本文概述
- CPP
- Java
- python
- C#
- PHP
例子:
Input : 1212112
Output : Yes
Explanation:-
the odd position element is 2+2+1=5
the even position element is 1+1+1+2=5
the difference is 5-5=0.so print yes.Input :12345
Output : No
Explanation:-
the odd position element is 1+3+5=9
the even position element is 2+4=6
the difference is 9-6=3 notequal
to zero. So print no.
方法1:一位一位地遍历数字并找到两个和。如果两个和之间的差为0, 则打印是, 否则打印否。
【奇数和偶数之和之间有什么区别()】方法2:
可以使用以下方法轻松解决11的除数。仅当数字可被11整除时, 才能满足此条件。因此, 请检查数字是否可被11整除。
CPP
//C++ program to check if difference between sum of
//odd digits and sum of even digits is 0 or not
#include <
bits/stdc++.h>
using namespace std;
bool isDiff0( long long int n)
{
return (n % 11 == 0);
}int main() {long int n = 1243
if (isDiff0(n))
cout <
<
"Yes" ;
else
cout <
<
"No" ;
return 0;
}
Java
//Java program to check if difference between sum of
//odd digits and sum of even digits is 0 or notimport java.io.*;
import java.util.*;
class GFG
{
public static boolean isDiff( int n)
{
return (n % 11 == 0 );
}
public static void main (String[] args)
{
int n = 1243 ;
if (isDiff(n))
System.out.print( "Yes" );
else
System.out.print( "No" );
}
}
python
# Python program to check if difference between sum of
# odd digits and sum of even digits is 0 or notdef isDiff(n):
return (n % 11 = = 0 )# Driver code
n = 1243 ;
if (isDiff(n)):
print ( "Yes" )
else :
print ( "No" )# Mohit Gupta_OMG <
0_o>
C#
//C# program to check if difference
//between sum of odd digits and sum
//of even digits is 0 or not
using System;
class GFG {public static bool isDiff( int n)
{
return (n % 11 == 0);
}public static void Main ()
{
int n = 1243;
if (isDiff(n))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}//This code is contributed by vt_m.
PHP
<
?php
//PHP program to check if
//difference between sum of
//odd digits and sum of
//even digits is 0 or notfunction isDiff0( $n )
{
return ( $n % 11 == 0);
}//Driver Code
$n = 1243;
if (isDiff0( $n ))
echo "Yes" ;
else
echo "No" ;
//This code is contributed by nitin mittal
?>
输出如下:
Yes
如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。
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