算术数

在数论中，算术数是一个整数，其正除数的平均值也是一个整数。换句话说，如果除数的数量除以除数之和，则数字N是算术数。
给定正整数n 。任务是检查n是否为算术数。

例子：

Input : n = 6
Output : Yes
Sum of divisor of 6 = 1 + 2 + 3 + 6 = 12.
Number of divisor of 6 = 4.
So, on dividing Sum of divisor by Number of divisor
= 12/4 = 3, which is an integer.

Input : n = 2
Output : No

算法

1. 找出一个数字的所有因子的总和，即总和。
2. 查找除数计数(例如Count)。
3. 检查总和是否可被计数整除。

C++

// CPP program to check if a number is Arithmetic
// number or not
#include 
using namespace std;
  
// Sieve Of Eratosthenes
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;
  
    // 1 is not a prime number
    prime[1] = false;
  
    for (int p = 2; p * p <= n; p++) {
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
            // Update all multiples of p
            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]) {
            // Storing primes in an array
            a[j] = p;
  
            // Update value in primesquare[p*p],
            // if p is prime.
            primesquare[p * p] = true;
            j++;
        }
    }
}
  
// Function to count divisors
int countDivisors(int n)
{
    // If number is 1, then it will have only 1
    // as a factor. So, total factors will be 1.
    if (n == 1)
        return 1;
  
    bool prime[n + 1], primesquare[n * n + 1];
  
    int a[n]; // for storing primes upto n
  
    // Calling SieveOfEratosthenes to store prime
    // factors of n and to store square of prime
    // factors of n
    SieveOfEratosthenes(n, prime, primesquare, a);
  
    // ans will contain total number of 
    // distinct divisors
    int ans = 1;
  
    // Loop for counting factors of n
    for (int i = 0;; i++) {
  
        // a[i] is not less than cube root n
        if (a[i] * a[i] * a[i] > n)
            break;
  
        // Calculating power of a[i] in n.
        // cnt is power of prime a[i] in n.
        int cnt = 1; 
  
        // if a[i] is a factor of n
        while (n % a[i] == 0) 
        {
            n = n / a[i];
            cnt = cnt + 1; // incrementing power
        }
  
        // Calculating number of divisors
        // If n = a^p * b^q then total 
        // divisors of n
        // are (p+1)*(q+1)
        ans = ans * cnt;
    }
  
    // if a[i] is greater than cube root of n
  
    // First case
    if (prime[n])
        ans = ans * 2;
  
    // Second case
    else if (primesquare[n])
        ans = ans * 3;
  
    // Third casse
    else if (n != 1)
        ans = ans * 4;
  
    return ans; // Total divisors
}
  
// Returns sum of all factors of n.
int sumofFactors(int n)
{
    // Traversing through all prime factors.
    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;
    }
  
    // This condition is to handle
    // the case when n is a prime
    // number greater than 2.
    if (n >= 2)
        res *= (1 + n);
  
    return res;
}
  
// Check if number is Arithmetic Number
// or not.
bool checkArithmetic(int n)
{
    int count = countDivisors(n);
    int sum = sumofFactors(n);
  
    return (sum  % count == 0);
}
  
// Driven Program
int main()
{
    int n = 6;
    (checkArithmetic(n)) ? (cout << "Yes") : 
                           (cout << "No");
    return 0;
}

Java

// Java program to check if a number is Arithmetic
// number or not
class GFG 
{
      
// Sieve Of Eratosthenes
static void SieveOfEratosthenes(int n, boolean prime[],
                        boolean primesquare[], int a[])
{
    for (int i = 2; i <= n; i++)
        prime[i] = true;
  
    for (int i = 0; i <= (n * n ); i++)
        primesquare[i] = false;
  
    // 1 is not a prime number
    prime[1] = false;
  
    for (int p = 2; p * p <= n; p++) 
    {
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) 
        {
            // Update all multiples of p
            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]) 
        {
            // Storing primes in an array
            a[j] = p;
  
            // Update value in primesquare[p*p],
            // if p is prime.
            primesquare[p * p] = true;
            j++;
        }
    }
}
  
// Function to count divisors
static int countDivisors(int n)
{
    // If number is 1, then it will have only 1
    // as a factor. So, total factors will be 1.
    if (n == 1)
        return 1;
  
    boolean prime[] = new boolean[n + 1],
            primesquare[] = new boolean[n * n + 1];
  
    int a[] = new int[n]; // for storing primes upto n
  
    // Calling SieveOfEratosthenes to store prime
    // factors of n and to store square of prime
    // factors of n
    SieveOfEratosthenes(n, prime, primesquare, a);
  
    // ans will contain total number of 
    // distinct divisors
    int ans = 1;
  
    // Loop for counting factors of n
    for (int i = 0;; i++) 
    {
  
        // a[i] is not less than cube root n
        if (a[i] * a[i] * a[i] > n)
            break;
  
        // Calculating power of a[i] in n.
        // cnt is power of prime a[i] in n.
        int cnt = 1; 
  
        // if a[i] is a factor of n
        while (n % a[i] == 0) 
        {
            n = n / a[i];
            cnt = cnt + 1; // incrementing power
        }
  
        // Calculating number of divisors
        // If n = a^p * b^q then total 
        // divisors of n
        // are (p+1)*(q+1)
        ans = ans * cnt;
    }
  
    // if a[i] is greater than cube root of n
  
    // First case
    if (prime[n])
        ans = ans * 2;
  
    // Second case
    else if (primesquare[n])
        ans = ans * 3;
  
    // Third casse
    else if (n != 1)
        ans = ans * 4;
  
    return ans; // Total divisors
}
  
// Returns sum of all factors of n.
static int sumofFactors(int n)
{
    // Traversing through all prime factors.
    int res = 1;
    for (int i = 2; i <= Math.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;
    }
  
    // This condition is to handle
    // the case when n is a prime
    // number greater than 2.
    if (n >= 2)
        res *= (1 + n);
  
    return res;
}
  
// Check if number is Arithmetic Number
// or not.
static boolean checkArithmetic(int n)
{
    int count = countDivisors(n);
    int sum = sumofFactors(n);
  
    return (sum % count == 0);
}
  
// Driver Program
public static void main(String[] args) 
{
    int n = 6;
    if(checkArithmetic(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code has been contributed by 29AjayKumar

Python3

# Python3 program to check if 
# a number is Arithmetic 
# number or not
import math
  
# Sieve Of Eratosthenes
def SieveOfEratosthenes(n, prime,primesquare, a):
  
    for i in range(2,n+1):
        prime[i] = True;
  
    for i in range((n * n + 1)+1):
        primesquare[i] = False;
  
    # 1 is not a
    # prime number
    prime[1] = False;
    p = 2;
  
    while (p * p <= n):
          
        # If prime[p] is 
        # not changed, then
        # it is a prime
        if (prime[p] == True):
            # Update all multiples of p
            for i in range(p * 2,n+1,p):
                prime[i] = False;
        p+=1;
  
    j = 0;
    for p in range(2,n+1):
        if (prime[p]): 
              
            # Storing primes in an array
            a[j] = p;
  
            # Update value in
            # primesquare[p*p],
            # if p is prime.
            primesquare[p * p] = True;
            j+=1;
  
# Function to count divisors
def countDivisors(n):
  
    # If number is 1, then it
    # will have only 1 as a 
    # factor. So, total factors 
    # will be 1.
    if (n == 1):
        return 1;
  
    prime = [False]*(n + 2);
    primesquare = [False]*(n *n + 3);
  
    # for storing primes upto n
    a = [0]*n; 
  
    # Calling SieveOfEratosthenes 
    # to store prime factors of 
    # n and to store square of 
    # prime factors of n
    SieveOfEratosthenes(n, prime,primesquare, a);
  
    # ans will contain 
    # total number of 
    # distinct divisors
    ans = 1;
  
    # Loop for counting
    # factors of n
    for i in range(0,True):
  
        # a[i] is not less 
        # than cube root n
        if (a[i] * a[i] * a[i] > n):
            break;
  
        # Calculating power of 
        # a[i] in n. cnt is power 
        # of prime a[i] in n.
        cnt = 1; 
  
        # if a[i] is a factor of n
        while (n % a[i] == 0): 
            n //= a[i];
              
            # incrementing power
            cnt = cnt + 1; 
  
        # Calculating number of 
        # divisors. If n = a^p * b^q 
        # then total divisors 
        # of n are (p+1)*(q+1)
        ans = ans * cnt;
  
    # if a[i] is greater 
    # than cube root of n
  
    # First case
    if (prime[n]):
        ans = ans * 2;
  
    # Second case
    elif (primesquare[n]):
        ans = ans * 3;
  
    # Third casse
    elif (n != 1):
        ans = ans * 4;
  
    return ans; # Total divisors
  
# Returns sum of
# all factors of n.
def sumofFactors(n):
  
    # Traversing through 
    # all prime factors.
    res = 1;
    for i in range(2,int(math.sqrt(n))+1): 
        count = 0; 
        curr_sum = 1;
        curr_term = 1;
        while (n % i == 0):
            count+=1;
            n //= i;
  
            curr_term *= i;
            curr_sum += curr_term;
  
        res *= curr_sum;
  
    # This condition is to handle
    # the case when n is a prime
    # number greater than 2.
    if (n >= 2):
        res *= (1 + n);
  
    return res;
  
# Check if number is 
# Arithmetic Number or not.
def checkArithmetic(n):
  
    count = countDivisors(n);
    sum = sumofFactors(n);
  
    return (sum % count == 0);
  
# Driver code
n = 6;
if(checkArithmetic(n)):
    print("Yes");
else:
    print("No");
  
# This code is contributed
# by mits

C#

// C# program to check if a number 
// is arithmetic number or not
using System;
  
class GFG 
{
      
// Sieve Of Eratosthenes
static 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 ); i++)
        primesquare[i] = false;
  
    // 1 is not a prime number
    prime[1] = false;
  
    for (int p = 2; p * p <= n; p++) 
    {
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) 
        {
            // Update all multiples of p
            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]) 
        {
            // Storing primes in an array
            a[j] = p;
  
            // Update value in primesquare[p*p],
            // if p is prime.
            primesquare[p * p] = true;
            j++;
        }
    }
}
  
// Function to count divisors
static int countDivisors(int n)
{
    // If number is 1, then it will have only 1
    // as a factor. So, total factors will be 1.
    if (n == 1)
        return 1;
  
    bool []prime = new bool[n + 1];
    bool []primesquare = new bool[n * n + 1];
  
    int []a = new int[n]; // for storing primes upto n
  
    // Calling SieveOfEratosthenes to store prime
    // factors of n and to store square of prime
    // factors of n
    SieveOfEratosthenes(n, prime, primesquare, a);
  
    // ans will contain total number of 
    // distinct divisors
    int ans = 1;
  
    // Loop for counting factors of n
    for (int i = 0;; i++) 
    {
  
        // a[i] is not less than cube root n
        if (a[i] * a[i] * a[i] > n)
            break;
  
        // Calculating power of a[i] in n.
        // cnt is power of prime a[i] in n.
        int cnt = 1; 
  
        // if a[i] is a factor of n
        while (n % a[i] == 0) 
        {
            n = n / a[i];
            cnt = cnt + 1; // incrementing power
        }
  
        // Calculating number of divisors
        // If n = a^p * b^q then total 
        // divisors of n
        // are (p+1)*(q+1)
        ans = ans * cnt;
    }
  
    // if a[i] is greater than cube root of n
  
    // First case
    if (prime[n])
        ans = ans * 2;
  
    // Second case
    else if (primesquare[n])
        ans = ans * 3;
  
    // Third casse
    else if (n != 1)
        ans = ans * 4;
  
    return ans; // Total divisors
}
  
// Returns sum of all factors of n.
static int sumofFactors(int n)
{
    // Traversing through all prime factors.
    int res = 1;
    for (int i = 2; i <= Math.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;
    }
  
    // This condition is to handle
    // the case when n is a prime
    // number greater than 2.
    if (n >= 2)
        res *= (1 + n);
  
    return res;
}
  
// Check if number is Arithmetic Number
// or not.
static bool checkArithmetic(int n)
{
    int count = countDivisors(n);
    int sum = sumofFactors(n);
  
    return (sum % count == 0);
}
  
// Driver code
public static void Main(String[] args) 
{
    int n = 6;
    if(checkArithmetic(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code contributed by Rajput-Ji

PHP

 $n)
            break;
  
        // Calculating power of 
        // a[i] in n. cnt is power 
        // of prime a[i] in n.
        $cnt = 1; 
  
        // if a[i] is a factor of n
        while ($n % $a[$i] == 0) 
        {
            $n = (int)($n / $a[$i]);
              
            // incrementing power
            $cnt = $cnt + 1; 
        }
  
        // Calculating number of 
        // divisors. If n = a^p * b^q 
        // then total divisors 
        // of n are (p+1)*(q+1)
        $ans = $ans * $cnt;
    }
  
    // if a[i] is greater 
    // than cube root of n
  
    // First case
    if ($prime[$n])
        $ans = $ans * 2;
  
    // Second case
    else if ($primesquare[$n])
        $ans = $ans * 3;
  
    // Third casse
    else if ($n != 1)
        $ans = $ans * 4;
  
    return $ans; // Total divisors
}
  
// Returns sum of
// all factors of n.
function sumofFactors($n)
{
    // Traversing through 
    // all prime factors.
    $res = 1;
    for ($i = 2; 
         $i <= sqrt($n); $i++) 
    {
        $count = 0; 
        $curr_sum = 1;
        $curr_term = 1;
        while ($n % $i == 0) 
        {
            $count++;
            $n = (int)($n / $i);
  
            $curr_term *= $i;
            $curr_sum += $curr_term;
        }
  
        $res *= $curr_sum;
    }
  
    // This condition is to handle
    // the case when n is a prime
    // number greater than 2.
    if ($n >= 2)
        $res *= (1 + $n);
  
    return $res;
}
  
// Check if number is 
// Arithmetic Number or not.
function checkArithmetic($n)
{
    $count = countDivisors($n);
    $sum = sumofFactors($n);
  
    return ($sum % $count == 0);
}
  
// Driver code
$n = 6;
echo (checkArithmetic($n)) ? 
                     "Yes" : "No";
  
// This code is contributed
// by mits
?>

输出：

Yes