Problem Description

Alice and Bob are playing a turn-based game on a circular field with flowers. There are x flowers in the clockwise direction between them and y flowers in the anti-clockwise direction.

The game rules are:

1. Alice always goes first
2. On each turn, a player must pick one flower from either the clockwise or anti-clockwise direction
3. The player who picks the last flower wins by capturing their opponent

Given two integers n and m, you need to find how many pairs (x, y) exist where:

• Alice wins the game
• 1 ≤ x ≤ n (clockwise flowers)
• 1 ≤ y ≤ m (anti-clockwise flowers)

The key insight is that Alice wins when the total number of flowers x + y is odd. This is because:

• With an odd total, Alice takes the last flower after all moves are made
• With an even total, Bob would take the last flower

The solution counts all valid pairs where x + y is odd:

• When x is odd and y is even: There are ⌈n/2⌉ odd values for x and ⌊m/2⌋ even values for y
• When x is even and y is odd: There are ⌊n/2⌋ even values for x and ⌈m/2⌉ odd values for y

The total count is: ⌈n/2⌉ × ⌊m/2⌋ + ⌊n/2⌋ × ⌈m/2⌉

In the code, this is calculated as:

• a1 = (n + 1) // 2 represents ⌈n/2⌉
• a2 = n // 2 represents ⌊n/2⌋
• b1 = (m + 1) // 2 represents ⌈m/2⌉
• b2 = m // 2 represents ⌊m/2⌋

Intuition

The key realization comes from analyzing who wins based on the total number of flowers. Since players alternate turns and each turn removes exactly one flower, the winner is determined by whether the total number of flowers is odd or even.

Think of it this way: if there are x + y total flowers, and players alternate picking one flower at a time, then:

• After 1 flower is picked, it's Bob's turn
• After 2 flowers are picked, it's Alice's turn
• After 3 flowers are picked, it's Bob's turn
• And so on...

Following this pattern, when x + y is odd, Alice will be the one to pick the last flower and win. When x + y is even, Bob picks the last flower and wins.

Since we want Alice to win, we need x + y to be odd. For a sum to be odd, one number must be odd and the other must be even.

Now we need to count how many valid pairs (x, y) satisfy this condition within the given bounds:

• Case 1: x is odd (ranges from 1, 3, 5, ... up to n) and y is even (ranges from 2, 4, 6, ... up to m)
• Case 2: x is even (ranges from 2, 4, 6, ... up to n) and y is odd (ranges from 1, 3, 5, ... up to m)

Counting the number of odd and even values in a range is straightforward:

• In range [1, n]: there are ⌈n/2⌉ odd numbers and ⌊n/2⌋ even numbers
• In range [1, m]: there are ⌈m/2⌉ odd numbers and ⌊m/2⌋ even numbers

The total number of valid pairs is simply the product of counts from each case, since we can pair any odd value from one range with any even value from the other range.

Learn more about Math patterns.

Solution Approach

The implementation follows directly from our mathematical analysis. We need to count pairs (x, y) where x + y is odd.

First, we calculate the count of odd and even numbers in each range:

• For range [1, n]:
  • Number of odd values: a1 = (n + 1) // 2
  • Number of even values: a2 = n // 2
• For range [1, m]:
  • Number of odd values: b1 = (m + 1) // 2
  • Number of even values: b2 = m // 2

The formula (n + 1) // 2 gives us ⌈n/2⌉ (ceiling division) for counting odd numbers starting from 1. For example:

• If n = 5: odd numbers are {1, 3, 5}, count = (5 + 1) // 2 = 3
• If n = 6: odd numbers are {1, 3, 5}, count = (6 + 1) // 2 = 3

The formula n // 2 gives us ⌊n/2⌋ (floor division) for counting even numbers. For example:

• If n = 5: even numbers are {2, 4}, count = 5 // 2 = 2
• If n = 6: even numbers are {2, 4, 6}, count = 6 // 2 = 3

Finally, we calculate the total number of valid pairs by considering both cases:

• Case 1: x is odd and y is even → a1 * b2 pairs
• Case 2: x is even and y is odd → a2 * b1 pairs

The answer is the sum: a1 * b2 + a2 * b1

This solution runs in O(1) time complexity since we're only performing a few arithmetic operations, and uses O(1) space complexity as we only store a few integer variables.

Example Walkthrough

Let's walk through a small example with n = 3 and m = 4.

We need to find all pairs (x, y) where Alice wins, with 1 ≤ x ≤ 3 and 1 ≤ y ≤ 4.

Step 1: Identify when Alice wins Alice wins when the total flowers x + y is odd. Let's check all possible pairs:

Alice wins in 6 cases where x + y is odd.

Step 2: Apply the formula approach

Count odd and even numbers in each range:

• For [1, 3]:
  • Odd numbers: {1, 3} → count = (3 + 1) // 2 = 2
  • Even numbers: {2} → count = 3 // 2 = 1
• For [1, 4]:
  • Odd numbers: {1, 3} → count = (4 + 1) // 2 = 2
  • Even numbers: {2, 4} → count = 4 // 2 = 2

Calculate valid pairs:

• Case 1 (x odd, y even): 2 × 2 = 4 pairs → (1,2), (1,4), (3,2), (3,4)
• Case 2 (x even, y odd): 1 × 2 = 2 pairs → (2,1), (2,3)

Total: 4 + 2 = 6 pairs where Alice wins.

This matches our manual count, confirming the formula works correctly!

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the algorithm performs a fixed number of arithmetic operations regardless of the input values n and m. The operations include:

• Two integer divisions with floor operation: (n + 1) // 2 and (m + 1) // 2
• Two integer divisions with floor operation: n // 2 and m // 2
• Two multiplications: a1 * b2 and a2 * b1
• One addition to combine the results

All these operations take constant time, making the overall time complexity O(1).

The space complexity is O(1) because the algorithm uses only a fixed amount of extra space to store four intermediate variables (a1, b1, a2, b2) regardless of the input size. No additional data structures that scale with input are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Winning Condition

Pitfall: Developers might incorrectly assume that the game involves strategy or optimal play, trying to implement complex game theory algorithms or dynamic programming solutions. They might think players need to choose which direction (clockwise or anti-clockwise) gives them an advantage.

Why it happens: The problem description mentions "circular field" and "choosing from clockwise or anti-clockwise direction," which can mislead developers into thinking the choice of direction matters strategically.

Solution: Recognize that regardless of which flowers players pick (clockwise or anti-clockwise), the total number of flowers determines the winner. The game is purely determined by parity - if the total x + y is odd, Alice wins (she takes the last flower); if even, Bob wins.

2. Incorrect Counting of Odd/Even Numbers

Pitfall: Using the wrong formula to count odd and even numbers in a range, such as:

• Using n // 2 for odd count (incorrect for odd values of n)
• Using (n + 1) // 2 for even count (incorrect)
• Forgetting that the range starts from 1, not 0

Example of incorrect implementation:

# WRONG: This incorrectly counts odd numbers
odd_count = n // 2  # This would give 2 for n=5, but there are 3 odd numbers (1,3,5)

# WRONG: Off-by-one error for even numbers
even_count = (n - 1) // 2  # This would give 1 for n=4, but there are 2 even numbers (2,4)

Solution: Use the correct formulas:

• Odd count in [1, n]: (n + 1) // 2
• Even count in [1, n]: n // 2

Verify with small examples: For n=5, odds are {1,3,5} = 3 numbers, evens are {2,4} = 2 numbers.

3. Integer Overflow in Large Inputs

Pitfall: When n and m are very large, the multiplication odd_count_n * even_count_m might cause integer overflow in languages with fixed integer sizes.

Solution: In Python, this isn't an issue due to arbitrary precision integers. However, in other languages like C++ or Java, consider:

• Using long/long long data types
• Checking for potential overflow before multiplication
• Using modular arithmetic if the problem requires it

4. Overthinking the Circular Field Aspect

Pitfall: Trying to model the circular arrangement or thinking that the relative positions of Alice and Bob matter for the game outcome.

Why it happens: The "circular field" description might lead developers to think about circular arrays, modular arithmetic, or that picking flowers changes the relative positions somehow.

Solution: The circular field is just flavor text. The key insight is that players simply pick flowers until none remain - the spatial arrangement doesn't affect the outcome, only the total count matters.