求前 N 个自然数的排列，使得 i % Pi 的和最大可能
原文:https://www . geesforgeks . org/find-第一个 n 个自然数的排列-这样 I-pi-的和就是最大可能/
给定一个数字 N 。任务是找到第一个 N 自然数的排列 P ，使得I % PIT9】的和最大可能。任务是找到最大可能的和，而不是它的排列。
示例:
输入: N = 5 输出: 10 可能的排列是 2 3 4 5 1。 模数值将为{1，2，3，4，0}。 1 + 2 + 3 + 4 + 0 = 10
输入:N = 8 T3】输出: 28
逼近:最大可能和为(N *(N–1))/2，由排列 2，3，4，5，…..n，1 。
下面是上述方法的实现:
C++
// 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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
// 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
java 描述语言
<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
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