Program to find greater value between a^n and b^n
Given three integers a, b and n, the task is to find out greater value between an and bn.
Examples:
Input: a = 3, b = 4, n = 5
Output: b^n is greater than a^n
Value of an is 243 and the value of bn is 1024.
So, bn is greater than an.
Input: a = -3, b = 2, n = 4
Output: a^n is greater than b^n
Value of an is 243 and the value of bn is 16.
So, an is greater than bn.
Basic Approach: For every value of a, b and n, calculate the values of an and bn. Then compare the result obtained and display the according to output.
The problem with this approach arises when there are large values of a, b and n. For large values of a, n, calculating an can exceed the limit of integer which will cause integer overflow.
Better approach is to check the value of n.
- If n is even then calculate the absolute value of a and b.
- If n is odd then take the given value as it is.
- Now check if a is equal to b. If yes, print 0.
- If a is greater than b, print 1.
- Otherwise, print 2.
C++
// C++ code for finding greater // between the a^n and b^n #include <bits/stdc++.h> using namespace std; // Function to find the greater value void findGreater( int a, int b, int n) { // If n is even if (!(n & 1)) { a = abs (a); b = abs (b); } if (a == b) cout << "a^n is equal to b^n" ; else if (a > b) cout << "a^n is greater than b^n" ; else cout << "b^n is greater than a^n" ; } // Driver code int main() { int a = 12, b = 24, n = 5; findGreater(a, b, n); return 0; } |
Java
// JAVA code for finding greater // between the a^n and b^n import java.io.*; class GFG { // Function to find // the greater value static void findGreater( int a, int b, int n) { // If n is even if (!((n & 1 ) > 0 )) { a = Math.abs(a); b = Math.abs(b); } if (a == b) System.out.println( "a^n is " + "equal to b^n" ); else if (a > b) System.out.println( "a^n is greater " + "than b^n" ); else System.out.println( "b^n is greater " + "than a^n" ); } // Driver code public static void main (String[] args) { int a = 12 , b = 24 , n = 5 ; findGreater(a, b, n); } } // This code is contributed // by shiv_bhakt. |
Python3
# Python3 code for finding greater # between the a^n and b^n import math # Function to find the greater value def findGreater(a, b, n): # If n is even if ((n & 1 ) > 0 ): a = abs (a); b = abs (b); if (a = = b): print ( "a^n is equal to b^n" ); elif (a > b): print ( "a^n is greater than b^n" ); else : print ( "b^n is greater than a^n" ); # Driver code a = 12 ; b = 24 ; n = 5 ; findGreater(a, b, n); # This code is contributed by mits |
C#
// C# code for finding greater // between the a^n and b^n using System; class GFG { // Function to find // the greater value static void findGreater( int a, int b, int n) { // If n is even if (!((n & 1) > 0)) { a = Math.Abs(a); b = Math.Abs(b); } if (a == b) Console.WriteLine( "a^n is " + "equal to b^n" ); else if (a > b) Console.WriteLine( "a^n is greater " + "than b^n" ); else Console.WriteLine( "b^n is greater " + "than a^n" ); } // Driver code public static void Main() { int a = 12, b = 24, n = 5; findGreater(a, b, n); } } // This code is contributed // by shiv_bhakt. |
PHP
<?php // PHP code for finding greater // between the a^n and b^n // Function to find // the greater value function findGreater( $a , $b , $n ) { // If n is even if (!( $n & 1)) { $a = abs ( $a ); $b = abs ( $b ); } if ( $a == $b ) echo "a^n is equal to b^n" ; else if ( $a > $b ) echo "a^n is greater than b^n" ; else echo "b^n is greater than a^n" ; } // Driver code $a = 12; $b = 24; $n = 5; findGreater( $a , $b , $n ); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript code for finding greater // between the a^n and b^n // Function to find // the greater value function findGreater(a , b , n) { // If n is even if (!((n & 1) > 0)) { a = Math.abs(a); b = Math.abs(b); } if (a == b) document.write( "a^n is " + "equal to b^n" ); else if (a > b) document.write( "a^n is greater " + "than b^n" ); else document.write( "b^n is greater " + "than a^n" ); } // Driver code var a = 12, b = 24, n = 5; findGreater(a, b, n); // This code is contributed by todaysgaurav </script> |
b^n is greater than a^n
Time Complexity:O(1)
Auxiliary Space: O(1)
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