Find the number of ways to divide number into four parts such that a = c and b = d
Given a number N. Find the number of ways to divide a number into four parts(a, b, c, d) such that a = c and b = d and a not equal to b.
Examples:
Input : N = 6
Output : 1
Expalantion : four parts are {1, 2, 1, 2}Input : N = 20
Output : 4
Expalantion : possible ways are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}.
Approach :
If the given N is odd the answer is 0, because the sum of four parts will not be even number.
If n is divisible by 4 then answer will be n/4 -1(here a equals to b can be possible so subtract that possibility) otherwise n/4.
Below is the implementation of the above approach :
C++
// C++ implementation for above approach#include <bits/stdc++.h>using namespace std; // Function to find the number of ways to divide// N into four parts such that a = c and b = dint possibleways(int n){ if (n % 2 == 1) return 0; else if (n % 4 == 0) return n / 4 - 1; else return n / 4;} // Driver codeint main(){ int n = 20; cout << possibleways(n); return 0;} |
Java
// Java implementation for above approachclass GFG{ // Function to find the number of ways to divide// N into four parts such that a = c and b = dstatic int possibleways(int n){ if (n % 2 == 1) return 0; else if (n % 4 == 0) return n / 4 - 1; else return n / 4;} // Driver codepublic static void main(String[] args){ int n = 20; System.out.println(possibleways(n));}} // This code contributed by Rajput-Ji |
Python3
# Python3 implementation for above approach # Function to find the number of ways # to divide N into four parts # such that a = c and b = ddef possibleways(n): if (n % 2 == 1): return 0; elif (n % 4 == 0): return n // 4 - 1; else: return n // 4; # Driver coden = 20;print(possibleways(n)); # This code is contributed by mits |
C#
// C# implementation for above approachclass GFG{ // Function to find the number of ways to divide// N into four parts such that a = c and b = dstatic int possibleways(int n){ if (n % 2 == 1) return 0; else if (n % 4 == 0) return n / 4 - 1; else return n / 4;} // Driver codestatic void Main(){ int n = 20; System.Console.WriteLine(possibleways(n));}} // This code is contributed by mits |
PHP
<?php// PHP implementation for above approach // Function to find the number of ways to divide// N into four parts such that a = c and b = dfunction possibleways($n){ if ($n % 2 == 1) return 0; else if ($n % 4 == 0) return $n / 4 - 1; else return $n / 4;} // Driver code$n = 20;echo possibleways($n); // This code is contributed by ajit?> |
Output:
4
