N 级六边形中彩色 0 的数量
原文:https://www . geesforgeks . org/number-of-color-0s-in-an-n-level-hexagon/
给定一个整数 N ，当 0s 按照以下方式着色时，任务是在 N 级六边形中找到着色的 0s 的计数:
例:
输入: N = 2 输出: 5 输入: N = 3 输出: 12
方法:对于 N = 1、2、3、… 的值，可以观察到一系列将形成为 1、5、12、22、35、… 。这是一个差异系列，差异在 AP 中是 4，7，10，13，… 。 因此 N 第 项将为 1 + {4 + 7 + 10 +13 +…..(n–1)条款} = 1+(n–1)(2 * 4+(n–1–1) 3)/2 = 1+(n–1)(8+(n–2) 3)/2 = 1+(n–1)(8+3n–6)/2 = 1+(n–1)(3n+2)/2 = n *(3 * n–1)/2 如下
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of
// coloured 0s in an n-level hexagon
int count(int n)
{
return n * (3 * n - 1) / 2;
}
// Driver code
int main()
{
int n = 3;
cout << count(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 count of
// coloured 0s in an n-level hexagon
static int count(int n)
{
return n * (3 * n - 1) / 2;
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.println(count(n));
}
}
// This code is contributed by Code_Mech
Python 3
# Python3 implementation of the approach
# Function to return the count of
# coloured 0s in an n-level hexagon
def count(n) :
return n * (3 * n - 1) // 2;
# Driver code
if __name__ == "__main__" :
n = 3;
print(count(n));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count of
// coloured 0s in an n-level hexagon
static int count(int n)
{
return n * (3 * n - 1) / 2;
}
// Driver code
public static void Main(String[] args)
{
int n = 3;
Console.WriteLine(count(n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count of
// coloured 0s in an n-level hexagon
function count(n)
{
return parseInt(n * (3 * n - 1) / 2);
}
// Driver code
var n = 3;
document.write(count(n));
</script>
Output:
12
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