To solve this problem, we need to find the maximum value of (A \mod B) where (A) and (B) are integers in the range ([L, R]) and (A > B).

Approach

The key insight to solve this problem efficiently is to recognize that for a given (B), the maximum possible remainder when dividing any number by (B) is (B-1) (since the remainder must be less than (B)). This remainder is achievable if there exists an (A) in the range ([B+1, R]) such that (A = 2B -1) (since (2B-1 \mod B = B-1)).

If (2B -1 > R), then the maximum remainder for (B) is (R \mod B) (since (R) is the largest number in the range, its remainder when divided by (B) will be the highest possible for any (A \leq R)).

Solution Code

L, R = map(int, input().split())

if L == R:
    print(0)
else:
    max_mod = 0
    for B in range(L, R):
        if 2 * B - 1 <= R:
            current = B - 1
        else:
            current = R % B
        if current > max_mod:
            max_mod = current
    print(max_mod)

Explanation

1. Edge Case Handling: If (L == R), there are no valid pairs where (A > B), so the result is 0.
2. Iterate Over Possible Values of (B): For each (B) from (L) to (R-1):
   • Check if (2B -1 \leq R). If yes, the maximum remainder for this (B) is (B-1).
   • If not, the maximum remainder is (R \mod B).
3. Track Maximum Remainder: Keep updating the maximum remainder found across all values of (B).

This approach efficiently computes the result with a time complexity of (O(R-L)), which is feasible for the given constraints (up to (10^6)). This ensures the solution is both optimal and easy to understand.

Example: For (L = 3) and (R =7):

• (B=3): (2*3-1=5 \leq 7), remainder is (2).
• (B=4): (2*4-1=7 \leq7), remainder is (3).
• (B=5): (2*5-1=9>7), remainder is (7\%5=2).
• (B=6): (2*6-1=11>7), remainder is (7\%6=1).

The maximum remainder is (3), which is the correct result.

This solution effectively handles all cases and provides the optimal result.

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