# 检查数字奇数位的数字总和是否可被K整除

2021年4月26日16:15:19 发表评论 703 次浏览

## 本文概述

• 在奇数个位置(从右到左)找到" N"的数字总和。
• 然后, 以" K"取模, 检查总和的可除性。
• 如果可以整除, 则输出"是", 否则输出"否"。

## C++

``````//C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

//function that checks the
//divisibility of the sum
//of the digits at odd places
//of the given number
bool SumDivisible( int n, int k)
{
int sum = 0, position = 1;
while (n> 0) {

//if position is odd
if (position % 2 == 1)
sum += n % 10;
n = n /10;
position++;
}

if (sum % k == 0)
return true ;
return false ;
}

//Driver code
int main()
{
int n = 592452;
int k = 3;

if (SumDivisible(n, k))
cout <<"YES" ;
else
cout <<"NO" ;
return 0;
}``````

## Java

``````//Java implementation of the approach
import java.util.*;

class solution
{

//function that checks the
//divisibility of the sum
//of the digits at odd places
//of the given number
static boolean SumDivisible( int n, int k)
{
int sum = 0 , position = 1 ;
while (n> 0 ) {

//if position is odd
if (position % 2 == 1 )
sum += n % 10 ;
n = n /10 ;
position++;
}

if (sum % k == 0 )
return true ;
return false ;
}

//Driver code
public static void main(String arr[])
{
int n = 592452 ;
int k = 3 ;

if (SumDivisible(n, k))
System.out.println( "YES" );
else
System.out.println( "NO" );

}
}
//This code is contributed by Surendra_Gangwar``````

## Python 3

``````# Python 3 implementation of the approach

# function that checks the divisibility
# of the sum of the digits at odd places
# of the given number
def SumDivisible(n, k):

sum = 0
position = 1
while (n> 0 ) :

# if position is odd
if (position % 2 = = 1 ):
sum + = n % 10
n = n //10
position + = 1

if ( sum % k = = 0 ):
return True
return False

# Driver code
if __name__ = = "__main__" :
n = 592452
k = 3

if (SumDivisible(n, k)):
print ( "YES" )
else :
print ( "NO" )

# This code is contributed
# by ChitraNayal``````

## C#

``````//C# implementation of the approach
using System;

class GFG
{
//function that checks the
//divisibility of the sum
//of the digits at odd places
//of the given number
static bool SumDivisible( int n, int k)
{
int sum = 0, position = 1;
while (n> 0)
{

//if position is odd
if (position % 2 == 1)
sum += n % 10;
n = n /10;
position++;
}

if (sum % k == 0)
return true ;
return false ;
}

//Driver code
static public void Main ()
{
int n = 592452;
int k = 3;

if (SumDivisible(n, k))
Console.WriteLine( "YES" );
else
Console.WriteLine( "NO" );
}
}

//This code is contributed by Sachin``````

## PHP

``````<?php
//PHP implementation of the approach

//function that checks the divisibility
//of the sum of the digits at odd places
//of the given number
function SumDivisible( \$n , \$k )
{
\$sum = 0;
\$position = 1;
while ( \$n> 0)
{

//if position is odd
if ( \$position % 2 == 1)
\$sum += \$n % 10;
\$n = (int) \$n /10;
\$position ++;
}

if ( \$sum % \$k == 0)
return true;
return false;
}

//Driver code
\$n = 592452;
\$k = 3;

if (SumDivisible( \$n , \$k ))
echo "YES" ;
else
echo "NO" ;

//This code is contributed
//by Sach_Code
?>``````

``YES``