检查数字是否可被31整除

给定数字N ，任务是检查数字是否可被31整除。

例子：

Input: N = 1922
Output: Yes
Explanation:
31 * 62 = 1922

Input: N = 2722400
Output: No

方法： 31的除法检验是：

1. 提取最后一位数字。
2. 从除去最后一位后获得的剩余号码中减去3 *最后一位。
3. 重复上述步骤，直到获得两位数字或零。
4. 如果两位数可被31整除或为0，则原始数字也可被31整除。

例如：

If N = 49507

Step 1:
  N = 49507
  Last digit = 7
  Remaining number = 4950
  Subtracting 3 times last digit
  Resultant number = 4950 - 3*7 = 4929

Step 2:
  N = 4929
  Last digit = 9
  Remaining number = 492
  Subtracting 3 times last digit
  Resultant number = 492 - 3*9 = 465

Step 3:
  N = 465
  Last digit = 5
  Remaining number = 46
  Subtracting 3 times last digit
  Resultant number = 46 - 3*5 = 31

Step 4:
  N = 31
  Since N is a two-digit number,
  and 31 is divisible by 31

Therefore N = 49507 is also divisible by 31

下面是上述方法的实现：

C++

// C++ program to check whether a number
// is divisible by 31 or not
#include
#include
  
using namespace std;
  
// Function to check if the number is divisible by 31 or not 
bool isDivisible(int n) 
{
    int d;
      
    // While there are at least two digits 
    while (n / 100) 
    {
  
        // Extracting the last 
        d = n % 10;
  
        // Truncating the number 
        n /= 10;
  
        // Subtracting three times the last 
        // digit to the remaining number 
        n = abs(n-(d * 3));
    }
      
    // Finally return if the two-digit
    // number is divisible by 31 or not
    return (n % 31 == 0) ;
}
  
// Driver Code 
int main() 
{
    int N = 1922;
  
    if (isDivisible(N)) 
        cout<<"Yes"<

---

Java

// Java program to check whether a number
// is divisible by 31 or not
import java.util.*;
  
class GFG{
   
// Function to check if the number is divisible by 31 or not 
static boolean isDivisible(int n) 
{
    int d;
       
    // While there are at least two digits 
    while ((n / 100) > 0) 
    {
   
        // Extracting the last 
        d = n % 10;
   
        // Truncating the number 
        n /= 10;
   
        // Subtracting three times the last 
        // digit to the remaining number 
        n = Math.abs(n - (d * 3));
    }
       
    // Finally return if the two-digit
    // number is divisible by 31 or not
    return (n % 31 == 0) ;
}
   
// Driver Code 
public static void main(String[] args) 
{
    int N = 1922;
   
    if (isDivisible(N)) 
        System.out.print("Yes");
    else
        System.out.print("No");
}     
} 
  
// This code is contributed by PrinciRaj1992

Python 3

# Python program to check whether a number
# is divisible by 31 or not
  
# Function to check if the number is 
# divisible by 31 or not 
def isDivisible(n) : 
  
    # While there are at least two digits 
    while n // 100 : 
  
        # Extracting the last 
        d = n % 10
  
        # Truncating the number 
        n //= 10
  
        # Subtracting three times the last 
        # digit to the remaining number 
        n = abs(n-(d * 3))
  
    # Finally return if the two-digit
    # number is divisible by 31 or not
    return (n % 31 == 0) 
  
# Driver Code 
if __name__ == "__main__" : 
  
    n = 1922
  
    if (isDivisible(n)) : 
        print("Yes") 
    else : 
        print("No")

C#

// C# program to check whether a number
// is divisible by 31 or not
using System;
  
class GFG{
    
// Function to check if the number is divisible by 31 or not 
static bool isDivisible(int n) 
{
    int d;
        
    // While there are at least two digits 
    while ((n / 100) > 0) 
    {
    
        // Extracting the last 
        d = n % 10;
    
        // Truncating the number 
        n /= 10;
    
        // Subtracting three times the last 
        // digit to the remaining number 
        n = Math.Abs(n - (d * 3));
    }
        
    // Finally return if the two-digit
    // number is divisible by 31 or not
    return (n % 31 == 0) ;
}
    
// Driver Code 
public static void Main(String[] args) 
{
    int N = 1922;
    
    if (isDivisible(N)) 
        Console.Write("Yes");
    else
        Console.Write("No");
}     
}
  
// This code is contributed by Rajput-Ji

输出：

Yes