CF 1331B - Limericks

CF 1331B - Limericks

Rating: -
Tags: *special, math, number theory
Solve time: 3m 36s
Verified: no

Solution

Problem Understanding

The problem asks us to compute a specific numeric property related to an integer input, denoted as a. While the problem statement is written in a poetic form, the underlying task is to find the number of integers less than a that are coprime to a. In other words, given a, we are asked to compute the value of Euler's totient function, commonly written as φ(a). This function counts the integers from 1 to a - 1 that share no common factors with a other than 1.

The input is a single integer a in the range 4 to 998, inclusive. Because the maximum a is less than 1000, we can afford to use simple factorization or iteration methods without worrying about hitting time limits. Since we only have one integer per test case, an O(sqrt(a)) approach is sufficient, as it will perform at most around 31 iterations in the worst case.

A non-obvious edge case arises when a is a power of a prime, such as 4, 8, or 9. For a = 4, the integers less than 4 that are coprime to 4 are 1 and 3, giving φ(4) = 2. A naive approach that only removes even numbers would fail for a = 9, which requires removing multiples of 3. Another potential pitfall is assuming a is prime or even; the input can be any integer in the given range.

Approaches

The brute-force approach to compute φ(a) would be to iterate through all integers from 1 to a - 1 and check for each whether the greatest common divisor with a is 1. This works because the GCD function correctly identifies coprime numbers. The operation count in the worst case is approximately a * log(a) if using Euclid's algorithm for GCD, which for a up to 998 is about 10^4 operations and acceptable for a single test case.

The optimal approach uses the multiplicative property of the Euler totient function. If a can be factorized as a = p1^e1 * p2^e2 * ... * pk^ek for distinct prime factors pi, then φ(a) = a * Π(1 - 1/pi). The observation that every integer less than a that is not divisible by any of its prime factors is coprime allows us to compute the answer in O(sqrt(a)) time, by trial division to find prime factors. This is faster for larger ranges, but with a < 1000, both brute-force and factorization approaches are feasible.

Algorithm Walkthrough

1. Read the integer a from input. This is the number for which we will compute the totient function.
2. Initialize a variable result to a. This will hold the running product for the totient formula.
3. Iterate over all integers i from 2 up to sqrt(a) (inclusive). Each i is a candidate