Cunningham Number is a number N of the form 
, where a, b >= 2.
Few Cunningham numbers are:
3, 5, 7, 8, 9, 10, 15, 17, 24, 26, 28…
Check if N is a Cunningham number
Given a number N, the task is to check if N is an Cunningham Number or not. If N is an Cunningham Number then print “Yes” else print “No”.
Examples:
Input: N = 126
Output: Yes
Explanation:
126 = 5^3+1
Input: N = 16
Output: No
Approach: The idea is to solve the equation in a desired form such that checking that the number is a Cunningham Number or not is easy.
// Cunningham Numbers are the // which can be represented as =>=>
=> 
Therefore, if 
or 
can be expressed in the form of 
, then the number is cunningham Number.
Below is the implementation of the above approach:
C++
// C++ implementation for the // above approach#include <bits/stdc++.h>using namespace std;// Function to check if a number// can be expressed as a^b.bool isPower(int a){ if (a == 1) return true; for (int i = 2; i * i <= a; i++) { double val = log(a) / log(i); if ((val - (int)val) < 0.00000001) return true; } return false;}// Function to check if N is a// Cunningham numberbool isCunningham(int n){ return isPower(n - 1) || isPower(n + 1);}// Driver Codeint main(){ // Given Number int n = 126; // Function Call if (isCunningham(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation for the above approach import java.util.*;class GFG{// Function to check if a number // can be expressed as a^b. static boolean isPower(int a) { if (a == 1) return true; for(int i = 2; i * i <= a; i++) { double val = Math.log(a) / Math.log(i); if ((val - (int)val) < 0.00000001) return true; } return false; } // Function to check if N is a // Cunningham number static boolean isCunningham(int n) { return isPower(n - 1) || isPower(n + 1); } // Driver Code public static void main (String[] args) { // Given Number int n = 126; // Function Call if (isCunningham(n)) System.out.print("Yes"); else System.out.print("No");} }// This code is contributed by Ritik Bansal |
Python3
# Python3 implementation for the # above approachimport math# Function to check if a number# can be expressed as a^b.def isPower(a): if (a == 1): return True i = 2 while(i * i <= a): val = math.log(a) / math.log(i) if ((val - int(val)) < 0.00000001): return True i += 1 return False # Function to check if N is a# Cunningham numberdef isCunningham(n): return isPower(n - 1) or isPower(n + 1) # Driver Code# Given Numbern = 126# Function Callif (isCunningham(n)): print("Yes")else: print("No")# This code is contributed by shubhamsingh10 |
C#
// C# implementation for the // above approachusing System;class GFG{ // Function to check if a number// can be expressed as a^b.static bool isPower(int a){ if (a == 1) return true; for(int i = 2; i * i <= a; i++) { double val = Math.Log(a) / Math.Log(i); if ((val - (int)val) < 0.00000001) return true; } return false;}// Function to check if N is a// Cunningham numberstatic bool isCunningham(int n){ return isPower(n - 1) || isPower(n + 1);}// Driver Codepublic static void Main (string[] args){ // Given number int n = 126; // Function Call if (isCunningham(n)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by rock_cool |
Javascript
<script>// Javascript implementation for the above approach // Function to check if a number // can be expressed as a^b. function isPower( a) { if (a == 1) return true; for ( let i = 2; i * i <= a; i++) { let val = Math.log(a) / Math.log(i); if ((val - parseInt( val) < 0.00000001)) return true; } return false; } // Function to check if N is a // Cunningham number function isCunningham(n) { return isPower(n - 1) || isPower(n + 1); } // Driver Code // Given Number let n = 126; // Function Call if (isCunningham(n)) document.write("Yes"); else document.write("No");// This code is contributed by Rajput-Ji</script> |
Output:
Yes
Time Complexity: O(n1/2)
Auxiliary Space: O(1)
Reference: https://oeis.org/A080262
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