Output $$$2 \times (L \times W + W \times H + L \times H)$$$.

import sys

def solve():
    input = sys.stdin.read
    data = input().split()
    if not data:
        return
    l = int(data[0])
    w = int(data[1])
    h = int(data[2])
    print(2 * (l * w + w * h + l * h))

if __name__ == "__main__":
    solve()

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int l, w, h;
    if (!(cin >> l >> w >> h)) return 0;
    cout << 2 * (l * w + w * h + l * h) << "\n";
    return 0;
}

Loop through the array to compute the sum, the maximum and minimum element of $$$[S_1, S_2, \dots , S_N]$$$. Output the sum minus the maximum and minimum.

Note that there are built-in functions for these operations. Refer to the code sections below.

import sys

def solve():
    input = sys.stdin.read
    data = input().split()
    if not data:
        return
    n = int(data[0])
    s = [int(x) for x in data[1 : n + 1]]
    print(sum(s) - min(s) - max(s))

if __name__ == "__main__":
    solve()

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> s(n);
    for (int i = 0; i < n; i++) cin >> s[i];
    int sum = accumulate(s.begin(), s.end(), 0);
    int min_s = *min_element(s.begin(), s.end());
    int max_s = *max_element(s.begin(), s.end());
    cout << sum - min_s - max_s << "\n";
    return 0;
}

A constructive approach is to place the sequence elements by alternating steps starting from the center:

• $$$A_1 = \lfloor \frac{N}{2} \rfloor + 1$$$.
• $$$A_2 = A_1 + 1$$$
• $$$A_3 = A_2 - 2$$$
• $$$A_4 = A_3 + 3$$$
• $$$\dots$$$

By alternating directions between adding and subtracting the step index $$$i$$$, the values trace out all distinct integers from $$$1$$$ to $$$N$$$ without ever falling out of bounds or repeating. This guarantees a valid sequence for any $$$N \ge 3$$$.

import sys

def solve():
    input = sys.stdin.read
    n = int(input().strip())
    ans = [0] * n
    ans[0] = n // 2 + 1
    sign = 1
    for i in range(1, n):
        ans[i] = ans[i - 1] + sign * i
        sign = -sign 
    print(*ans)

if __name__ == "__main__":
    solve()

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> a(n);
    a[0] = n / 2 + 1;
    int sign = 1;
    for (int i = 1; i < n; i++) {
        a[i] = a[i - 1] + sign * i;
        sign = -sign;
    }
    for (int i = 0; i < n; ++i) cout << a[i] << (i == n - 1 ? "" : " ");
    cout << "\n";
    return 0;
}

First, convert the current time into total minutes elapsed from midnight:

Since a day has $$$1440$$$ minutes ($$$24 \times 60$$$), we can find the time on the clock after any number of days by taking the modulo:

Finally, extract the new hour and minute from the time:

• $$$H' = \lfloor \frac{\text{final_minutes}}{60} \rfloor$$$
• $$$M' = \text{final_minutes} \pmod{60}$$$

import sys

def solve():
    input = sys.stdin.read
    data = input().split()
    if not data:
        return
    h = int(data[0])
    m = int(data[1])
    k = int(data[2])
    total_minutes = h * 60 + m + k
    final_minutes = total_minutes % 1440
    new_h = final_minutes // 60
    new_m = final_minutes % 60
    print(new_h, new_m)

if __name__ == "__main__":
    solve()

#include <iostream>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    long long h, m, k;
    if (!(cin >> h >> m >> k)) return 0;
    long long total_minutes = h * 60 + m + k;
    long long final_minutes = total_minutes % 1440;
    long long new_h = final_minutes / 60;
    long long new_m = final_minutes % 60;
    cout << new_h << " " << new_m << "\n";
    return 0;
}

~~~~~

~~~~~

~~~~~

~~~~~