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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