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 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 . 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
Sample Input 0
5 4
Sample Output 0
2
3
3
Explanation 0
All possible values of and are:
The maximum possible value of that is also is , so we print on first line.
The maximum possible value of that is also is , so we print on second line.
The maximum possible value of that is also is , so we print on third line.
Solution
#include <stdio.h>
int main()
{
int n, k;
scanf("%d %d", &n, &k);
int mxAnd = 0, mxOr = 0, mxXor = 0;
for(int i = 1; i <= n; i++){
for(int j = i + 1; j <= n; j++){
if(mxAnd < (i & j) && (i & j) < k)
mxAnd = i & j;
if(mxOr < (i | j) && (i | j) < k)
mxOr = i | j;
if(mxXor < (i ^ j) && (i ^ j) < k)
mxXor = i ^ j;
}
}
printf("%d\n", mxAnd);
printf("%d\n", mxOr);
printf("%d\n", mxXor);
return 0;
}