CareerCup

Adding 1 to all 1s will produce a perfect power of 2
=> n & (n+1) == 0

n & (n+1) == 0

Well, unsigned. Like 7 for instance is 111.

If a number j is n bits in it's binary form, then do this:
j & (first n bits of j+1) == 0.

If that's true, then the number was all ones.

step1: negate the number (~q say number is q)
step 2: find xor of the two number(Q xor ~Q)
step 3: IF step2 result == Q all the every digit in binary is 1 else false

if(~num == 0) {printf("the num contains all ones\n");}

Cant we just do an XOR with all 1's ???

public class Allones {

public static void main(String args[]){

int no = Integer.parseInt("10", 2);

System.out.println((no&(no+1))==0);

}

}

if( (n& ~0)== ~0)
printf("\nAll bits set to 1\n");
else
printf("\nNot set to 1 all bits\n");

Sorry it shud be
if((n&~0)==(unsigned)(~0))
printf("\nAll bits set to 1\n");
else
printf("\nNot all bits set to 1\n");

Forgot to Typecast with Unsigned ;-)

i asked why im here when reading all these answers. here actually is the valid numbers,
1, 11, 111, 1111, 11111, ..., 11111111

return ~x == 0;

i have 2 similar answers. the 1st one is

if((x^o) == 1){       //x is the supplied number
        system.out.println("this number "+x+" has all 1's in its bin.representation");

}
the second one is:

if((x^1) == 0){       //x is the supplied number
        system.out.println("this number "+x+" has all 1's in its bin.representation");

}

The above algorithms can be solved by following that on every right shift of the binary number it's & operation with 1 should not be 0 if 0 then return true else till the number is greater than 0 return true if not found
Implementation:

#include<bits/stdc++.h>
using namespace std;
int findsetbits(int n){
    while(n > 0){
        if(n&1 == 0)
            return 0;
        n = n>>1;
    }
    return 1;
}
int main(){
    int t, n;
    cin>>t;
    while(t--){
        cin>>n;
        if(findsetbits(n) == 0)
            cout<<"NO"<<endl;
        else
            cout<<"YES"<<endl;
    }
	return 0;

}

if (signednum == -1)

I guess n^n == 0 will also work.