第 n 个顺序图中的单元数量
给定一个整数 N ，任务是找到给定类型的 N 第T5】阶图中的单元格数:
例:
输入: N = 2 输出: 5 输入: N = 3 输出: 13
方法:可以观察到，对于 N = 1、2、3、… 的值，一个系列将形成为 1、5、13、25、41、61、85、113、145、181、… ，其 N th 项将为N2+(N–1)2 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number
// of cells in the nth order
// figure of the given type
int cntCells(int n)
{
int cells = pow(n, 2) + pow(n - 1, 2);
return cells;
}
// Driver code
int main()
{
int n = 3;
cout << cntCells(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 number
// of cells in the nth order
// figure of the given type
static int cntCells(int n)
{
int cells = (int)Math.pow(n, 2) +
(int)Math.pow(n - 1, 2);
return cells;
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.println(cntCells(n));
}
}
// This code is contributed by Code_Mech
Python 3
# Python3 implementation of the approach
# Function to return the number
# of cells in the nth order
# figure of the given type
def cntCells(n) :
cells = pow(n, 2) + pow(n - 1, 2);
return cells;
# Driver code
if __name__ == "__main__" :
n = 3;
print(cntCells(n));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the number
// of cells in the nth order
// figure of the given type
static int cntCells(int n)
{
int cells = (int)Math.Pow(n, 2) +
(int)Math.Pow(n - 1, 2);
return cells;
}
// Driver code
public static void Main(String[] args)
{
int n = 3;
Console.WriteLine(cntCells(n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the number
// of cells in the nth order
// figure of the given type
function cntCells(n)
{
var cells = Math.pow(n, 2) + Math.pow(n - 1, 2);
return cells;
}
// Driver code
var n = 3;
document.write(cntCells(n));
</script>
Output:
13
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