We can solve this problem greedily. Let's choose the first element to be $$$l$$$. The second element will be $$$l + 1$$$, the third will be $$$l + 3$$$, and so on. In general, the $$$i$$$-th element is given by $$$l + \frac{i \cdot (i + 1)}{2}$$$.

Why does this work? Assume the array $$$a$$$ is the one built by our greedy algorithm, and suppose there's a better array $$$b$$$ such that $$$\text{len}(b) \gt \text{len}(a)$$$. Because of how we construct $$$a$$$, there must exist some index $$$i$$$ such that for all $$$j \lt i$$$, we have $$$a_j = b_j$$$, and at index $$$i$$$, $$$a_i \lt b_i$$$ (since we always pick the smallest valid number possible).

Without loss of generality, assume $$$\text{len}(b) = \text{len}(a) + 1 = n$$$. Then, observe that the gap between the last two elements of $$$b$$$ must grow: $$$b_n - b_{n-1} \gt b_{n-1} - b_{n-2} \ge a_{n-1} - a_{n-2}$$$. But this means we could have added $$$b_n$$$ to the array $$$a$$$, which contradicts our assumption that $$$a$$$ was the longest valid array. So, the greedy approach gives us the optimal sequence.

Now the goal is to find the largest $$$x$$$ such that $$$l + \frac{x \cdot (x + 1)}{2} \le r$$$. We can do this efficiently using binary search, by solving the inequality directly using the discriminant of the corresponding quadratic equation, or even by brute force if the bounds allow.

t = int(input())
for _ in range(t):
    l, r = map(int, input().split())
    low, high = 1, int(2e5)
    ans = 1
    while low <= high:
        mid = (low + high) // 2
        total = mid * (mid - 1) // 2
        if l + total <= r:
            ans = mid
            low = mid + 1
        else:
            high = mid - 1
    print(ans)

To solve the problem of finding the $$$k$$$-th positive integer that is not divisible by $$$n$$$, we need to understand how the count of such numbers behaves. If we simply count up to $$$k$$$, we might mistakenly include numbers that are divisible by $$$n$$$, which we must skip. For example, if $$$n = 3$$$ and $$$k = 7$$$, then just taking the first 7 numbers gives us: $$$1, 2, 3, 4, 5, 6, 7$$$, but 3 and 6 are divisible by 3, so we actually have only 5 valid numbers among them. So, we need to find a number $$$x$$$ such that there are exactly $$$k$$$ numbers from $$$1$$$ to $$$x$$$ that are not divisible by $$$n$$$.

The number of integers from $$$1$$$ to $$$x$$$ that are divisible by $$$n$$$ is given by $$$\left\lfloor \frac{x}{n} \right\rfloor$$$. Hence, the number of integers in that range not divisible by $$$n$$$ is $$$x - \left\lfloor \frac{x}{n} \right\rfloor$$$. Our goal is to find the smallest $$$x$$$ for which this expression equals $$$k$$$. Since the function $$$x - \left\lfloor \frac{x}{n} \right\rfloor$$$ increases as $$$x$$$ increases, we can use binary search to efficiently find the minimum such $$$x$$$.

We start the binary search with a reasonable upper limit like $$$2k$$$, since in the worst case every $$$n$$$-th number is divisible and we're skipping one number every $$$n$$$ steps. In each step of the binary search, we compute how many non-divisible numbers exist up to the mid-point using the formula $$$mid - \left\lfloor \frac{mid}{n} \right\rfloor$$$. If this number is less than $$$k$$$, we need to search in the right half (i.e., higher values of $$$x$$$). Otherwise, we search in the left half, trying to find a smaller valid $$$x$$$. Once we find the smallest $$$x$$$ that satisfies the condition, it will be the $$$k$$$-th number that is not divisible by $$$n$$$.

This approach ensures that we solve each test case in logarithmic time, making it efficient even for large values of $$$k$$$ and $$$n$$$ up to $$$10^9$$$.

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    low, high = 1, 2 * k
    ans = -1
    while low <= high:
        mid = (low + high) // 2
        count = mid - (mid // n)
        if count >= k:
            ans = mid
            high = mid - 1
        else:
            low = mid + 1
    print(ans)

To solve the "Place of the Olympiad" problem efficiently, we aim to minimize the length of the longest bench while seating all the participants in the available rows and columns. The key observation is that the number of desks that can be placed in one row increases as the length of the longest bench increases. This means we can use a binary search over possible bench lengths to find the minimum possible value of the longest bench that still allows seating all the participants.

In each test case, we are given n rows, m columns, and k participants. The objective is to find the minimum value of x, the maximum bench length, such that we can fit all k participants within the constraints. The number of desks we can fit in one row depends on how many benches of length x can be formed in a row, and the total number of desks across all rows is the sum of the desks in each row. Using binary search, we determine the smallest x where the total number of desks fits k.

The binary search approach works by checking the number of desks that can be placed in n rows with a given maximum bench length x. If the total number of desks is sufficient, we attempt to reduce x; otherwise, we increase x. This process continues until we find the smallest x that satisfies the condition.

def is_valid(n, m, k, x):
    row_capacity = x * (m // (x + 1)) + (m % (x + 1))
    total_capacity = n * row_capacity
    return total_capacity >= k

def solve(n, m, k):
    low, high = 1, m
    answer = m
    
    while low <= high:
        mid = (low + high) // 2
        if is_valid(n, m, k, mid):
            answer = mid
            high = mid - 1
        else:
            low = mid + 1
    
    return answer

t = int(input())
for _ in range(t):
    n, m, k = map(int, input().split())
    print(solve(n, m, k))

Problem Explanation:

We need to find a sequence of ( n ) distinct integers ( a_1, a_2, \dots, a_n ) such that the difference between the maximum and minimum of the sequence equals the sum of all integers in the sequence. Specifically, we need to satisfy the following condition:

This means the largest number in the sequence minus the smallest number should equal the sum of all the elements in the sequence.

Approach:

Case 1: Odd ( n )

For odd values of ( n ), we begin by constructing a sequence of ( n ) consecutive integers centered around ( n ). This means that we take the range from ( \frac{n}{2} ) to ( \frac{3n}{2} ), creating a symmetric sequence. For example, if ( n = 5 ), the sequence would be:

This sequence is symmetric around the middle value and has a maximum difference of ( n — 1 ) between the smallest and largest numbers.

Next, we increase each of the integers by 2, ensuring that the difference between the smallest and largest element remains the same. So, the sequence ( [3, 4, 5, 6, 7] ) becomes:

This change increases the sum, and now the sum becomes ( n^2 + 2n ).

At this stage, we need to ensure the sum matches the target value of ( (n + 1)^2 ). The equation for this is:

To adjust the sequence to match the target sum, we make a slight modification. We increase the second-to-last element by 1. This gives us the final sequence that satisfies the condition:

For ( n = 5 ), after this adjustment, the sequence becomes:

This sequence now satisfies the condition because:

• The difference between the maximum (10) and minimum (4) is ( 10 — 4 = 6 ).
• The sum of the elements is ( 4 + 6 + 7 + 9 + 10 = 36 ), and ( 36 = 6 ).

Case 2: Even ( n )

For even values of ( n ), we can use a similar approach. The sequence is split into two parts. The first part is from ( \frac{n}{2} ) to ( n — 1 ), and the second part is from ( n + 1 ) to ( \frac{3n}{2} ). This ensures that the difference between the largest and smallest element is exactly ( n ), and the sum of the elements is ( n^2 ), making it a valid solution.

Conclusion:

The approach involves constructing a sequence of consecutive integers, adjusting the values slightly to match the required sum, and ensuring that the difference between the maximum and minimum elements equals the sum of all elements in the sequence. By carefully adjusting the sequence for both odd and even values of ( n ), we can generate a sequence that satisfies the given condition.

for _ in range(int(input())):
    n = int(input())
    if n % 2 == 0:
        arr = []
        for i in range(n // 2, n // 2 + n + 1):
            if i != n:
                arr.append(i)
        print(*arr)
    else:
        arr = []
        for i in range(n // 2 + 3, n // 2 + 3 + n):
            arr.append(i)
        arr[0] -= 1
        arr[-1] += 1
        arr[-2] += 1
        print(*arr)