GCD等于给定范围内1的配对

给定一个范围,即L和R,任务是找出我们是否可以形成成对,使每对的GCD为1。L-R范围内的每个数字都应该正好包含在一对中。 例如:

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Input: L = 1, R = 8Output: Yes{2, 7}, {4, 1}, {3, 8}, {6, 5}All pairs have GCD as 1. Input: L = 2, R = 4Output: No

方法 :因为(L,R)范围内的每个数字必须在每对中包含一次。因此,如果L-R是偶数,那么这是不可能的。如果L-R是奇数,则打印所有相邻对,因为相邻对的GCD始终为1。 以下是上述方法的实施情况:

C++

// C++ program to print all pairs
#include <bits/stdc++.h>
using namespace std;
// Function to print all pairs
bool checkPairs( int l, int r)
{
// check if even
if ((l - r) % 2 == 0)
return false ;
/* We can print all adjacent pairs
for (int i = l; i < r; i += 2) {
cout << "{" << i << ", " << i + 1 << "}, ";
} */
return true ;
}
// Driver Code
int main()
{
int l = 1, r = 8;
if (checkPairs(l, r))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}


JAVA

// Java program to print all pairs
class GFG
{
// Function to print all pairs
static boolean checkPairs( int l, int r)
{
// check if even
if ((l - r) % 2 == 0 )
return false ;
/* We can print all adjacent pairs
for (int i = l; i < r; i += 2)
{
System.out.print("{"+i+", "+i + 1+"}, ");
} */
return true ;
}
// Driver Code
public static void main(String[] args)
{
int l = 1 , r = 8 ;
if (checkPairs(l, r))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
// This code is contributed by mits


Python 3

# Python 3 program to print all pairs
# Function to print all pairs
def checkPairs(l, r) :
# check if even
if (l - r) % 2 = = 0 :
return False
""" we can print all adjacent pairs
for i in range(l,r,2) :
print("{",i,",",i + 1, "},")
"""
return True
# Driver Code
if __name__ = = "__main__" :
l, r = 1 , 8
if checkPairs(l, r) :
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by ANKITRAI1


C#

// C# program to print all pairs
using System;
class GFG
{
// Function to print all pairs
static bool checkPairs( int l, int r)
{
// check if even
if ((l - r) % 2 == 0)
return false ;
/* We can print all adjacent pairs
for (int i = l; i < r; i += 2)
{
System.out.print("{"+i+", "+i + 1+"}, ");
} */
return true ;
}
// Driver Code
static public void Main ()
{
int l = 1, r = 8;
if (checkPairs(l, r))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
}
// This code is contributed by Raj


PHP

<?php
// PHP program to print all pairs
// Function to print all pairs
function checkPairs( $l , $r )
{
// check if even
if (( $l - $r ) % 2 == 0)
return false;
/* We can print all adjacent pairs
for (int i = l; i < r; i += 2)
{
cout << "{" << i << ", " << i + 1 << "}, ";
} */
return true;
}
// Driver Code
$l = 1;
$r = 8;
if (checkPairs( $l , $r ))
echo ( "Yes" );
else
echo ( "No" );
// This code is contributed
// by Shivi_Aggarwal
?>


Javascript

<script>
// JavaScript program to print all pairs
// Function to print all pairs
function checkPairs(l , r) {
// check if even
if ((l - r) % 2 == 0)
return false ;
/*
We can print all adjacent pairs
for (i = l; i < r; i += 2) {
document.write("{"+i+", "+i + 1+"], "); }
*/
return true ;
}
// Driver Code
var l = 1, r = 8;
if (checkPairs(l, r))
document.write( "Yes" );
else
document.write( "No" );
// This code is contributed by todaysgaurav
</script>


输出:

Yes

时间复杂性: O(N) 辅助空间: O(1)

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