使用 n 边正多边形的 3 个顶点形成的给定角度的出现次数

原文:https://www . geeksforgeeks . org/给定角度正多边形使用 3 个顶点形成的出现次数/

给定一个 n 边正多边形和一个角度θ，任务是在一个顶点标记为 A 1 、A 2 、…、A n 的正 n 边形(有 n 个顶点的正多边形)中找到角度(A i 、A j 、A k ) = θ ( i < j < k)的出现次数。 示例:

Input: n = 4, ang = 90
Output: 4

Input: n = 6, ang = 50
Output: 0

进场:

  1. 首先我们检查这样一个角度是否存在。
  2. 考虑顶点为 x 、 y 和 z ，要寻找的角度为 xyz。
  3. x 与 y 之间的边数为 a ，而 y 与 z 之间的边数为 b 。
  4. 然后∠XYZ = 180 –( 180 *(a+b))/n。
  5. 因此≈XYZ * n(mod 180)= 0。
  6. 接下来我们需要找到这些角度的计数。
  7. 由于多边形是规则的，我们只需要计算一个顶点的角度，就可以直接将结果乘以 n(顶点的数量)。
  8. 在每个顶点处，可以在 n-1-freq 次处找到角度，其中 freq = (n*ang)/180 并描绘了创建所需角度后剩余的边数，即 z 和 x 之间的边数

以下是上述方法的实现:

C++

// C++ implementation of the approach

#include <bits/stdc++.h>
using namespace std;

// Function that calculates occurrences
// of given angle that can be created
// using any 3 sides
int solve(int ang, int n)
{

    // Maximum angle in a regular n-gon
    // is equal to the interior angle
    // If the given angle
    // is greater than the interior angle
    // then the given angle cannot be created
    if ((ang * n) > (180 * (n - 2))) {
        return 0;
    }

    // The given angle times n should be divisible
    // by 180 else it cannot be created
    else if ((ang * n) % 180 != 0) {
        return 0;
    }

    // Initialise answer
    int ans = 1;

    // Calculate the frequency
    // of given angle for each vertex
    int freq = (ang * n) / 180;

    // Multiply answer by frequency.
    ans = ans * (n - 1 - freq);

    // Multiply answer by the number of vertices.
    ans = ans * n;

    return ans;
}

// Driver code
int main()
{
    int ang = 90, n = 4;

    cout << solve(ang, 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 that calculates occurrences
// of given angle that can be created
// using any 3 sides
static int solve(int ang, int n)
{

    // Maximum angle in a regular n-gon
    // is equal to the interior angle
    // If the given angle
    // is greater than the interior angle
    // then the given angle cannot be created
    if ((ang * n) > (180 * (n - 2)))
    {
        return 0;
    }

    // The given angle times n should be divisible
    // by 180 else it cannot be created
    else if ((ang * n) % 180 != 0)
    {
        return 0;
    }

    // Initialise answer
    int ans = 1;

    // Calculate the frequency
    // of given angle for each vertex
    int freq = (ang * n) / 180;

    // Multiply answer by frequency.
    ans = ans * (n - 1 - freq);

    // Multiply answer by the number of vertices.
    ans = ans * n;

    return ans;
}

// Driver code
public static void main (String[] args)
{
    int ang = 90, n = 4;
    System.out.println(solve(ang, n));
}
}

// This code is contributed by Rajput-Ji

Python 3

# Python3 implementation of the approach

# Function that calculates occurrences
# of given angle that can be created
# using any 3 sides
def solve(ang, n):

    # Maximum angle in a regular n-gon
    # is equal to the interior angle
    # If the given angle
    # is greater than the interior angle
    # then the given angle cannot be created
    if ((ang * n) > (180 * (n - 2))):
        return 0

    # The given angle times n should be divisible
    # by 180 else it cannot be created
    elif ((ang * n) % 180 != 0):
        return 0

    # Initialise answer
    ans = 1

    # Calculate the frequency
    # of given angle for each vertex
    freq = (ang * n) // 180

    # Multiply answer by frequency.
    ans = ans * (n - 1 - freq)

    # Multiply answer by the number of vertices.
    ans = ans * n

    return ans

# Driver code
ang = 90
n = 4

print(solve(ang, n))

# This code is contributed by Mohit Kumar

C

// C# implementation of the approach
using System;

class GFG
{

// Function that calculates occurrences
// of given angle that can be created
// using any 3 sides
static int solve(int ang, int n)
{

    // Maximum angle in a regular n-gon
    // is equal to the interior angle
    // If the given angle
    // is greater than the interior angle
    // then the given angle cannot be created
    if ((ang * n) > (180 * (n - 2)))
    {
        return 0;
    }

    // The given angle times n should be divisible
    // by 180 else it cannot be created
    else if ((ang * n) % 180 != 0)
    {
        return 0;
    }

    // Initialise answer
    int ans = 1;

    // Calculate the frequency
    // of given angle for each vertex
    int freq = (ang * n) / 180;

    // Multiply answer by frequency.
    ans = ans * (n - 1 - freq);

    // Multiply answer by the
    // number of vertices.
    ans = ans * n;

    return ans;
}

// Driver code
public static void Main (String[] args)
{
    int ang = 90, n = 4;
    Console.WriteLine(solve(ang, n));
}
}

// This code is contributed by Princi Singh

java 描述语言

<script>

// JavaScript implementation of the approach
// Function that calculates occurrences
// of given angle that can be created
// using any 3 sides
function solve(ang , n)
{

    // Maximum angle in a regular n-gon
    // is equal to the interior angle
    // If the given angle
    // is greater than the interior angle
    // then the given angle cannot be created
    if ((ang * n) > (180 * (n - 2)))
    {
        return 0;
    }

    // The given angle times n should be divisible
    // by 180 else it cannot be created
    else if ((ang * n) % 180 != 0)
    {
        return 0;
    }

    // Initialise answer
    var ans = 1;

    // Calculate the frequency
    // of given angle for each vertex
    var freq = (ang * n) / 180;

    // Multiply answer by frequency.
    ans = ans * (n - 1 - freq);

    // Multiply answer by the number of vertices.
    ans = ans * n;

    return ans;
}

// Driver code
var ang = 90, n = 4;
document.write(solve(ang, n));

// This code contributed by shikhasingrajput 

</script>

Output: 

4

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