纸牌号码

给定数字N ，任务是检查N是否为Droll Number 。如果N是纸卷号码，则打印“是”，否则打印“否” 。

A droll Number is a number such that the sum of even prime divisors of N is equaled to the sum of odd prime divisors of N.

例子：

Input: N = 6272
Output: Yes
Explanation:
6272 = 2*2*2*2*2*2*2*7*7 is droll
since 2+2+2+2+2+2+2 = 14 = 7+7.

Input: N = 10
Output: No

方法：这个想法是找到偶数素数和和奇数素数和，并检查它们是否相等。如果它们相等，则N是纸卷号码，并为其打印“是” ，否则打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to check droll numbers
bool isDroll(int n)
{
    if (n == 1)
        return false;
  
    // To store sum of even prime factors
    int sum_even = 0;
  
    // To store sum of odd prime factors
    int sum_odd = 0;
  
    // Add the number of 2s
    // that divide n in sum_even
    while (n % 2 == 0) {
        sum_even += 2;
        n = n / 2;
    }
  
    // N must be odd at this point.
    // So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i + 2) {
  
        // While i divides n,
        // print i and divide n
        while (n % i == 0) {
            sum_odd += i;
            n = n / i;
        }
    }
  
    // This condition is to handle
    // the case when n is a prime
    // number greater than 2
    if (n > 2)
        sum_odd += n;
  
    // Condition to check droll number
    return sum_even == sum_odd;
}
  
Driver Code int main()
{
    // Given Number N
    int N = 72;
  
    // Function Call
    if (isDroll(N))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program for the above approach
class GFG{ 
  
// Function to check droll numbers
static boolean isDroll(int n)
{
    if (n == 1)
        return false;
  
    // To store sum of even prime factors
    int sum_even = 0;
  
    // To store sum of odd prime factors
    int sum_odd = 0;
  
    // Add the number of 2s
    // that divide n in sum_even
    while (n % 2 == 0)
    {
        sum_even += 2;
        n = n / 2;
    }
  
    // N must be odd at this point.
    // So we can skip
    // one element (Note i = i +2)
    for(int i = 3; i <= Math.sqrt(n); 
                   i = i + 2)
    {
         
       // While i divides n, 
       // print i and divide n 
       while (n % i == 0)
       { 
           sum_odd += i; 
           n = n / i; 
       } 
    }
      
    // This condition is to handle 
    // the case when n is a prime 
    // number greater than 2 
    if (n > 2) 
        sum_odd += n;
  
    // Condition to check droll number
    return sum_even == sum_odd;
}
  
// Driver code 
public static void main(String[] args) 
{ 
      
    // Given Number N
    int n = 72;
  
    // Function Call
    if (isDroll(n))
        System.out.println("Yes");
    else
        System.out.println("No");
} 
} 
  
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
import math;
  
# Function to check droll numbers
def isDroll(n):
  
    if (n == 1):
        return False;
  
    # To store sum of even prime factors
    sum_even = 0;
  
    # To store sum of odd prime factors
    sum_odd = 0;
  
    # Add the number of 2s
    # that divide n in sum_even
    while (n % 2 == 0):
        sum_even += 2;
        n = n // 2;
      
    # N must be odd at this point.
    # So we can skip
    # one element (Note i = i +2)
    for i in range(3, int(math.sqrt(n)) + 1, 2):
  
        # While i divides n,
        # print i and divide n
        while (n % i == 0):
            sum_odd += i;
            n = n // i;
          
    # This condition is to handle
    # the case when n is a prime
    # number greater than 2
    if (n > 2):
        sum_odd += n;
  
    # Condition to check droll number
    return sum_even == sum_odd;
  
# Driver Code
  
# Given Number N
N = 72;
  
# Function Call
if (isDroll(N)):
    print("Yes");
else:
    print("No");
  
# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{ 
  
// Function to check droll numbers
static bool isDroll(int n)
{
    if (n == 1)
        return false;
  
    // To store sum of even prime factors
    int sum_even = 0;
  
    // To store sum of odd prime factors
    int sum_odd = 0;
  
    // Add the number of 2s
    // that divide n in sum_even
    while (n % 2 == 0)
    {
        sum_even += 2;
        n = n / 2;
    }
  
    // N must be odd at this point.
    // So we can skip
    // one element (Note i = i +2)
    for(int i = 3; i <= Math.Sqrt(n); 
                   i = i + 2)
    {
         
       // While i divides n, 
       // print i and divide n 
       while (n % i == 0)
       { 
           sum_odd += i; 
           n = n / i; 
       } 
    }
      
    // This condition is to handle 
    // the case when n is a prime 
    // number greater than 2 
    if (n > 2) 
        sum_odd += n;
  
    // Condition to check droll number
    return sum_even == sum_odd;
}
  
// Driver code 
public static void Main() 
{ 
      
    // Given Number N
    int n = 72;
  
    // Function Call
    if (isDroll(n))
        Console.Write("Yes");
    else
        Console.Write("No");
} 
} 
  
// This code is contributed by Code_Mech

输出：

Yes

时间复杂度： O(sqrt(N))