Delhi's Odd Even

#1

My code is not passing all the test cases.

#include
using namespace std ;

int main()
{
int n,no,sum=0,d;
cin>>n;

while(n!=0)
{
cin>>no;

while(no!=0)

{
d=no%10;
no/=10;
sum+=d;
}

if( (sum%2==0 && sum%4==0) )
{
	cout<<"Yes"<<endl;
}

else if( (sum%2==1 && sum%3==0) )

{
cout<<“Yes”<<endl;
}

 else
{
 cout<<"No"<<endl;
}

n--;

}

return 0;

}

#2

@mvganeshk15
the logic of this problem
For ex if you are given a no.:

982452

then maintain 2 variables evesum=0, oddsum = 0 ;
and they will be:

9   +   8   +   2   +   4   +   5   +   2

evesum = 8 + 2 + 4 + 2 = 16
oddsum = 9 + 5 = 14
if((evesum%4==0)||(oddsum%3==0)){
cout<<“Yes”<<endl;
}else{
cout<<“No”<<endl;
}

1 Like
#3

Respected Sir,

Thanks for solution now i can understand. Actually i misunderstood the problem ,i thought the sum of the digits are even and divisible by 4 instead of sum of even digits are divisible by 4.

Thanks
Ganesh

#4

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