PROBLEM LINK:
Author:Arkapravo Ghosh
Tester:Aparup Ghosh
Editorialist:Aparup Ghosh
DIFFICULTY:
EASY
PREREQUISITES:
Primality test, factorization
PROBLEM:
Given a positive integer N, you need to find the number of factors of N except 1 and itself. If N is a prime number then simply output -1.
QUICK EXPLANATION:
We can find the number of factors of N except 1 and itself by iterating from 2 upto sqrt(N). If the no of factors in zero(0), then output -1(i.e. N is a prime) otherwise output the number of factors.
EXPLANATION:
According to the given problem we need to find the number of factors of N except 1 and itself, as it is mentioned that the job cannot be performed in 1 day(according to the instructions) or N days(as Chef will leave). So we iterate from 2 to sqrt(N) and keep a count of the number of factors of N. The idea of iterating from 2 to sqrt(N) is simple, you can identify it with a simple observation. For every number (< sqrt(N) ) which is a factor of N, there is another number(> sqrt(N) ) which is a factor of N too. So for every number(< sqrt(N) ), if it is a factor of N, we increase the count by 2. If the factor is equal to sqrt(N) (in case N is perfect square), then we increase the count by 1. After counting the number of factors of N, if the count is equal to 0(i.e. N is prime), we print -1, else we print the number of factors of N. You can find the solutions below.
AUTHOR'S AND EDITORIALIST'S SOLUTIONS:
Author's solution can be found here.
Editorialist’s solution can be found here.