Problem
In this challenge, you will use logical bitwise operators. All data is stored in its binary representation. The logical operators, and C language, use to represent true and to represent false. The logical operators compare bits in two numbers and return true or false, or, for each bit compared.
Bitwise AND operator &
The output of bitwise AND is 1 if the corresponding bits of two operands is 1. If either bit of an operand is 0, the result of the corresponding bit is evaluated to 0. It is denoted by &.Bitwise OR operator |
The output of bitwise OR is 1 if at least one corresponding bit of two operands is 1. It is denoted by |.Bitwise XOR (exclusive OR) operator ^
The result of bitwise XOR operator is 1 if the corresponding bits of two operands are opposite. It is denoted by .
For example, for integers 3 and 5,
3 = 00000011 (In Binary)
5 = 00000101 (In Binary)
AND operation OR operation XOR operation
00000011 00000011 00000011
& 00000101 | 00000101 ^ 00000101
________ ________ ________
00000001 = 1 00000111 = 7 00000110 = 6
You will be given an integer , and a threshold, i1nnik$. Print the results of the and, or and exclusive or comparisons on separate lines, in that order.
Example
The results of the comparisons are below:
a b and or xor
1 2 0 3 3
1 3 1 3 2
2 3 2 3 1
For the and
comparison, the maximum is 2. For the or
comparison, none of the values is less than, so the maximum is . For the xor
comparison, the maximum value less than is . The function should print:
2
0
2
Function Description
Complete the calculate_the_maximum function in the editor below.
calculate_the_maximum has the following parameters:
- int n: the highest number to consider
- int k: the result of a comparison must be lower than this number to be considered
Prints
Print the maximum values for the and
, or
and xor
comparisons, each on a separate line.
Input Format
The only line contains space-separated integers, and.
Constraints
- 2 <= n <= 10^3
- 2 <= k <= n
Sample Input 0
5 4
Sample Output 0
2
3
3
Explanation 0
All possible values of and are:
- 1 2
- 1 3
- 1 4
- 1 5
- 2 3
- 2 4
- 2 5
- 3 4
- 3 5
- 4 5
The maximum possible value of AND 2 that is also is 2 < k, so we print 2 on first line.
The maximum possible value of OR that is also is 3 < k , so we print 3 on second line.
The maximum possible value of XOR that is also is 3< k , so we print 3 on third line.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
//Complete the following function.
void calculate_the_maximum(int n, int k) {
//Write your code here.
int i, j, max_and=0, max_or=0, max_exor=0, and_op, or_op, xor_op;
for(i = 1; i< n; i++){
for(j = (i+1); j<=n; j++){
and_op = i & j;
or_op = i | j;
xor_op = i ^ j;
if( and_op > max_and && and_op < k){
max_and = and_op;
}
if( or_op > max_or && or_op < k){
max_or = or_op;
}
if( xor_op > max_exor && xor_op < k){
max_exor = xor_op;
}
}
}
printf("%d\n%d\n%d", max_and, max_or, max_exor);
}
int main() {
int n, k;
scanf("%d %d", &n, &k);
calculate_the_maximum(n, k);
return 0;
}
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