I mean its almost impossible if you think about it. Solving these many questions in 3 hrs with 1 PC.

First Contest going to start in almost 1 hour Link for First Contest (6pm — 11pm) :

https://codeforces.me/group/wlb0UYQSQF/contest/710855 (Div 1)

https://codeforces.me/group/wlb0UYQSQF/contest/710866 (Div 2)

Majorly they will be in the evening only from 6-8. Rest will be updated on the whatsapp group

We will be live on Youtube

Orzz!

Yes

Just saw the AK prediction come true today haha XD.

contest is not opening from registered mail id

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

#ifdef LOCAL
// #include "..\debugging.h"
#endif

#define int long long
using vi = vector<int>;
using vc = vector<char>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using pi = pair<int, int>;
using vvi = vector<vector<int>>;
using vcb = vector<vector<bool>>;
using vvc = vector<vector<char>>;
using mi = map<int, int>;
using si = set<int>;


#define endl "\n"
#define all(x) begin(x), end(x)
#define sz(x) int((x).size())
#define fmap(name, x) mi name; for(auto it:x) ++name[it]
#define F first
#define S second
#define elif else if
#define PB emplace_back
#define MP make_pair
#define REPi(a, b) for (int i = a; i < b; i++)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define REPO(a) for (int i = 0; i < a; i++)
#define vvl(ab, r, c, v)  vvi ab(r, vector<int>(c, v))
#define dbg(v)  cout << #v << " = " << (v) << endl;

const long long mod = 1000000007ll, mod2 = 998244353ll;

int dist(vvi &tree, int p, int node, int p2 = -1) {
    int ans = 1;
    for(int c: tree[node])
        if(c != p && c != p2) {
            ans = max(ans, dist(tree, node, c )+1);
        }
    return ans;
}

void path(vvi &tree, int node, int target, int p, vi &ans) {
    for(int i: tree[node]) {
        if(i == p)
            continue;
        ans.PB(i);
        if(i == target) {
            return;
        }
        path(tree, i, target, node, ans);
        if(ans.back() != target)
            ans.pop_back();
    }
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;

        vvi tree(n);
        for(int i=1; i<n; ++i) {
            int u, v;
            cin >> u >> v;
            --u, --v;
            tree[u].PB(v);
            tree[v].PB(u);
        }

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

        vi adist(n, 0), bdist(n, 0), p = {-1, 0};
        path(tree, 0, u, -1, p);
        p.PB(-1);

        int m = p.size()-2;
        vi dp(m);

        for(int i=1; i<m+1; ++i)
            dp[i-1] = dist(tree, p[i-1], p[i], p[i+1]);


        vi a(m), b(m);
        for(int i=m-1; i>=0; --i) {
            a[i] = dp[i] + i;
            b[i] = dp[i] + m-i-1;
        }

        vi suffdp(m, b.back());
        vi predp(m,a.front());
        for(int i =1 ;i<m-1;i++){
            predp[i] = max(a[i],predp[i-1]);
        }
        for(int i=m-2; i>=0; --i)
            suffdp[i] = max(suffdp[i+1], b[i]);

        // dbg(p);
        // dbg(a);
        // dbg(b);
        // dbg(dp);
        // dbg(suffdp);
        // dbg(predp);


        int start = 0, end = m-1;
        int counter = 1;
        while(start < end){
            if(counter && a[start] > suffdp[start+1] || !counter && b[end] > predp[end-1] ){
                counter = !counter;
                break;
            }
            if(counter)
                start++;
            else
                end--;

            counter = !counter;
        }

        if(start == end)
            counter = !counter;
        cout << (counter?"Bob" : "Alice") << endl;
    }

    return 0;
}

can someone explain why this is wrong for F1.

Basically I tried to find the path from 1 to u. For every path value I found Dp(max depth not on path)

then I checked for early exit. If this makes sense

what does this mean?

Is the round rated...... been waiting for the update since morning

you can find the sum of all the positive number on odd indices and same should be done for even

we are then finding max of them and printing

in your case when all numbers are negetive suppose -1 -1 -2 5 -9

we know the ans should be 5

we know that the sum at odd indices is more now we will choose the number on odd indices that are negetive and remove them -> -3 5 -9 now we will remove -3 and -9

I got -2 because judge is not checking pretest 1 . Please help!! I got penaly because of printing unecessary things T_T