#include
using namespace std;
int main(int argc, char const *argv[])
{
int n, x, sum=0, l;
cin>>n;
while(n--){
cin>>x;
if (x>=0 && x <=1000000000)
{
while(x!=0){
l = x%10;
sum = sum + l;
x = x/10;
}
if(sum%2==0){
if (sum%4==0)
{
cout<<"Yes"<<endl;
}
else{
cout<<"No"<<endl;
}
}
else
{
if (sum%3 == 0)
{
cout<<"Yes"<<endl;
}
else
{
cout<<"No"<<endl;
}
}
sum = 0;
}
}
return 0;
}
@pranjalsoni You logic is going wrong. Do a dry run and check it yourself.
What actually you are doing is that you are finding the sum of all digits and then checking its divisibility with 4 and 3 which is wrong. 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 
#include
using namespace std;
int main(int argc, char const *argv[])
{
int n, x, sumofeven=0, sumofodd=0, l1, l2, a, b;
cin>>n;
while(n--){
cin>>x;
a = x;
b = x;
if (x>=0 && x <=1000000000)
{
while(a!=0){
l1 = a%10;
if (l1%2==0)
{
sumofeven = sumofeven + l1;
}
a = a/10;
}
while(b!=0){
l2 = b%10;
if (l2%2!=0)
{
sumofodd = sumofodd + l2;
}
b = b/10;
}
if (sumofeven%4==0 || sumofodd%3==0)
{
cout<<"Yes"<<endl;
}
else{
cout<<"No"<<endl;
}
sumofodd=0;
sumofeven=0;
}
}
return 0;
}
What problem are you facing. Mention that. Also save the code on codingblocks ide and share its link.