Print N lines of 4 numbers such that every pair among 4 numbers has a GCD K
Given N and K, the task is to print N lines where each line contains 4 numbers such that every among those 4 numbers has a GCD K and the maximum number used in N*4 should be minimized.
Note: In case of multiple outputs, print any one.
Examples:
Input: N = 1, K = 1
Output: 1 2 3 5
Every pair among 1, 2, 3 and 5 gives a GCD K and the largest number among these is 5 which the minimum possible.
Input: 2 2
Output:
2 4 6 22
14 18 10 16
In the above input, the maximum number is 22, which is the minimum possible to make 2 lines of 4 numbers.
Approach: The first observation is that if we can solve the given problem for K=1, we can solve the problem with GCD K by simply multiplying the answers with K. We know that any three consecutive odd numbers have a GCD 1 always when paired, so three numbers of every line can be easily obtained. Hence the lines will look like:
1 3 5 _
7 9 11 _
13 15 17 _
.
.
.
An even number cannot be inserted always, because inserting 6 in third line will give GCD(6, 9) as 3. So the best number that can be inserted is a number between the first two off numbers of every line. Hence the pattern looks like:
1 2 3 5
7 8 9 11
13 14 15 17
.
.
.
To obtain given GCD K, one can easily multiply K to the obtained numbers. Hence for i-th line:
- the first number will be k * (6*i+1)
- the second number will be k * (6*i+1)
- the third number will be k * (6*i+3)
- the fourth number will be k * (6*i+5)
The maximum number among N*4 numbers will be k * (6*i – 1)
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
void printLines( int n, int k)
{
for ( int i = 0; i < n; i++) {
cout << k * (6 * i + 1) << " "
<< k * (6 * i + 2) << " "
<< k * (6 * i + 3) << " "
<< k * (6 * i + 5) << endl;
}
}
int main()
{
int n = 2, k = 2;
printLines(n, k);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
static void printLines( int n, int k)
{
for ( int i = 0 ; i < n; i++) {
System.out.println ( k * ( 6 * i + 1 ) + " "
+ k * ( 6 * i + 2 ) + " "
+ k * ( 6 * i + 3 ) + " "
+ k * ( 6 * i + 5 ) );
}
}
public static void main(String args[])
{
int n = 2 , k = 2 ;
printLines(n, k);
}
}
|
Python3
def printLines(n, k) :
for i in range (n) :
print ( k * ( 6 * i + 1 ),
k * ( 6 * i + 2 ),
k * ( 6 * i + 3 ),
k * ( 6 * i + 5 ))
if __name__ = = "__main__" :
n, k = 2 , 2
printLines(n, k)
|
PHP
<?php
function printLines( $n , $k )
{
for ( $i = 0; $i < $n ; $i ++)
{
echo ( $k * (6 * $i + 1));
echo ( " " );
echo ( $k * (6 * $i + 2));
echo ( " " );
echo ( $k * (6 * $i + 3));
echo ( " " );
echo ( $k * (6 * $i + 5));
echo ( "\n" );
}
}
$n = 2;
$k = 2;
printLines( $n , $k );
?>
|
C#
using System;
class GFG
{
static void printLines( int n, int k)
{
for ( int i = 0; i < n; i++)
{
Console.WriteLine ( k * (6 * i + 1) + " " +
k * (6 * i + 2) + " " +
k * (6 * i + 3) + " " +
k * (6 * i + 5) );
}
}
public static void Main()
{
int n = 2, k = 2;
printLines(n, k);
}
}
|
Javascript
<script>
function printLines(n , k)
{
for (i = 0; i < n; i++)
{
document.write(k * (6 * i + 1)
+ " " + k * (6 * i + 2) + " "
+ k * (6 * i + 3) + " "
+ k * (6 * i + 5)+ "<br/>" );
}
}
var n = 2, k = 2;
printLines(n, k);
</script>
|
Output:
2 4 6 10
14 16 18 22
Time Complexity: O(N*4), as we are using a loop to traverse N times and doing 4 operations in each traversal.
Auxiliary Space: O(1), as we are not using any extra space.
Last Updated :
09 Jun, 2022
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