Delhi's Odd Even Problem

#1

#include

using namespace std;

int main() {

int N;

cin>>N;

int num;

int even, odd;


while(N!=0){

	int sum = 0;

	cin>>num;

  while(num!=0){

	int rem = num%10;

	sum = sum +rem;

	num = num/10;

   }

   if(sum%2==0){

	   even = sum;

   }

   else{

	    odd = sum;

   }

   if(even%4==0 and odd%3==0){

	   cout<<"Yes"<<endl;

   }

   else{

	   cout<<"No"<<endl;

   }

	N = N-1;

}

	return 0;

}
/where is the problem in my code/
/* when i run this code i got the desirable output */
/but when i submit this code i got TESTCASE # 1 : wrong-answer (Time: 0 s) as a output/

#2

@rajsaxena.personal Your logic is going wrong.Please dry run you code keeping this in mind:
You have to find the sum of all digits which are even and the sum of all digits which are odd separately. If the sum of all even digits is divisible by 4 or sum of all odd digits is divisible by 3, then only print “Yes”. In all other cases print “No”.

Hope this helps :slightly_smiling_face: