The numbers we consider are "boring" numbers, meaning they consist of the same digit (e.g., (1, 11, 111, 1111, \dots)).

Instead of checking every number, note that numbers are called in order: all 1s first, then all 2s, and so on.

To find the total keypresses, we count: - All previous digits' keypresses (full sequences for numbers before (x)). - The keypresses needed for numbers up to (x) using a simple sum formula.

Each "boring" number is called in a structured order: all numbers made of 1s first, then 2s, then 3s, and so on. Instead of iterating through every number, we can break the total keypresses into two parts: 1. Counting keypresses for previous digits: — Before reaching (x), all numbers made of smaller digits (e.g., (1, 11, 111, 1111) if (x) is in the 2s) are fully called. — Each digit contributes (10) keypresses for its full sequence. Thus, for a digit (d), the total keypresses before it is: [ 10 *(d-1) ] 2. Counting keypresses for numbers leading up to (x): — If (x) has (n) digits (e.g., (x = 222) has (n = 3)), we count keypresses for forming each prefix: [ 1 + 2 + ... + n ] — This is directly computed using a simple loop summing (1) to (n). By combining these two values, we get the total keypress count in (O(1)) time.

t = int(input())
for _ in range(t):
    x = int(input())
    val = x % 10  # Get the digit (since x consists of the same digit)
    ans = 10 * (val - 1)  # Count keypresses for previous digits
    sz = len(str(x))  # Get the number of digits in x
    ans += sum(range(1, sz + 1))  # Add keypresses for the current digit
    print(ans)

Sort the bases by their defensive power and maintain a running sum of gold. For each spaceship, find the maximum number of bases it can attack efficiently.

The problem requires us to determine how much gold each spaceship can steal based on its attacking power.

Approach:

1. Sort the Bases by Defensive Power

• This allows us to efficiently find all bases a spaceship can attack using a two-pointer technique.
• Keep track of the cumulative gold collected as we iterate through the bases.

1. Sort the Spaceships and Process Them in Order

• Sorting spaceships ensures that we process weaker spaceships first, maintaining efficiency in tracking available bases.
• Maintain the original indices of spaceships so that we can return the results in the correct order.

1. Use Two-Pointer Technique

• Iterate through the sorted spaceships, and for each spaceship, move a pointer (l) to include all bases it can attack.
• Keep a running sum of gold to quickly determine the total amount available.

1. Store Results and Print in Original Order

• Since we modified the order of spaceships during sorting, we store results in an array indexed by their original positions.

Example Walkthrough:

Input:

5 4  
1 3 5 2 4  
0 1  
4 2  
2 8  
9 4  

Processing:

• Sort bases: [(0,1), (2,8), (4,2), (9,4)]
• Sort spaceships with indices: [(1,0), (2,3), (3,1), (4,4), (5,2)]
• Maintain a cumulative gold sum:
• Spaceship (1): Can attack base (0,1), steals 1.
• Spaceship (2): Can attack bases (0,1), (2,8), steals 9.
• Spaceship (3): Can attack bases (0,1), (2,8), steals 9.
• Spaceship (4): Can attack bases (0,1), (2,8), (4,2), steals 11.
• Spaceship (5): Can attack bases (0,1), (2,8), (4,2), steals 11.

Output:

1 9 11 9 11  

Complexity Analysis:

• Sorting bases: O(b log b)
• Sorting spaceships: O(s log s)
• Two-pointer traversal: O(s + b)
• Overall complexity: O((s + b) log(s + b))

def solve():
    # Read input values: number of spaceships (s) and number of bases (b)
    s, b = map(int, input().split())
    
    # Read the attacking power of each spaceship
    ships = list(map(int, input().split()))
    
    # Read the defensive power and gold of each base
    bases = [list(map(int, input().split())) for _ in range(b)]
    
    # Step 1: Sort bases by defensive power to process them efficiently
    bases.sort()
    
    # Step 2: Sort spaceships by attacking power while keeping track of original indices
    ships = sorted((ships[i], i) for i in range(s))
    
    # Array to store the maximum gold each spaceship can steal
    ans = [0] * s
    
    # Two-pointer approach to efficiently determine how much gold each spaceship can steal
    l, max_gold = 0, 0
    
    # Step 3: Process each spaceship in increasing order of attacking power
    for i in range(s):
        # Move the pointer 'l' to include all bases that the spaceship can attack
        while l < b and bases[l][0] <= ships[i][0]:
            max_gold += bases[l][1]  # Add the gold from the base
            l += 1  # Move to the next base
        
        # Store the total gold that this spaceship can steal
        ans[ships[i][1]] = max_gold  # Store result at original index
    
    # Step 4: Print the results in the original order of spaceships
    print(*ans)

if __name__ == "__main__":
    solve()

Consider that the player's position after each move must lie within the intersection of the reachable area from the previous position (given ( k )) and the current segment's bounds.

Use binary search to efficiently determine the minimal ( k ). For each candidate ( k ), simulate the player's movement through all segments to check feasibility.

1. Binary Search Setup: The minimal ( k ) lies between 0 and a sufficiently large upper bound.
2. Feasibility Check: For each candidate ( k ):

• Start at position 0.
• For each segment, compute the reachable interval after moving at most ( k ) units from the previous position.
• Intersect this interval with the current segment's bounds.
• If the intersection is empty, ( k ) is too small.

1. Binary Search Adjustment: Adjust the search range based on feasibility results to find the minimal ( k ).

import sys def can_complete_level(n, segments, k): L, R = 0, 0 # The range of reachable positions for l, r in segments: # Compute the new reachable range L = max(l, L — k) R = min(r, R + k) if L > R: # If no valid landing position exists return False return True def find_min_k(n, segments): left, right = 0, 10**9 # Binary search for k while left <= right: mid = (left + right) // 2 if can_complete_level(n, segments, mid): right = mid — 1 else: left = mid + 1 return left def main(): t = int(input()) results = [] for _ in range(t): n = int(input()) segments = [] for _ in range(n): l, r = map(int, inpit().split()) segments.append((l, r)) results.append(str(find_min_k(n, segments))) sys.stdout.write(" ".join(results) + "\n") if name == "__main__": main()

• Binary Search: The search runs in ( O(\log(\text{max_distance})) ) iterations, where (\text{max_distance}) is the upper bound (( 10^9 )).
• Feasibility Check: Each check processes ( n ) segments in ( O(n) ) time.
• Total Complexity: ( O(t \cdot n \cdot \log(10^9)) ), which is efficient given the constraints (sum of ( n ) across test cases is ( 2 \cdot 10^5 )).

Consider selecting Farmer John's room first and then placing the cows in the nearest available unoccupied rooms.

If the farthest distance is ( x ), can you calculate the maximum number of cows that can be placed?

Try using prefix sum to efficiently calculate how many cows can be placed within a given range.

This problem might have another solution where we can calculate the answer directly. But for a moment, let's think about whether this can be solved without binary search.

The first thing we do when solving output binary search problems is check: Can we easily verify if a given answer is valid or not? If someone told us the answer in advance, could we efficiently check its validity? If so, we can use binary search to guess the answer. For this to work, the function must be monotonic (i.e., if a certain ( x ) is possible, all larger values are also possible). Since this function is monotonic, we can use binary search to efficiently find the minimum ( x ) that satisfies the conditions.

The binary search must start with ( l = 0 ) and ( r = n ), as this ensures we explore the full possible range of distances. To check feasibility for a given distance ( d ), we calculate how many available rooms exist in the range ( left, right ), where: - ( left = max(0, i — d) ) - ( right = min(n, i + d + 1) )

We use the prefix sum array to efficiently compute the number of unoccupied rooms within this range.

Complexity Analysis

• Binary Search: The search runs in ( O(\log n) ) iterations since we are halving the search space each time.
• Feasibility Check: Each check processes ( n ) rooms in ( O(n) ) time.
• Prefix Sum Computation: Constructing the prefix sum array takes ( O(n) ) time.
• Total Complexity: The overall complexity is ( O(n log n) ), which is efficient given the constraints.

import sys

def solve():
    n, k = map(int, sys.stdin.readline().strip().split())
    s = sys.stdin.readline().strip()
    prefix = [0]
    
    # Compute prefix sum for counting free rooms quickly
    for val in s:
        prefix.append(prefix[-1] + (1 if val == '0' else 0))
    
    def max_cows_in_range(d):
        """Given max distance d, calculate the maximum number of cows that can be placed"""
        max_cows = 0
        for i in range(n):
            if s[i] == '0':  # Consider only unoccupied rooms
                left = max(0, i - d)
                right = min(n, i + d + 1)
                max_cows = max(max_cows, prefix[right] - prefix[left])
        return max_cows
    
    # Binary search for the minimum possible maximum distance
    l, r = 0, n
    while l < r:
        mid = (l + r) // 2
        if max_cows_in_range(mid) >= k + 1:
            r = mid  # Try a smaller maximum distance
        else:
            l = mid + 1
    
    print(l)

if __name__ == "__main__":
    solve()

The problem asks for the k-th largest number in the multiplication table. However, listing all numbers and sorting them in descending order is too slow. Instead, think about how to find the k-th smallest number first—since the k-th largest in a sorted list is the same as the (n*m — k + 1)-th smallest.

To efficiently find the k-th smallest number without generating all elements, use binary search on possible values (from 1 to n*m). For a given mid-value, count how many numbers in the table are ≤ mid by checking each row.

The problem requires finding the k-th smallest number in an n × m multiplication table where each cell (i, j) contains i * j. Since generating and sorting all n * m elements is too slow for large n and m, we use binary search on the possible values (from 1 to n * m).

For a given mid-value during binary search, we count how many numbers in the table are ≤ mid by iterating through each row i and calculating min(mid // i, m). This gives the count of valid numbers in that row.

If the total count is less than k, the k-th smallest number must be larger than mid, so we adjust the search to higher values.

If the count is ≥ k, the k-th smallest number could be mid or smaller, so we search lower values.

The binary search efficiently narrows down the correct value in O(n log(n * m)) time, making it suitable for large constraints.

def solve():
    n, m, k = map(int, input().split())

    def find(mid):
        count = 0
        for i in range(1, n + 1):
            add = min(mid // i, m)
            if add == 0:
                break
            count += add
        return count < k

    low, high = 1, n * m
    answer = 0

    while low <= high:
        mid = (low + high) // 2
        if find(mid):
            low = mid + 1
        else:
            answer = mid
            high = mid - 1

    print(answer)

t = 1
for _ in range(t):
    solve()