Pattern Printing

 Print a pattern of numbers from to  as shown below. Each of the numbers is separated by a single space.

                            4 4 4 4 4 4 4  
                            4 3 3 3 3 3 4   
                            4 3 2 2 2 3 4   
                            4 3 2 1 2 3 4   
                            4 3 2 2 2 3 4   
                            4 3 3 3 3 3 4   
                            4 4 4 4 4 4 4   

Input Format

The input will contain a single integer .

Constraints

Sample Input 0

2

Sample Output 0

2 2 2
2 1 2
2 2 2

Sample Input 1

5

Sample Output 1

5 5 5 5 5 5 5 5 5 
5 4 4 4 4 4 4 4 5 
5 4 3 3 3 3 3 4 5 
5 4 3 2 2 2 3 4 5 
5 4 3 2 1 2 3 4 5 
5 4 3 2 2 2 3 4 5 
5 4 3 3 3 3 3 4 5 
5 4 4 4 4 4 4 4 5 
5 5 5 5 5 5 5 5 5

Sample Input 2

7

Sample Output 2

7 7 7 7 7 7 7 7 7 7 7 7 7 
7 6 6 6 6 6 6 6 6 6 6 6 7 
7 6 5 5 5 5 5 5 5 5 5 6 7 
7 6 5 4 4 4 4 4 4 4 5 6 7 
7 6 5 4 3 3 3 3 3 4 5 6 7 
7 6 5 4 3 2 2 2 3 4 5 6 7 
7 6 5 4 3 2 1 2 3 4 5 6 7 
7 6 5 4 3 2 2 2 3 4 5 6 7 
7 6 5 4 3 3 3 3 3 4 5 6 7 
7 6 5 4 4 4 4 4 4 4 5 6 7 
7 6 5 5 5 5 5 5 5 5 5 6 7 
7 6 6 6 6 6 6 6 6 6 6 6 7 
7 7 7 7 7 7 7 7 7 7 7 7 7 

SOLUTION:

#include <stdio.h>

int main() 

{

    int n,min;

    scanf("%d", &n);

    int len = n*2 - 1;

    for(int i=0;i<len;i++){

        for(int j=0;j<len;j++){

            if(i<j)

            {

                min=i;

            }

            else {

                min=j;

            }

            if(min<len-i)

            {

                min=min;

            }

            else {

                min=len-i-1;

            }

            if(min<len-j-1){

                min=min;

            }

            else {

                min=len-j-1;

            }

            

            printf("%d ", n-min);

        }

        printf("\n");

    }

    return 0;

}

Bitwise Operators and Solutions in C

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:

10. • 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>

#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 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);

}

int main() {

    int n, k;

  

    scanf("%d %d", &n, &k);

    calculate_the_maximum(n, k);

 

    return 0;

}