Then I will solve all their problems one by one, and in return they will cook food for me and me and my friends would laugh the same way as your pfp

As long as the chefs bring snacks, I'm ready

let them cook

what s the equivalent for 1700-1800 cf problems in codechef ?

can you find any issue for the code below: GCD and Xor

ll t;
cin >> t;

while (t--) {
    ll n, k;
    cin >> n >> k;
    ll a[n];
    ll cnt = 0;
    ll eve = 0;
    ll divs = 0;

    for (ll i = 0; i < n; i++) {
        cin >> a[i];
        if (a[i] == k) cnt++;
        if (a[i] % k == 0) divs++;
    }

    // If all elements are equal to k
    if (cnt == n) {
        cout << 0 << endl;
        continue;
    }

    // If all elements are divisible by k
    if (divs == n) {
        cout << 1 << endl;
        continue;
    }

    ll niraj = 0;
    int flag = 0;

    // Check how many adjacent pairs are not equal
    for (ll i = 1; i < n; i++) {
        if (a[i] != a[i - 1]) {
            niraj++;
        }
    }

    ll c = -1, d = -1;

    // Check if there are more than two distinct values
    for (ll i = 0; i < n; i++) {
        if (a[i] == c || a[i] == d) continue;

        if (c == -1) c = a[i];
        else if (d == -1) d = a[i];
        else {
            flag = 1;
            break;
        }
    }

    // If all elements are the same
    if (niraj == 0) {
        // case like  2 2 2 2 2 and k != 2
        cout << 1 << endl;
        continue;
    }

    // If at most two distinct values and conditions are met
    if (niraj <= 1 && flag == 0) {
        //case like :  2 2 2 2 3 3 3 3 or 3 3 3 3 2 2 2  and k = 3
        cout << 1 << endl;
    } else if (niraj == 2 && flag == 0 && a[0] == a[n - 1] && a[0] == k) {
         // case like : 2 2 2 2 3 3 3 3 3 2 2 2 2 , k = 2
        cout << 1 << endl;
    } else {
        cout << 2 << endl;
    }
}

Let $$$x$$$ be the maximum integer, such that $$$2^x \le n-1$$$, then maximum value of $$$X$$$ can be $$$2^{x+1} -1$$$.

Let do operations in the following way:

Observe that we reach $$$2^x, 2^{x-1}, ... ,4, 2, 1$$$ while deleting leaves one by one

If the number of leaves cut are odd at tree with size $$$4$$$, remove edges $$$1,2,1$$$(sub-tree size) otherwise remove $$$3,1,1$$$

For a given value of $$$N$$$, the maximum value that $$$X$$$ can be xorred with is $$$N-1$$$. That gives an upper bound of $$$2^{\lfloor log_2(N-1)+1 \rfloor}-1$$$ on $$$X$$$. Consider $$$N = 4$$$. Note that any value of $$$0 \leq X \leq 3$$$ is obtainable. For $$$N > 4$$$, keep removing leaves till $$$N = 4$$$. After the removal of each leaf, let $$$siz$$$ denote the size of the remaining tree. Do $$$X = X \oplus siz$$$ if $$$siz$$$ is a power of $$$2$$$; $$$X = X \oplus 1$$$ otherwise. By this method, we can set all possible bits $$$\geq 2$$$ in $$$X$$$. Once $$$N = 4$$$, any combination of the last $$$2$$$ bits can be obtained. Thus, the upper bound is achievable for $$$N \geq 4$$$. The answers for $$$N = 2$$$ and $$$N = 3$$$ are $$$1$$$ and $$$3$$$ respectively.

Yeah think about this testcase: 0 0 0 0 0 1 1

Didn't think you'd participate. I guess holiday today :)

$$$a \cdot x + b \cdot y \ge min(a, b) \cdot (x + y) > min(a, b) \cdot max(x, y) $$$