第 n 个五边形数字
给定一个整数 n，求第 n 个五边形数。前三个五边形数字是 1、5 和 12(请看下图)。 第 n 个五边形数字 P n 是由规则五边形的轮廓组成的点图案中的不同点的数量，当五边形重叠时，边长可达 n 个点【来源Wiki T6】示例:T8】
Input: n = 1
Output: 1
Input: n = 2
Output: 5
Input: n = 3
Output: 12
一般来说，多边形数(三角形数、正方形数等)是以正多边形形状排列的点或卵石表示的数。前几个五边形数字是:1、5、12 等。 如果 s 是多边形的边数，则第 n 个边数 P (s，n)的公式为
nth s-gonal number P(s, n) = (s - 2)n(n-1)/2 + n
If we put s = 5, we get
n'th Pentagonal number Pn = 3*n*(n-1)/2 + n
示例:
五边形数字
下面是上述思想在不同编程语言中的实现。
C++
// C++ program for above approach
#include<bits/stdc++.h>
using namespace std;
// Finding the nth pentagonal number
int pentagonalNum(int n)
{
return (3 * n * n - n) / 2;
}
// Driver code
int main()
{
int n = 10;
cout << "10th Pentagonal Number is = "
<< pentagonalNum(n);
return 0;
}
// This code is contributed by Code_Mech
C
// C program for above approach
#include <stdio.h>
#include <stdlib.h>
// Finding the nth Pentagonal Number
int pentagonalNum(int n)
{
return (3*n*n - n)/2;
}
// Driver program to test above function
int main()
{
int n = 10;
printf("10th Pentagonal Number is = %d \n \n",
pentagonalNum(n));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for above approach
class Pentagonal
{
int pentagonalNum(int n)
{
return (3*n*n - n)/2;
}
}
public class GeeksCode
{
public static void main(String[] args)
{
Pentagonal obj = new Pentagonal();
int n = 10;
System.out.printf("10th petagonal number is = "
+ obj.pentagonalNum(n));
}
}
计算机编程语言
# Python program for finding pentagonal numbers
def pentagonalNum( n ):
return (3*n*n - n)/2
#Script Begins
n = 10
print "10th Pentagonal Number is = ", pentagonalNum(n)
#Scripts Ends
C
// C# program for above approach
using System;
class GFG {
static int pentagonalNum(int n)
{
return (3 * n * n - n) / 2;
}
public static void Main()
{
int n = 10;
Console.WriteLine("10th petagonal"
+ " number is = " + pentagonalNum(n));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for above approach
// Finding the nth Pentagonal Number
function pentagonalNum($n)
{
return (3 * $n * $n - $n) / 2;
}
// Driver Code
$n = 10;
echo "10th Pentagonal Number is = ",
pentagonalNum($n);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program for above approach
function pentagonalNum(n)
{
return (3 * n * n - n) / 2;
}
// Driver code to test above methods
let n = 10;
document.write("10th petagonal"
+ " number is = " + pentagonalNum(n));
// This code is contributed by avijitmondal1998.
</script>
输出:
10th Pentagonal Number is = 145
时间复杂度: O(1) 辅助空间: O(1) 参考: https://en.wikipedia.org/wiki/Polygonal_number 本文由马扎伊玛目汗供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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