检查 N 边多边形是否可能从 N 个给定角度

给定一个由N 个元素组成的数组arr[] ，其中每个元素代表一个多边形的一个角度（以度为单位），任务是检查是否可以用所有给定的角度制作一个N 边多边形。如果可能，则打印Yes否则打印No 。

例子：

Input: N = 3, arr[] = {60, 60, 60}
Output: Yes
Explanation: There exists a triangle(i.e. a polygon) satisfying the above angles. Hence the output is Yes.

Input: N = 4, arr[] = {90, 90, 90, 100}
Output: No
Explanation: There does not exist any polygon satisfying the above angles. Hence the output is No.

方法：仅当所有给定角度的总和等于180*(N-2) 时，才可能出现N 边多边形。因此，ide 是找到数组arr[] 中给定的所有角度的总和，如果总和等于180*(N-2)则打印Yes ，否则打印No 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if the polygon
// exists or not
void checkValidPolygon(int arr[], int N)
{
    // Initialize a variable to
    // store the sum of angles
    int sum = 0;
 
    // Loop through the array and
    // calculate the sum of angles
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }
 
    // Check the condition for
    // an N-side polygon
    if (sum == 180 * (N - 2))
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver Code
int main()
{
    int N = 3;
 
    // Given array arr[]
    int arr[] = { 60, 60, 60 };
 
    // Function Call
    checkValidPolygon(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to check if the polygon
// exists or not
static void checkValidPolygon(int arr[], int N)
{
     
    // Initialize a variable to
    // store the sum of angles
    int sum = 0;
 
    // Loop through the array and
    // calculate the sum of angles
    for(int i = 0; i < N; i++)
    {
        sum += arr[i];
    }
 
    // Check the condition for
    // an N-side polygon
    if (sum == 180 * (N - 2))
        System.out.println("Yes");
    else
        System.out.println("No");
}
     
// Driver code
public static void main(String[] args)
{
    int N = 3;
     
    // Given array arr[]
    int arr[] = { 60, 60, 60 };
 
    // Function call
    checkValidPolygon(arr, N);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to check if the polygon
# exists or not
def checkValidPolygon(arr, N):
 
    # Initialize a variable to
    # store the sum of angles
    Sum = 0
 
    # Loop through the array and
    # calculate the sum of angles
    for i in range(N):
        Sum += arr[i]
 
    # Check the condition for
    # an N-side polygon
    if Sum == 180 * (N - 2):
        print("Yes")
    else:
        print("No")
         
# Driver Code
N = 3
 
# Given array arr[]
arr = [ 60, 60, 60 ]
 
# Function Call
checkValidPolygon(arr, N)
 
# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if the polygon
// exists or not
static void checkValidPolygon(int []arr, int N)
{
     
    // Initialize a variable to
    // store the sum of angles
    int sum = 0;
 
    // Loop through the array and
    // calculate the sum of angles
    for(int i = 0; i < N; i++)
    {
        sum += arr[i];
    }
 
    // Check the condition for
    // an N-side polygon
    if (sum == 180 * (N - 2))
        Console.Write("Yes");
    else
        Console.Write("No");
}
     
// Driver code
public static void Main(string[] args)
{
    int N = 3;
     
    // Given array arr[]
    int []arr = { 60, 60, 60 };
 
    // Function call
    checkValidPolygon(arr, N);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript program for the above approach
 
// Function to check if the polygon
// exists or not
function checkValidPolygon(arr, N)
{
     
    // Initialize a variable to
    // store the sum of angles
    var sum = 0;
 
    // Loop through the array and
    // calculate the sum of angles
    for(var i = 0; i < N; i++)
    {
        sum += arr[i];
    }
 
    // Check the condition for
    // an N-side polygon
    if (sum == 180 * (N - 2))
        document.write("Yes");
    else
        document.write("No");
}
 
// Driver code
var N = 3;
     
// Given array arr[]
var arr = [ 60, 60, 60 ];
 
// Function call
checkValidPolygon(arr, N);
            
// This code is contributed by Kirti
 

输出：

Yes

时间复杂度： O(N)，其中 N 是数组的长度。
辅助空间： O(1)