Xor is Mad

QUESTION

The problem is straight and simple.\nGiven a number X ,find how many positive A ( A>0) exists, such that \n\n1 . A OR X =A + X \n2. A<X\n\nInput:\n\nThe first line of the input will contain T , the number of test-cases. \n\nNext T lines will contain integer X . \n\nOutput:\n\nFor each test-case , output the answer in a separate line.\n\nConstraints:\n\n1<=T<=10^5\n1<=X<=10^7\n\nNote: \nOR is the bitwise Exclusive-OR operator( XOR ). and + is the usual addition symbol\n.

ANSWER


             #include<stdio.h>
    int main()
    {long t,n,count;
    scanf("%ld",&t);
    while(t--)
    {count=1;
    scanf("%ld",&n);
    while(n)
    {if(n%2==0)
    count*=2;
    n=n/2;
    }
    printf("%ld\n",count-1);
    }
    }
 
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