# Arithmetic Number in C++

The arithmetic number is a number which has the average of all positive divisors is an integer i.e. for a number n if the number of divisors can divide the sum of divisors then n is an arithmetic number.

Let’s take an example to understand the concept better,

Input : n = 6
Output : YES
Explanation :
Divisors are 1 , 2 , 3 , 6
Sum = 1+2+3+6 = 12
Number of divisors = 4
Sum of divisors / number of divisor = 12 / 4 = 3

## Algorithm

Step 1 : Calculate the sum of divisors and store into sum variable.
Step 2 : Find the total number of divisors.
Step 3 : Check if the remainder when sum of divisors is divided by the total number of divisors is equal to 0.
Step 4 : If remainder is equal 0. Print YES. Else print NO.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(int n, bool prime[],bool primesquare[], int a[]);
int countDivisors(int n) ;
int sumofFactors(int n) ;
int main(){
int n = 46;
int divcount = countDivisors(n);
int divsum = sumofFactors(n);
if(divsum % divcount == 0 ){
cout<<"YES";
}
else
cout<<"NO";
return 0;
}
void SieveOfEratosthenes(int n, bool prime[],bool primesquare[], int a[]){
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int i = 0; i <= (n * n + 1); i++)
primesquare[i] = false;
prime[1] = false;
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * 2; i <= n; i += p)
prime[i] = false;
}
}
int j = 0;
for (int p = 2; p <= n; p++) {
if (prime[p]) {
a[j] = p;
primesquare[p * p] = true;
j++;
}
}
}
int countDivisors(int n){
if (n == 1)
return 1;
bool prime[n + 1], primesquare[n * n + 1];
int a[n];
SieveOfEratosthenes(n, prime, primesquare, a);
int ans = 1;
for (int i = 0;; i++) {
if (a[i] * a[i] * a[i] > n)
break;
int cnt = 1;
while (n % a[i] == 0){
n = n / a[i];
cnt = cnt + 1;
}
ans = ans * cnt;
}
if (prime[n])
ans = ans * 2;
else if (primesquare[n])
ans = ans * 3;
else if (n != 1)
ans = ans * 4;
return ans;
}
int sumofFactors(int n){
int res = 1;
for (int i = 2; i <= sqrt(n); i++) {
int count = 0, curr_sum = 1;
int curr_term = 1;
while (n % i == 0) {
count++;
n = n / i;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}

## Output

YES

Updated on: 24-Oct-2019

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