PCTprobability orz

This user is spamming his blogs in the feed.

Look there are tons of senseless blogs by him.

I request admins and MikeMirzayanov to disable/ban this user for all our mental peace.

Thanks

xerox_07

yamani_kobayashi

musa10th

kaptaan_Aditya_deshmukh

Ab151823

Thanh_Do

wlu28

They submit the leaked code first to check if it is correct, then resubmit it after changing var names, etc to avoid plagiarism.

I went through each skipped D submission and manually caught these.

Their submissions speak for themselves, I don't need to explain anything. (Just check their skipped submission of D, it is 100% same as the leaked D code I shared in my previous blog)

I request Mike to ban them all.

Thanks

#include<bits/stdc++.h>

using namespace std;


const int N = 1e5 + 10;

int w[N];

vector<int> edges[N];

bool visited[N];

int cnt = 0;

int f[N];


void dfs(int u) {

        int c = 0;

        visited[u] = true;

        for(int v : edges[u]) {

                if(visited[v]) continue;

                ++c; dfs(v);

        }


        if(!c) ++f[w[u]];

        else if(u != 1 && w[u] == 2) ++cnt;

}


void solve() {

        int n; cin >> n;

        cnt = 0;

        for(int i = 1; i <= n; i++) edges[i].clear(), visited[i] = false;

        for(int i = 1; i < n; i++) {

                int u, v; cin >> u >> v;

                edges[u].push_back(v);

                edges[v].push_back(u);

        }

        string s; cin >> s;

        for(int i = 0; i < n; i++) {

                if(s[i] == '?') w[i + 1] = 2;

                else w[i + 1] = s[i] - '0';

        }


        f[0] = f[1] = f[2] = 0;

        dfs(1);


        int ans = 0;

        if(w[1] == 0) ans = f[1] + (f[2] + 1) / 2;

        if(w[1] == 1) ans = f[0] + (f[2] + 1) / 2;

        if(w[1] == 2 && f[0] == f[1]) ans = f[1] + (f[2] + (cnt % 2)) / 2;

        if(w[1] == 2 && f[0] != f[1]) ans = max(f[0], f[1]) + f[2] / 2;

        cout << ans << '\n';

}


int main() {

        int t; cin >> t;

        while(t--) {

                solve();

        }

}

#include<bits/stdc++.h>

using namespace std;


int w[100010];


void solve() {

        int n, a, b; cin >> n >> a >> b;

        int c = __gcd(a, b);


        for(int i = 1; i <= n; i++) {

                cin >> w[i];

                w[i] %= c;

        }

        sort(w + 1, w + 1 + n);

        int ans = w[n] - w[1];

        for(int i = 1; i < n; i++) {

                ans = min(ans, w[i] + c - w[i + 1]);

        }

        cout << ans << '\n';

}


int main() {

        int t; cin >> t;

        while(t--) {

                solve();

        }

selmon__bhoi check his submissions (has cheated 5 times already)

AbhiKundu2002 Pro cheater, check his submissions, you will find nonsensical similarities.

(has cheated 3 times)

MikeMirzayanov Please ban these cheaters asap.

I saw many people in the telegram group that I have joined sharing solution for D1 (~30% AC), D2 (~60% AC), E(~15% AC).

Not to mention the cheating for ABCs is also prominent.

Its sad to see so many cheaters doing their thing to cheat. Its insane.

I want to know the experience of other people like how difficult do you think D1,D2,E was ?

(personally i was only able to do till D2, and i think d1 was pretty ez as a D, D2 was kinda good as a D, I don’t know about E)

You are given an integer N and a prime P. Count the number, modulo P, of undirected connected graphs G of N vertices numbered 1 to N that satisfy the following conditions:

There are no self-loops in G. Note that multiple edges are allowed.

For all edges (u,v) in G, if we delete (u,v) from G, G remains connected. In other words, G is edge-biconnected.

For all edges (u,v) in G, if we delete (u,v) from G, G is no longer edge-biconnected. Two graphs are considered different if and only if there exists a pair of distinct vertices (u,v) such that the numbers of edges connecting u and v in the two graphs are different.

N<=50, P<=1e9

Input : Two integers N and P

Output: Print the answer.

Problem A :

1,2,3,4,.......n is an intutive and easy answer.

Difficulty: 5/10 for a A

Problem B

Search the whole matrix a on 2x2 small grids and keep changing the values accordingly till i=0,n-1. j=0,m-1. In the end you will check if the two matrices are now equal or not.

Difficulty: 9/10 for a B

Problem C

There are nine cases totally ABC, ACB, BAC, BCA, CAB, CBA now check for each case seprately and apply binary search to find the lower bound for total_sum/3. Implementation is kinda tough.

Difficulty: 8/10 for a C

Problem D

Just count number of inversions using merge sort/segment tree. If (count1-count2)%2==0 YES else NO. Corner case: check if a & b are a permutation of each other or not.

Difficulty: 5/10 for a D