给定一个数字N ,任务是找到第N个Icositrigonal 数字或不。
An Icositrigonal number is class of figurate number. It has 23 – sided polygon called Icositrigon. The N-th Icositrigonal number count’s the 23 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icositrigonol numbers are 1, 23, 66, 130, 215, 321, 448 …
例子:
Input: N = 2
Output: 23
Explanation:
The second Icositrigonol number is 66.
Input: N = 6
Output: 321
方法:第 N 个 Icositrigonal 数由以下公式给出:
下面是上述方法的实现:
C++
// C++ program to find nth
// Icositrigonal number
#include
using namespace std;
// Function to find N-th
// Icositrigonal number
int Icositrigonal_num(int n)
{
// Formula to calculate nth
// Icositrigonal number
return (21 * n * n - 19 * n) / 2;
}
// Driver Code
int main()
{
int n = 3;
cout << Icositrigonal_num(n) << endl;
n = 10;
cout << Icositrigonal_num(n);
return 0;
}
Java
// Java program to find nth
// Icositrigonal number
class GFG{
// Function to find N-th
// Icositrigonal number
static int IcositrigonalNum(int n)
{
// Formula to calculate nth
// Icositrigonal number
return (21 * n * n - 19 * n) / 2;
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.print(IcositrigonalNum(n) + "\n");
n = 10;
System.out.print(IcositrigonalNum(n));
}
}
// This code is contributed by spp____
Python3
# Python3 program to find nth
# Icositrigonal number
# Function to find N-th
# Icositrigonal number
def IcositrigonalNum(n):
# Formula to calculate nth
# Icositrigonal number
return (21 * n * n - 19 * n) / 2;
# Driver code
n = 3
print(IcositrigonalNum(n))
n = 10
print(IcositrigonalNum(n))
# This code is contributed by spp____
C#
// C# program to find nth
// Icositrigonal number
using System;
class GFG{
// Function to find N-th
// Icositrigonal number
static int IcositrigonalNum(int n)
{
// Formula to calculate nth
// Icositrigonal number
return (21 * n * n - 19 * n) / 2;
}
// Driver code
public static void Main()
{
int n = 3;
Console.WriteLine(IcositrigonalNum(n));
n = 10;
Console.WriteLine(IcositrigonalNum(n));
}
}
// This code is contributed by spp____
Javascript
输出:
66
955
参考: https : //en.wikipedia.org/wiki/Polygonal_number
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