Find the permutation of first N natural numbers such that sum of i % P<sub>i</sub> is maximum possible

Given a number N. The task is to find the permutation P of first N natural numbers such that sum of i % Pi is maximum possible. The task is to find the maximum possible sum not it’s permutation.

Examples:

Input: N = 5
Output: 10
Possible permutation is 2 3 4 5 1.
Modulus values will be {1, 2, 3, 4, 0}.
1 + 2 + 3 + 4 + 0 = 10

Input: N = 8
Output: 28

Approach: Maximum possible sum is (N * (N – 1)) / 2 and it is formed by the permutation 2, 3, 4, 5, ….. N, 1.



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the permutation of
// the first N natural numbers such that
// the sum of (i % Pi) is maximum possible
// and return the maximum sum
int Max_Sum(int n)
{
    return (n * (n - 1)) / 2;
}
  
// Driver code
int main()
{
    int n = 8;
  
    // Function call
    cout << Max_Sum(n);
  
    return 0;
}
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// Java implementation of the approach
class GFG
{
  
// Function to find the permutation of
// the first N natural numbers such that
// the sum of (i % Pi) is maximum possible
// and return the maximum sum
static int Max_Sum(int n)
{
    return (n * (n - 1)) / 2;
}
  
// Driver code
public static void main (String[] args)
{
    int n = 8;
  
    // Function call
    System.out.println(Max_Sum(n));
}
}
  
// This code is contributed by Rajput-Ji
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# Python3 implementation of the approach
  
# Function to find the permutation of
# the first N natural numbers such that
# the sum of (i % Pi) is maximum possible
# and return the maximum sum
def Max_Sum(n) :
      
    return (n * (n - 1)) // 2;
  
# Driver code
if __name__ == "__main__" :
      
    n = 8;
  
    # Function call
    print(Max_Sum(n));
      
# This code is contributed by AnkitRai01
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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to find the permutation of
// the first N natural numbers such that
// the sum of (i % Pi) is maximum possible
// and return the maximum sum
static int Max_Sum(int n)
{
    return (n * (n - 1)) / 2;
}
  
// Driver code
public static void Main (String[] args)
{
    int n = 8;
  
    // Function call
    Console.WriteLine(Max_Sum(n));
}
}
  
// This code is contributed by Princi Singh
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Output:
28

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