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C Program for Difference between sums of odd and even digits

  • Last Updated : 05 Dec, 2018

Given a long integer, we need to find if the difference between sum of odd digits and sum of even digits is 0 or not. The indexes start from zero (0 index is for leftmost digit).

Examples:

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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 not  equal
to zero. So print no.

Method 1: One by one traverse digits and find the two sums. If difference between two sums is 0, print yes, else no.

Method 2 : This can be easily solved using divisibility of 11. This condition is only satisfied if the number is divisible by 11. So check the number is divisible by 11 or not.




// 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;
}
Output:
Yes

Please refer complete article on Difference between sums of odd and even digits for more details!

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