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Find the permutation of first N natural numbers such that sum of i % Pi 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:  




// 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;
}




// Java implementation of the approach
import java.io.*;
public 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




# 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




// 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




<script>
 
// Javascript 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
function Max_Sum(n)
{
    return parseInt((n * (n - 1)) / 2);
}
 
// Driver code
let n = 8;
 
// Function call
document.write(Max_Sum(n));
 
// This code is contributed by rishavmahato348
 
</script>

Output: 
28

 

Time Complexity: O(1)

Auxiliary Space: O(1)


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