Does the exact temperature to each officer matter??

What if every officer wore an infinite amount of jackets??

In that case only the Mod matters.

We first initialize the answer we are going to print to infinity (1e18), then we make an Array $$$B$$$, containing the set of $$$A_i mod T$$$ and then we itterate over every $$$B_i$$$ and set the answer to $$$min(ans, max(a[i - 1] - a[i] + k, a[n - 1] - a[i]))$$$, we add that $$$T$$$ as if (s)he took a jacket off. Time complexity: $$$O(N).$$$

Think of 2 dijkstras.

We first assign two arrays, $$$DistanceFromS$$$ and $$$DistanceFromT$$$, then we do two dijkstras filling the two arrays, then for every element in the $$$DistanceFromS$$$ we do binary search to find the lower bound of $$$K \geq DistanceFromS_i + L$$$ if there exists one, that means we can make a road connecting the two points. Time complexity: $$$O(N*Log(N)).$$$

didn't solve this lol

For problem 3, first consider how to answer one question. Balls at the same position can be merged, then consider the entire process in reverse order, The last ball collected must be one of the balls on left or right side of the end point, and the last $$$K$$$ balls collected must form an interval. Enumerate the last ball collected(at most 2 different balls possible), the observation leads a DP that $$$fl,r$$$ indicates if we collect the balls between $$$l-th$$$ to the $$$r-th$$$ position at the end of entire process, what is the minimal time to collect them. It answers a question in $$$O(n^2)$$$time, where n is the different places that have balls. If we calculate all $$$n$$$ beginning position, we can answer a question in $$$O(Log(N))$$$ time.

Another observation is that if we have n different positions that have balls, the minimal time is at least $$$N^2/2$$$, so we only need to solve the task for $$$N \leq 1000$$$, previous $$$O(N)$$$ solution works.

(DK How to link comments sorry), wasa855 Dominater069

This only gets you the first 4 subtasks (50 grades)

Brute force approach works.

Think of sorting the arrays.

We make another two arrays of pairs, to keep track of the element and it's index, and sort both of them. Now to building the last answer, we itterate through every element in the arrays and check if (after sorted) there exists someone giving a gift to himself, since that's not allowed we check if we can exchange gifts with someone 1 index lower or higher, if we can't then the answer is no, and if we can we swap them. After that we itterate in the final two arrays and check if $$$A_i \geq B_i$$$, if it isn't then answer is No otherwise we print Yes. Time complexity $$$O(N*Log(N)*Q)$$$

int n, q, a[500010], b[500010];

void solve() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    cin >> q;
    while (q--) {
        vector<pair<int, int>> v, v2;
        int l, r;
        cin >> l >> r;
        --l;
        for (int i = l; i < r; ++i) v.pb({ b[i],i }), v2.pb({ a[i],i });
        sort(all(v)); sort(all(v2));
        bool flag = true;
        for (int i = 0; i < v.size(); ++i) {
            if (v[i].ss == v2[i].ss) {
                if (i && v2[i - 1].ff >= v[i].ff) swap(v[i - 1], v[i]);
                else if (i < v.size() - 1 && v2[i].ff >= v[i + 1].ff)  swap(v[i], v[i + 1]);
                else { flag = false; break; }
            }
        }
        for (int i = 0;i < v.size();i++) {
            if (v2[i].ff < v[i].ff) { flag = false; break; }
        }
        if (flag) cout << "Yes\n";
        else cout << "No\n";
    }
}

I think greedy approach works