AND product (hint bit manipulation )

Consider two non-negative long integers, and , where . The bitwise AND of all long integers in the inclusive range between and can be expressed as , where is the bitwise AND operator.
Given pairs of long integers, and , compute and print the bitwise AND of all natural numbers in the inclusive range between and .

Input Format

The first line contains a single integer, , denoting the number of intervals to calculate results for.
Each line of the subsequent lines contains two space-separated long integers describing the respective values of and .

Constraints

• 
  

Output Format

For each pair of long integers, print the bitwise AND of all numbers in the inclusive range between and on a new line.

Sample Input

3 
12 15 
2 3 
8 13

Sample Output

12 
2 
8

Explanation

There are three pairs to compute results for:

1. and
   , so we print on a new line.
2. and
   , so we print on a new line.
3. and
   , so we print on a new line.

#include <stdio.h>
unsigned long one=1;
int which_msb1(long unsigned int *z,long unsigned int *z1,int flag )
{
    while( ((*z & (one<<63) )== 0 )&&flag>0 )
        {

           (*z)=(*z)<<1;
           (*z1)=(*z1)<<1;
            flag--;
           // printf("flag is %d",flag);
        }
    return flag ;
}
int main() {
    int test ;
    scanf("%d",&test);
    while (test >0){

     unsigned long x;
     unsigned long y;
        scanf("%ld",&x);
        scanf("%ld",&y);
     unsigned int tmp=64;
     //unsigned long prod= x;
     unsigned long z=x;
        unsigned long z1=y;
     int flag=64;
   
      
         //printf("%ld\n",z);
         //printf("%lu/n",z & one<<63);
        flag= which_msb1(&z,&z1,flag);
        //printf("%u\n",flag);
        while (  (z & (one<<63))==(z1 &(one<<63)) &&flag>0 )
        {
            z= z<<1;
            z1=z1<<1;
            flag--;
        }
        //printf("%d",flag);
        int temp2= flag;
        z=x;

        z=z>>(temp2);

        z=z<<(temp2);

        printf("%ld\n",z);

     test--;
 
    }
 return 0;

}

Left rotation in array

A left rotation operation on an array of size shifts each of the array's elements unit to the left. For example, if left rotations are performed on array , then the array would become .
Given an array of integers and a number, , perform left rotations on the array. Then print the updated array as a single line of space-separated integers.

Input Format

The first line contains two space-separated integers denoting the respective values of (the number of integers) and (the number of left rotations you must perform).
The second line contains space-separated integers describing the respective elements of the array's initial state.

Constraints

• 
  

Output Format

Print a single line of space-separated integers denoting the final state of the array after performing left rotations.

Sample Input

5 4
1 2 3 4 5

Sample Output

5 1 2 3 4

Explanation

When we perform left rotations, the array undergoes the following sequence of changes:

Thus, we print the array's final state as a single line of space-separated values, which is 5 1 2 3 4.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int n,r ;
    scanf("%d %d",&n,&r);
    int a[n],b[n];
    for (int i =0; i<n;i++)
    {
        scanf("%d",&a[i]);
        int l;
        if ((i-r)%n<0)
        {
            l=(i-r)%n+n;
        }
        else
            l=(i-r)%n;
        b[l]= a[i];
    }
   for(int i=0;i<n;i++)
   {
       printf("%d ",b[i]);
   }
   
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}