📜  检查 N 是否为 Icositrigonal 数的程序

📅  最后修改于: 2021-10-23 08:57:58             🧑  作者: Mango

给定一个整数N ,任务是检查它是否是一个 Icositrigonal 数。

例子:

方法:

  1. 第i个的icositrigonal数项K被给定为
    K^{th} Term = \frac{21*K^{2} - 19*K}{2}
  2. 因为我们必须检查给定的数字是否可以表示为正弦三角数。这可以检查如下 –
  1. 最后,检查使用这个公式计算的值是一个整数,这意味着 N 是一个正弦三角数。

下面是上述方法的实现:

C++
// C++ implementation to check that
// a number is a icositrigonal number or not
 
#include 
 
using namespace std;
 
// Function to check that the
// number is a icositrigonal number
bool isicositrigonal(int N)
{
    float n
        = (19 + sqrt(168 * N + 361))
          / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}


Java
// Java implementation to check that
// a number is a icositrigonal number or not
import java.util.*;
class GFG{
 
// Function to check that the
// number is a icositrigonal number
static boolean isicositrigonal(int N)
{
    float n = (float)(19 + Math.sqrt(168 * N + 361)) / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String args[])
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Akanksha_Rai


Python3
# Python3 implementation to check that a 
# number is a icositrigonal number or not
import math
 
# Function to check that the number
# is a icositrigonal number
def isicositrigonal(N):
 
    n = (19 + math.sqrt(168 * N + 361)) / 42
 
    # Condition to check if the number 
    # is a icositrigonal number
    return (n - int(n)) == 0
 
# Driver Code
i = 23
 
# Function call
if (isicositrigonal(i)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by divyamohan123


C#
// C# implementation to check that
// a number is a icositrigonal number or not
using System;
class GFG{
 
// Function to check that the
// number is a icositrigonal number
static bool isicositrigonal(int N)
{
    float n = (float)(19 + Math.Sqrt(168 * N + 361)) / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Nidhi_Biet


Javascript


输出:
Yes

时间复杂度: O(1)

辅助空间: O(1)