is obviously NO, whereas your algorithm prints YES. Seems like an issue with dealing with consecutive equal elements.

→ Ответить

»

4 года назад, скрыть # |

← Rev. 3 →

0

History repeats itself. Four hacking attempts (all are of the same submission) are now stuck either in "Waiting" or "In queue" state. MikeMirzayanov, can we do something about this?

UPD: still in queue after the final testing. Déjà vu...

→ Ответить

»

_aka5h

4 года назад, скрыть # ^ |

← Rev. 3 →

-13

Ok cool

→ Ответить

»

4 года назад, скрыть # ^ |

+11

Bro, why have you decided to use unordered_set, which has $$$O(N)$$$ worst case time complexity? It's like you made a special effort to be hacked.

→ Ответить

»

_aka5h

4 года назад, скрыть # ^ |

-8

Ohh.. is that why my solution got hacked? Damn, I thought they have O(1) in all cases. Well, according to the approach I have used what would have been a more preferred data structure ?

→ Ответить

»

4 года назад, скрыть # ^ |

0

set would perfectly do it. Or even simpler: we may just sort all hashes and binary search over them.

→ Ответить

»

4 года назад, скрыть # ^ |

+11

I would argue that sort and binary search isn't simpler. This requires a bit more code to be written and there is a bit higher chance to mess it up.

→ Ответить

»

strokeme

4 года назад, скрыть # |

+8

BTW what's a counterexample for binary inversions if I only flip either the first 0 bit, last 1 bit, or none?

→ Ответить

»

lady_b1rd

4 года назад, скрыть # ^ |

+8

there is none. thats the solution

→ Ответить

»

Aizensolo

4 года назад, скрыть # |

0

Hi everybody! How can i register for the competition?

→ Ответить

»

4 года назад, скрыть # ^ |

0

Go to https://codeforces.me/contests and you will find the list of future contests accepting registration.

→ Ответить

»

satyabratojha

4 года назад, скрыть # |

0

why the rating is not increasing? I wanna see myself cyan

→ Ответить

»

3.14159...

4 года назад, скрыть # |

← Rev. 2 →

0

What is the difficulty of problems in this contest ??? I don't see any tag.

→ Ответить

»

deephudka005

4 года назад, скрыть # |

0

@Down Though my rating less than 1400 in whole codeforces rating history and I have given more than 5 contests still my rating has not been increased after securing 2000 rank.

→ Ответить

»

qwexd

4 года назад, скрыть # ^ |

0

I'm looking into it

→ Ответить

»

aadarshsinha

4 года назад, скрыть # |

0

For the G Question it is giving me wrong answer on test case 2 . Whats wrong in this code . I am first storing the xor of path till all the node from a and b as root respectively , Then checking for common elements . https://codeforces.me/contest/1760/submission/182136945

→ Ответить

»

div4only

4 года назад, скрыть # ^ |

← Rev. 2 →

0

When starting from a, the dfs cannot cross b. Either arrive at b directly, or block at b because of the nonzero xor.

→ Ответить

»

aadarshsinha

4 года назад, скрыть # ^ |

0

Thank you so much . Now i got what i was doing wrong . ;)

→ Ответить

»

parav_kamboz

4 года назад, скрыть # |

0

Have the ratings updated ?

→ Ответить

»

srabons_thinkLab

4 года назад, скрыть # ^ |

+8

No, not yet.

→ Ответить

»

Kapil4457

4 года назад, скрыть # |

+4

qwexd I have given 7 contest in which I have solve more than 1 problems in almost all..Can you please check why my rating is not increasing.I am grey rated which means I have rating less than 1400.

→ Ответить

»

The_fallen_emperor

4 года назад, скрыть # ^ |

+1

Assuming you have no competitive programming exp prior to coming to codeforces. Solving 1 problem in div 2 rounds places u somewhere between 9k-12k in rank, for getting a better rating, you need to have a better rank, for getting better rank you have to solve around 3-4 problems depending on round difficulty, A and B are generally ad-hoc/implementation type. However, from C you need to practice problems of specific tags.

→ Ответить

»

Kapil4457

4 года назад, скрыть # ^ |

0

bro I am asking him why has my rating not changed from the last Div 4 round.In simple words I am asking him to check why under my contest section , it is not showing the rating change from last contest..

→ Ответить

»

4 года назад, скрыть # ^ |

0

It's been shown under unrated contest list for all users maybe...it could be under the unrated list until cf figures lut ratings for all users...maybe, not an official answer

→ Ответить

»

DaAbyad

4 года назад, скрыть # |

0

I solved the same problems with another account because the other one made a lot of stupid mistakes, but the rating didn't increased.

can you please accept it from this account and ignore the other

→ Ответить

»

4 года назад, скрыть # ^ |

0

No

→ Ответить

»

Mbappe

4 года назад, скрыть # |

+1

Please Codeforces send Me a message said that "Your solution 181995500 for the problem 1760F significantly coincides with solutions Morad/181995500, Hussam-alwan/182013671, Rakhimov_Ans/182023461 If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. " and skipped all My Solutions And I Have not Any conclusive evidence But Only i submitted My Solution on 18:58 and anthor Person submitted on 19:26 So I Haven't an account on Idone and didn't Use it and I should Be a Specialist in this Contest !!!! please Help Me! Thanks For Your Time

→ Ответить

»

puspendra_09

4 года назад, скрыть # |

0

In the last div 4, my solution for the problem E Binary Inversions link to problemgot skipped due to the similarity of code from the user merahijalwa/181973274. first of all I don't know him and the similarity in code is due to code available in the geeks for geeks platform for counting the number of inversions in the array link to code from geeks for geeks website .
please consider this as this is a legit case and this is fist time i am going to reach pupil rank. please MikeMirzayanov consider this case as my case is completely legit and nor i have use any online ides. If anyone know how to contact codeforces team..please comment.

→ Ответить

»

4 года назад, скрыть # |

← Rev. 3 →

0

→ Ответить

»

4 года назад, скрыть # |

0

Ah screw my luck. I'm 1399 now

→ Ответить

»

mahir1

4 года назад, скрыть # ^ |

0

You will be specialist by next contest hopefully!

→ Ответить

»

4 года назад, скрыть # ^ |

0

Thanks. I hope so

→ Ответить

»

moaforlife

4 года назад, скрыть # ^ |

+1

Well, maybe you get added by one after the rating is rolled back =)

→ Ответить

»

GalaxyXing

4 года назад, скрыть # |

← Rev. 2 →

0

Why am I not rated for div4

→ Ответить

»

4 года назад, скрыть # ^ |

← Rev. 2 →

0

You have to participate in atleast n contest and solving atleast x problems in them to be eligible for being a rated participant, I don't remember n and x values, but you can find it in this post's description itself

→ Ответить

»

chromate00

4 года назад, скрыть # ^ |

0

That's the prerequisite of being a trusted participant. You will be rated regardless of being trusted, if your rating is <1400.

→ Ответить

»

4 года назад, скрыть # |

0

Nice Contest, Quality problemset and Good tutorial!! Kudos to the CF team & contest team

→ Ответить

»

JesseChan

4 года назад, скрыть # |

-9

I didn't share my code to ANYONE!!!!!!! What will happen if someone locked and try to hack me,and share my code to other? YOU HAVE WRONGED ME

→ Ответить

»

onebit1024

4 года назад, скрыть # ^ |

0

Maybe try to share your code and prove that you've not plagiarized rather than screaming from next time on.

→ Ответить

»

llc5pg

4 года назад, скрыть # ^ |

0

Blue coder is already out of competition for Div 4. Why bother?

→ Ответить

»

JesseChan

4 года назад, скрыть # ^ |

0

Because i did't do such thing.i dont care the rating but i do care credict.

→ Ответить

»

semwin2008

4 года назад, скрыть # |

-8

ю орн срунц

→ Ответить

»

vantaablackk

4 года назад, скрыть # |

+15

:sadge:

→ Ответить

»

goodeath

4 года назад, скрыть # ^ |

0

Same for me. I Solved 5 problems and yet got -2 rating

→ Ответить

»

34z12000

4 года назад, скрыть # ^ |

+8

They were easy af

→ Ответить

»

4 года назад, скрыть # |

← Rev. 2 →

-27

This is a BAD round. Problems aren't FUN at all. They are too STANDARD and ROUTINIZED. PLEASE refrain from setting such rounds from now on.

UPD: PLEASE don't downvote me! I am just saying my opinion.

→ Ответить

»

dalex

4 года назад, скрыть # ^ |

+10

For Div 4 these problems are very good

→ Ответить

»

4 года назад, скрыть # ^ |

-27

No, THIS ROUND IS BAD.

→ Ответить

»

4 года назад, скрыть # ^ |

+8

No, I find Div4 problems better than Div2. Of course first ones are pretty standard because they are very easy, but EFGH and sometimes D are really good

→ Ответить

»

4 года назад, скрыть # ^ |

-26

Nah, Div.2 problems are really BETTER than Div.4. You are JUST A PUPIL and why do you have the right to speak?

→ Ответить

»

4 года назад, скрыть # ^ |

+9

No, many Div2s have really bad problems and Div4 and Div3 are always made by the same people so they always have the same good quality. And yes I'm a pupil and I will speak as much as I want

→ Ответить

»

4 года назад, скрыть # ^ |

← Rev. 2 →

+8

You are allowed to speak as much as you want.

→ Ответить

»

4 года назад, скрыть # ^ |

0

Then how am I speaking

→ Ответить

»

aadarshsinha

4 года назад, скрыть # |

-8

Why my rating is not updated . yesterday i saw it was there but now its gone :(

→ Ответить

»