Sum of N Integers – Even Reversed

                     Sum of N Integers - Even Reversed 

                                    
The program must accept N integers and print the sum S of all POSITIVE integers with the even positive integers reversed.

Input Format:
The first line will contain N.
The second line will contain N integers separated by a space.

Output Format: 
The first line will contain S.

Boundary Conditions: 1 <= N <= 100000

Example Input/Output 1:
Input:
4 39 -8 57 24
Output:
138
Explanation:
The sum = 39+57+42 = 138 (The even number 24 is reversed)

Example Input/Output 2:
Input:
3 -23 -11 -445
Output: 0

Code:


#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

int n,i,temp=0,sum=0;

cin>>n;

int arr[n];

    for(i=0;i<n;i++)

    {

        cin>>arr[i];

        if(arr[i]>0)

        {

             if((arr[i]%2==0))

             while(arr[i]>0)

            {   temp*=10;

                temp+=arr[i]%10;

                arr[i]=arr[i]/10;

            }

           sum+=arr[i]+temp;

           temp=0;

        }

    }

    cout<<sum;

}

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