Objective
Welcome to the last day! Today, we're discussing bitwise operations. Check out the Tutorial tab for learning materials and an instructional video!
Welcome to the last day! Today, we're discussing bitwise operations. Check out the Tutorial tab for learning materials and an instructional video!
Task
Given set . Find two integers, and (where ), from set such that the value of is the maximum possible and also less than a given integer, . In this case, represents the bitwise AND operator.
Given set . Find two integers, and (where ), from set such that the value of is the maximum possible and also less than a given integer, . In this case, represents the bitwise AND operator.
Input Format
The first line contains an integer, , the number of test cases.
Each of the subsequent lines defines a test case as space-separated integers, and , respectively.
Each of the subsequent lines defines a test case as space-separated integers, and , respectively.
Constraints
Output Format
For each test case, print the maximum possible value of on a new line.
Sample Input
3
5 2
8 5
2 2
Sample Output
1
4
0
Explanation
All possible values of and are:
The maximum possible value of that is also is , so we print on a new line.
Solution :
#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int T, n, k, i, j, val;
scanf("%d",&T);
while (T > 0)
{
int maximum = 0;
scanf("%d%d",&n, &k);
int max_val = 0;
int a = 0, b = k - 1;
for (a = n; a > 2; a--)
{
if (a == b)
continue;
if ((a & b) > max_val)
{
max_val = a & b;
}
}
printf("%d\n",max_val );
T--;
}
return 0;
}
