Codeforces Round #633 (Div. 2)

Contest link: https://codeforces.com/contest/1339

A. Filling Diamonds

Putting it upright will cause a chain, making all other positions have to adapt to this upright position, so just output how many cases there are.

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

void solve() {
    int n; cin >> n;
    cout << n << "\n";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

B. Sorted Adjacent Differences

Just output from the middle to both sides.

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

void solve() {
    int n; cin >> n;
    int a[n]; for (int & i : a) cin >> i;
    sort(a, a + n);
    int l, r;
    if (n % 2) {
        cout << a[n / 2] << ' ';
        l = n / 2 - 1;
        r = n / 2 + 1;
    } else {
        l = n / 2 - 1;
        r = n / 2;
    }
    while (l >= 0) {
        cout << a[l] << ' ' << a[r] << ' ';
        --l, ++r;
    }
    cout << "\n";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

C. Powered Addition

If there is $ a_i> a_ {i + 1} $, then the latter must be transformed to be equal to the former, because the difference between the two can be expressed as a binary number, we only need to take the power of $ 1 $ each time, The answer sought is the highest power of all differences.

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

void solve() {
    int n; cin >> n;
    ll a[n]; for (ll & i : a) cin >> i;
    if (n == 1 || (a[0] <= a[1] && is_sorted(a, a + n))) {
        cout << "0\n";
        return;
    } 
    int ans = 0;
    for (int i = 0; i < n - 1; i++) {
        if (a[i] > a[i + 1]) {
            int sub = a[i] - a[i + 1];
            int cnt = 0;
            while (sub) {
                ++cnt;
                sub >>= 1;
            }
            years = max (years, cnt);
            a[i + 1] = a[i];
        }
    }
    cout << ans << "\n";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

D. Edge Weight Assignment

The picture is still not very familiar, it took almost an hour and a half , to be filled.