Tom and jerry solution

JUNE LONG CHALLENGE:                 Problem code: EOEO


The Tom and Jerry Game! 

logic:
For odd case: you just have to find all the values which are even.


For even case: Jerry only wins when Jerry strength is even and tom strength is odd so we have to find only those values of JS which can divided by 2 more times than times who in simpler terms (or in bit terms ) we have to to find all that values which have mores zeroes in the end than TS



CODE:

 #include<bits/stdc++.h>
using namespace std;
int main(){
    long long t;
    cin>>t;
    while(t--){
        long long tom,jerry,count=0;
        cin>>tom;
        if(tom%2!=0)cout<<(tom-1)/2<<"\n"; /* if tom strength is odd  */
        else{                               /* if to strength is even */
            jerry = __builtin_ctz (tom);  /* this gcc function returns no. of zeroes at the end of the number in binary representation */
            jerry++;                    /* for finding all the no. which have more zeroes at the end of the number in binary representation */
            jerry=1<<jerry;             /* make the smallest no. which have more zeroes */
            cout<<tom/jerry<<"\n"; /* and by simply diving you will get the required result  */
        }                 
    }
    return 0;
}