In this challenge, you will use logical bitwise operators. All data is stored in its binary representation. The logical operators, and C language, use 1 to represent true and 0 to represent false. The logical operators compare bits in two numbers and return true or false, 0 or 1, 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 = 6You will be given an integer n, and a threshold, k. For each number i from 1 through n, find the maximum value of the logical and, or and xor when compared against all integers through n that are greater than i. Consider a value only if the comparison returns a result less than k [else 0].
Print the results of the and, or and exclusive or comparisons on separate lines, in that order.Example
n = 3
k = 3The 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 1For the and comparison, the maximum is 2. For the or comparison, none of the values is less than k, so the maximum is 0. For the xor comparison, the maximum value less than k is 2. The function should print:
2 0 2Function 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 2 space-separated integers, n and k.
Constraints
- 2 <= n <= 103
- 2 <= k <= n
Sample Input 0
5 4Sample Output 0
2 3 3
So This is My Answer
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
void calculate_the_maximum(int n, int k) {
int m1=0,m2=0,m3=0;
for (int x=1; x<n; x++){
for (int y=2; y<=n; y++){
//and
int a=x&y;
if((a>m1) && (a<k)){
m1=a;
};
//or
int b=x|y;
if((b>m2) && (b<k)){
m2=b;
};
//xor
int c=x^y;
if((c>m3) && (c<k)){
m3=c;
};
}
}
printf("%d \n",m1);
printf("%d \n",m2);
printf("%d \n",m3);
}
int main() {
int n, k;
scanf("%d %d", &n, &k);
calculate_the_maximum(n, k);
return 0;
}
And My Output is
3
3
3
Expected Output
2
3
3
What is the mistake of my code?
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
void calculate_the_maximum(int n, int k) {
int m1=0,m2=0,m3=0;
for (int x=1; x<n; x++){
for (int y=x+1; y<=n; y++){
//and
int a= x & y;
if((a>m1) && (a<k)){
m1=a;
};
//or
int b=x|y;
if((b>m2) && (b<k)){
m2=b;
};
//xor
int c=x^y;
if((c>m3) && (c<k)){
m3=c;
};
}
}
printf("%d \n",m1);
printf("%d \n",m2);
printf("%d \n",m3);
}
int main() {
int n, k;
scanf("%d %d", &n, &k);
calculate_the_maximum(n, k);
return 0;
}