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.
HackerRank Bitwise Operators Solution in C (Sample-1)
#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 maxAnd = 0; int maxOr = 0; int maxXor = 0; for (int i=1; i<=n; i++) { for (int j=i+1; j<=n; j++) { if (((i&j) > maxAnd) && ((i&j) < k)) { maxAnd = i&j; } if (((i|j) > maxOr) && ((i|j) < k)) { maxOr = i|j; } if (((i^j) > maxXor) && ((i^j) < k)) { maxXor = i^j; } } } printf("%d\n%d\n%d\n", maxAnd, maxOr, maxXor); } int main() { int n, k; scanf("%d %d", &n, &k); calculate_the_maximum(n, k); return 0; }
HackerRank Bitwise Operators Solution in C (Sample-2)
#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. unsigned int max1=0, max2=0, max3=0; unsigned int *S = (unsigned int *) malloc( n*sizeof(unsigned int) ); for (size_t i=0; i<n; i++) S[i]=i+1; for (size_t i=0; i<n-1; i++) for (size_t j=i+1; j<n; j++) { if ( ( (S[i]&S[j]) > max1 ) && ( (S[i]&S[j]) < k ) ) max1 = (S[i]&S[j]); if ( ( (S[i]|S[j]) > max2 ) && ( (S[i]|S[j]) < k ) ) max2 = (S[i]|S[j]); if ( ( (S[i]^S[j]) > max3 ) && ( (S[i]^S[j]) < k ) ) max3 = (S[i]^S[j]); } printf("%u\n%u\n%u\n",max1,max2,max3); } int main() { int n, k; scanf("%d %d", &n, &k); calculate_the_maximum(n, k); return 0; }
HackerRank Bitwise Operators Solution in C (Sample-3)
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> //Complete the following function. void calculate_the_maximum(int n, int k) { int maxAnd = 0; int maxOr = 0; int maxXor = 0; for (int i=1; i<=n; i++) { for (int j=i+1; j<=n; j++) { if (((i&j) > maxAnd) && ((i&j) < k)) { maxAnd = i&j; } if (((i|j) > maxOr) && ((i|j) < k)) { maxOr = i|j; } if (((i^j) > maxXor) && ((i^j) < k)) { maxXor = i^j; } } } printf("%d\n%d\n%d\n", maxAnd, maxOr, maxXor); } int main() { int n, k; scanf("%d %d", &n, &k); calculate_the_maximum(n, k); return 0; }
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