Given three positive integers n, m, and t, where n represents the upper limit of a sequence of numbers (1 to n), m is a divisor, and t is a target position, determine the number that would be found at the t-th position after sorting the numbers 1 to n based on their remainders when divided by m. The sorting order is: first by remainder (0, 1, 2, ..., m-1), and within each remainder group, in ascending order of the original number." Example: "If n = 10, m = 4, and t = 8, the sorted sequence would be [4, 8, 1, 5, 9, 2, 6, 10, 3, 7]. Because: Number that can be divided by m: 4, 8 Number that when divide by m has a remaining of 1: 1, 5, 9 Number that when divide by m has a remaining of 3: 3, 7 Therefore, the number at the 8th position is 10." "How can I efficiently find the number at the t-th position in a sequence sorted by remainders? I need a solution that works for large values of n (up to 10^16) and m (up to 10^6) in Python, while respecting a time limit of 8 seconds. The sorting criteria are: first by the remainder when divided by m, and then in ascending order within each remainder group." Ok so i have try another the one and it's pretty fast but it fail one test case: Input: 768767 123 6250 Output: 768750 And my output was 768727 This is the code def specialEvent(n, m, t): counts = [0] * m for i in range(1, m + 1): counts[i % m] = (n - i) // m + 1 group_index = 0 while t > counts[group_index]: t -= counts[group_index] group_index += 1 result = group_index + (t - 1) * m return result Continue reading...
