This new start time is very good.

Hope this round won't be as bad as the last one...

qp

Bro I cannot solve problem D.

Today's problems after C were better than yesterday's ARC imo. Especially; the problem F >>

I guess they have swapped ARC and ABC :))

Has anybody solved using divide and conquer for E?? If successful, please share it.

The solution of C is always surprisingly short.

Is there any particular way to solve problem E without using RNG?

why this giving error , this problems like c are so bad just one line code if someone guess 10 sec to solve if someone not no matter what he do always wrong i first utilise the b then simply take contri of a and c , any case you think about ??

void solve() {
    // executing code from here 
    int t;cin>>t;
    while (t--) {
        int a,b,c;cin>>a>>b>>c;
        int ans=0;
        ans=min({a,b,c});
        a-=ans;c-=ans;
        // dbg(a);dbg(b);
        if(a>c){
            // dbg(min(c,a/2));
            ans+=min(c,a/2);
        }else{
            ans+=min(c/2,a);
        }
        cout<<ans<<endl;
    }
}

Why problem F couldn't be solved with Dijkstra?

#include <bits/stdc++.h>

using namespace std;

constexpr long long INF = LLONG_MAX / 2;

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

    int n, m;
    cin >> n >> m;

    vector<int> W(n);
    for (int i = 0; i < n; i++) {
        cin >> W[i];
    }

    vector<vector<int>> G(n);
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        --u, --v;
        G[u].push_back(v);
        G[v].push_back(u);
    }

    vector<array<long long, 2>> dis(n, {INF, INF});
    dis[0] = {0, 0};

    priority_queue<array<long long, 3>, vector<array<long long, 3>>, greater<array<long long, 3>>> pque;
    pque.push({0, 0, 0});

    while (not pque.empty()) {
        auto [d, w, u] = pque.top();
        pque.pop();

        long long nw = w + W[u];
        for (auto v : G[u]) {
            long long nd = d + nw;
            if ((nd < dis[v][0]) or (nd == dis[v][0] and nw < dis[v][1])) {
                dis[v] = {nd, nw};
                pque.push({nd, nw, v});
            }
        }
    }

    for (int i = 0; i < n; i++) {
        cout << dis[i][0] << "\n";
    }

    return 0;
}

Can anyone prove my solution for F? I claim that every vertex only have $$$m$$$ pairs of (current weight, fuel used) being useful , but I can't prove it.

my solution

all nice problem！！！ althought I just solve 3，but solve the rest problem with Editorial help me learn a lot！！：)

Is it fair to have random in E? The same solution can be judged differently just by random.

For problem D, first fill vector by $$$\frac{k}{2^n}$$$. Then keep incrementing numbers in even position by 1 alternating between left and right, continue till you have n%{2^k} remaining. If there is still some left then do the same for odd position. The imbalance will always be 1 when there is remainder other than 0.

Why this solution is incorrect??

problem D spj maybe wrong? Without considering K.

https://atcoder.jp/contests/abc422/submissions/69146833

There is a discuss of D's checker was wrong :Portal