从 1 到 n 的数字的最接近和划分(分成两个子集)

原文:https://www . geeksforgeeks . org/最接近的从 1 到 n 的两个数字子集的和划分/

给定一个整数序列 1，2，3，4，…，n 。任务是将它分成两组 A 和 B ，每个元素恰好属于一组，| sum(A)–sum(B)|是最小可能。打印| sum(A)–sum(B)|的值。 举例:

输入:3 T3】输出: 0 A = {1，2}和 B = { 3 } ans | sum(A)–sum(B)| = | 3–3 | = 0。 输入: 6 输出: 0 A = {1，3，4}和 B = {2，5 } ans | sum(A)–sum(B)| = | 3–3 | = 0。 输入: 5 输出: 1

进场:取mod = n % 4

  1. 如果 mod = 0 或 mod = 3 ，则打印 0 。
  2. 如果 mod = 1 或 mod = 2 ，则打印 1 。

这是因为这些组将被选择为 A = {N，N–3，N–4，N–7，N–8，…..}，B = { N–1，N–2，N–5，N–6，…..} 从 N 到 1 开始，将第一个元素放在 A 组，然后每隔 2 个元素在 B，A，B，A，…..

  • 当 n % 4 = 0 时: N = 8，A = {1，4，5，8}和 B = {2，3，6，7}
  • 当 n % 4 = 1: N = 9，A = {1，4，5，8，9}和 B = {2，3，6，7}
  • 当 n % 4 = 2: N = 10 时，A = {1，4，5，8，9}和 B = {2，3，6，7，10}
  • 当 n % 4 = 3: N = 11 时，A = {1，4，5，8，9}和 B = {2，3，6，7，10，11}

以下是上述方法的实现:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum required
// absolute difference
int minAbsDiff(int n)
{
    int mod = n % 4;

    if (mod == 0 || mod == 3)
        return 0;

    return 1;
}

// Driver code
int main()
{
    int n = 5;
    cout << minAbsDiff(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 return the minimum required
// absolute difference

    static int minAbsDiff(int n)
    {
        int mod = n % 4;
        if (mod == 0 || mod == 3)
        {
            return 0;
        }
        return 1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(minAbsDiff(n));
    }
}

// This code is contributed by Rajput-JI

Python 3

# Python3 implementation of the approach

# Function to return the minimum required
# absolute difference
def minAbsDiff(n) :
    mod = n % 4;

    if (mod == 0 or mod == 3) :
        return 0;

    return 1;

# Driver code
if __name__ == "__main__" :

    n = 5;
    print(minAbsDiff(n))

# This code is contributed by Ryuga

C

// C# implementation of the
// above approach
using System;

class GFG
{

    // Function to return the minimum 
    // required absolute difference
    static int minAbsDiff(int n)
    {
        int mod = n % 4;
        if (mod == 0 || mod == 3)
        {
            return 0;
        }
        return 1;
    }

    // Driver code
    static public void Main ()
    {
        int n = 5;
        Console.WriteLine(minAbsDiff(n));
    }
}

// This code is contributed by akt_mit

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

<?php
// PHP implementation of the approach

// Function to return the minimum
// required absolute difference
function minAbsDiff($n)
{
    $mod = $n % 4;

    if ($mod == 0 || $mod == 3)
        return 0;

    return 1;
}

// Driver code
$n = 5;
echo minAbsDiff($n);

// This code is contributed by Tushil.
?>

java 描述语言

<script>
    // Javascript implementation of the above approach

    // Function to return the minimum 
    // required absolute difference
    function minAbsDiff(n)
    {
        let mod = n % 4;
        if (mod == 0 || mod == 3)
        {
            return 0;
        }
        return 1;
    }

    let n = 5;
      document.write(minAbsDiff(n));

</script>

Output: 

1

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