Given an integer N ≥ 2, you can split the number as a sum of k integers i.e. N = k1 + k2 + … + kn where each kth element is ≥ 2 then the cost of splitting is calculated as maxDiv(k1) + maxDiv(k2) + … + maxDiv(kn) where maxDiv(x) is the maximum divisor of x which is < x.
The task is to split the number in such a way that the cost is minimized, print the minimized cost in the end.
Examples:
Input: N = 6
Output: 2
6 can be represented as (3 + 3) and the cost will be 1 + 1 = 2.Input: N = 5
Output: 1
Approach:
- When n is prime then the cost will be 1 as we don’t have to split n and the greatest divisor of n less than itself will be 1.
- If n is odd and n – 2 is prime then n can be split into (2 + prime) which will cost 1 + 1 = 2.
- If n is even then the cost will be 2 as according to Goldbach’s conjecture, every even number greater than 2 can be expressed as sum of two primes which is proven till 4 * 1018.
- If all of the above conditions are not satisfied then n must be odd now and if 3 is subtracted from n then it will become even which can be expressed as (3 + even) = (3 + prime + prime) which will cost 3.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // check if a number is prime or not bool isPrime(int x) { // run a loop upto square root of x for (int i = 2; i * i <= x; i++) { if (x % i == 0) return 0; } return 1; } // Function to return the minimized cost int minimumCost(int n) { // If n is prime if (isPrime(n)) return 1; // If n is odd and can be // split into (prime + 2) // then cost will be 1 + 1 = 2 if (n % 2 == 1 && isPrime(n - 2)) return 2; // Every non-prime even number // can be expressed as the // sum of two primes if (n % 2 == 0) return 2; // n is odd so n can be split into (3 + even) // further even part can be // split into (prime + prime) // (3 + prime + prime) will cost 3 return 3; } // Driver code int main() { int n = 6; cout << minimumCost(n); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { // check if a number is prime or not static boolean isPrime(int x) { // run a loop upto square root of x for (int i = 2; i * i <= x; i++) { if (x % i == 0) return false; } return true; } // Function to return the minimized cost static int minimumCost(int n) { // If n is prime if (isPrime(n)) return 1; // If n is odd and can be // split into (prime + 2) // then cost will be 1 + 1 = 2 if (n % 2 == 1 && isPrime(n - 2)) return 2; // Every non-prime even number // can be expressed as the // sum of two primes if (n % 2 == 0) return 2; // n is odd so n can be split into (3 + even) // further even part can be // split into (prime + prime) // (3 + prime + prime) will cost 3 return 3; } // Driver code public static void main(String args[]) { int n = 6; System.out.println(minimumCost(n)); } } // This code is contributed by // Surendra_Gangwar |
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Python3
# Python3 implementation of the approach from math import sqrt # check if a number is prime or not def isPrime(x) : # run a loop upto square root of x for i in range(2, int(sqrt(x)) + 1) : if (x % i == 0) : return 0; return 1; # Function to return the minimized cost def minimumCost(n) : # If n is prime if (isPrime(n)) : return 1; # If n is odd and can be # split into (prime + 2) # then cost will be 1 + 1 = 2 if (n % 2 == 1 and isPrime(n - 2)) : return 2; # Every non-prime even number # can be expressed as the # sum of two primes if (n % 2 == 0) : return 2; # n is odd so n can be split into # (3 + even) further even part can be # split into (prime + prime) # (3 + prime + prime) will cost 3 return 3; # Driver code if __name__ == "__main__" : n = 6; print(minimumCost(n)); # This code is contributed by Ryuga |
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C#
// C# implementation of the approach using System; public class GFG { // check if a number is prime or not static bool isPrime(int x) { // run a loop upto square root of x for (int i = 2; i * i <= x; i++) { if (x % i == 0) return false; } return true; } // Function to return the minimized cost static int minimumCost(int n) { // If n is prime if (isPrime(n)) return 1; // If n is odd and can be // split into (prime + 2) // then cost will be 1 + 1 = 2 if (n % 2 == 1 && isPrime(n - 2)) return 2; // Every non-prime even number // can be expressed as the // sum of two primes if (n % 2 == 0) return 2; // n is odd so n can be split into (3 + even) // further even part can be // split into (prime + prime) // (3 + prime + prime) will cost 3 return 3; } // Driver code public static void Main(String []args) { int n = 6; Console.WriteLine(minimumCost(n)); } } // This code is contributed by // Rajput-Ji |
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PHP
<?php // PHP implementation of the approach // check if a number is prime or not function isPrime($x) { // run a loop upto square root of x for ($i = 2; $i * $i <= $x; $i++) { if ($x % $i == 0) return 0; } return 1; } // Function to return the minimized cost function minimumCost($n) { // If n is prime if (isPrime($n)) return 1; // If n is odd and can be // split into (prime + 2) // then cost will be 1 + 1 = 2 if ($n % 2 == 1 && isPrime($n - 2)) return 2; // Every non-prime even number // can be expressed as the // sum of two primes if ($n % 2 == 0) return 2; // n is odd so n can be split into (3 + even) // further even part can be // split into (prime + prime) // (3 + prime + prime) will cost 3 return 3; } // Driver code $n = 6; echo(minimumCost($n)); // This code is contributed by Code_Mech. |
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Output:
2
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