Convert the given number in to binary and stored into every bit into
array
now compare the a[i]==0 if true then print that value that is nithing
but zero else number doesn't has zero in its binary form.
e.g code is given below
int binary_zero(int n)
{
for(int i=0;i<array>length;i++)
{
a[i]=n%2;
if(ai[i]==0)
true;//take action
count_zero++
else
false//take action..
count_one++;
n=n/2;
}
}
Regards
Shashank Mani
Birla Istitue of Technology,Mesra
Computer Science &Engg.
Cell No.+91-9740852296
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