Find the sum of digits of a number at even and odd places | C ++ implementation of the approach ; Function to return the reverse of a number ; Function to find the sum of the odd and even positioned digits in a number ; If c is even number then it means digit extracted is at even place ; Driver code
#include <bits/stdc++.h> NEW_LINE using namespace std ; int reverse ( int n ) { int rev = 0 ; while ( n != 0 ) { rev = ( rev * 10 ) + ( n % 10 ) ; n /= 10 ; } return rev ; } void getSum ( int n ) { n = reverse ( n ) ; int sumOdd = 0 , sumEven = 0 , c = 1 ; while ( n != 0 ) { if ( c % 2 == 0 ) sumEven += n % 10 ; else sumOdd += n % 10 ; n /= 10 ; c ++ ; } cout << " Sum ▁ odd ▁ = ▁ " << sumOdd << " STRNEWLINE " ; cout << " Sum ▁ even ▁ = ▁ " << sumEven ; } int main ( ) { int n = 457892 ; getSum ( n ) ; return 0 ; }