确定给定的整数N是否为奇数

给定一个整数N，我们的任务是确定整数N是否为奇数。如果是，则打印“是”，否则输出“否”。

特殊数字是数字的数字总和的三倍。

例子：

Input: N = 27
Output: Yes
Explanation:
Digit sum for 27 is 9 and 3 * 9 = 27 which is equal to N. Hence the output is yes.

Input: N = 36
Output: No
Explanation:
Digit sum for 36 is 9 and 3 * 9 = 27 which is not equal to N. Hence the output is no.

方法：

为了解决上述问题，我们必须首先找到数字N的位数之和。然后检查该数字的位数之和乘以3是否实际上是数字N本身。如果是，则打印“是”，否则输出为“否”。

下面是上述方法的实现：

C++

// C++ implementation to check if the
// number is peculiar
  
#include 
using namespace std;
  
// Function to find sum of digits
// of a number
int sumDig(int n)
{
    int s = 0;
  
    while (n != 0) {
        s = s + (n % 10);
  
        n = n / 10;
    }
  
    return s;
}
  
// Function to check if the 
// number is peculiar
bool Pec(int n)
{
    // Store a duplicate of n
    int dup = n;
  
    int dig = sumDig(n);
  
    if (dig * 3 == dup)
        return true;
  
    else
        return false;
}
  
// Driver code
int main()
{
    int n = 36;
  
    if (Pec(n) == true)
        cout << "Yes" << endl;
  
    else
        cout << "No" << endl;
  
    return 0;
}

Java

// Java implementation to check if the
// number is peculiar
import java.io.*;
  
class GFG{
  
// Function to find sum of digits
// of a number
static int sumDig(int n)
{
    int s = 0;
  
    while (n != 0)
    {
        s = s + (n % 10);
        n = n / 10;
    }
    return s;
}
  
// Function to check if number is peculiar
static boolean Pec(int n)
{
      
    // Store a duplicate of n
    int dup = n;
    int dig = sumDig(n);
  
    if (dig * 3 == dup)
        return true;
    else
        return false;
}
  
// Driver code
public static void main (String[] args) 
{
    int n = 36;
  
    if (Pec(n) == true)
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by shubhamsingh10

Python3

# Python3 implementation to check if the
# number is peculiar
  
# Function to get sum of digits
# of a number 
def sumDig(n):
      
    s = 0
    while (n != 0): 
        s = s + int(n % 10) 
        n = int(n / 10) 
      
    return s
      
# Function to check if the  
# number is peculiar     
def Pec(n):
      
    dup = n
    dig = sumDig(n)
  
    if(dig * 3 == dup):
        return "Yes"
    else :
        return "No"
          
# Driver code 
n = 36
  
if Pec(n) == True:
    print("Yes")
else:
    print("No")
  
# This code is contributed by grand_master

C#

// C# implementation to check if the
// number is peculiar
using System;
  
class GFG{
  
// Function to find sum of digits
// of a number
static int sumDig(int n)
{
    int s = 0;
  
    while (n != 0)
    {
        s = s + (n % 10);
        n = n / 10;
    }
    return s;
}
  
// Function to check if the number is peculiar
static bool Pec(int n)
{
      
    // Store a duplicate of n
    int dup = n;
    int dig = sumDig(n);
  
    if (dig * 3 == dup)
        return true;
    else
        return false;
}
  
// Driver code
public static void Main() 
{
    int n = 36;
  
    if (Pec(n) == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by Akanksha_Rai

输出：

No