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Clumsy Factorial - The factorial of a positive integer n is the product of all positive integers less than or equal to n. * For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. We make a clumsy factorial using the integers in decreasing order by swapping out the multiply operations for a fixed rotation of operations with ...
Factorial Trailing Zeroes - Given an integer n, return the number of trailing zeroes in n!. Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1. Example 1: Input: n = 3 Output: 0 Explanation: 3! = 6, no trailing zero.
For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. We instead make a clumsy factorial: using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.
class Solution: def clumsy (self, n: int)-> int: if n == 1: return 1 if n == 2: return 2 if n == 3: return 6 if n == 4: return 7 if n % 4 == 1: return n + 2 if n % 4 == 2: return n + 2 if n % 4 == 3: return n-1 return n + 1
18 lut 2024 · You can assume there are 3 possible input functions: sum, fib, and factorial. sum accepts two integers a and b and returns a + b . fib accepts a single integer n and returns 1 if n <= 1 or fib(n - 1) + fib(n - 2) otherwise.
24 wrz 2024 · Given two numbers N and M. Find the number of ways in which factorial N can be expressed as a sum of two or more consecutive numbers. Print the result modulo M.Examples: Input : N = 3, M = 7 Output : 1 Explanation: 3! can be expressed in one way, i.e. 1 + 2 + 3 = 6. Hence 1 % 7 = 1 Input : N = 4, M = 7 Output : 1 Explanation: 4! can be expressed in
I just thought that someone might know a real solution to this problem. I was on a programming contest back in 2004, and there was this problem: Given n, find sum of digits of n!. n can be from 0 to 10000. Time limit: 1 second.
