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C Program to Check if count of divisors is even or odd?
Given a number “n” as an input, this program is to find the total number of divisors of n is even or odd. An even number is an integer that is exactly divisible by 2. Example: 0, 8, -24
An odd number is an integer that is not exactly divisible by 2. Example: 1, 7, -11, 15
Input: 10 Output: Even
Explanation
Find all the divisors of the n and then check if the total number of divisors are even or odd. To do this find all divisor and count the number and then divide this number by 2 to check if it is even or odd.
Example
#include <iostream>
#include <math.h>
using namespace std;
int main() {
int n=10;
int count = 0;
for (int i = 1; i <= sqrt(n) + 1; i++) {
if (n % i == 0)
count += (n / i == i) ? 1 : 2;
}
if (count % 2 == 0)
printf("Even
");
else
printf("Odd
");
return 0;
}
Updated on: 20-Aug-2019
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