给定角度 哪里
.任务是检查是否有可能使正多边形的所有内角都等于
.如果可能,则打印“是”,否则打印“否”(不带引号)。 例如:
null
Input: angle = 90Output: YESPolygons with sides 4 ispossible with angle 90 degrees.Input: angle = 30Output: NO
方法: 内角定义为正多边形任意两个相邻边之间的角度。 它由 哪里 N 是多边形中的边数。 这可以写成
. 根据我们得到的重新安排条件,
. 因此,如果 N 是一个 整数 答案是“是”,否则答案是“否”。 以下是上述方法的实施情况:
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // Function to check whether it is possible // to make a regular polygon with a given angle. void makePolygon( float a) { // N denotes the number of sides // of polygons possible float n = 360 / (180 - a); if (n == ( int )n) cout << "YES" ; else cout << "NO" ; } // Driver code int main() { float a = 90; // function to print the required answer makePolygon(a); return 0; } |
JAVA
class GFG { // Function to check whether // it is possible to make a // regular polygon with a given angle. static void makePolygon( double a) { // N denotes the number of // sides of polygons possible double n = 360 / ( 180 - a); if (n == ( int )n) System.out.println( "YES" ); else System.out.println( "NO" ); } // Driver code public static void main (String[] args) { double a = 90 ; // function to print // the required answer makePolygon(a); } } // This code is contributed by Bilal |
Python3
# Python 3 implementation # of above approach # Function to check whether # it is possible to make a # regular polygon with a # given angle. def makePolygon(a) : # N denotes the number of sides # of polygons possible n = 360 / ( 180 - a) if n = = int (n) : print ( "YES" ) else : print ( "NO" ) # Driver Code if __name__ = = "__main__" : a = 90 # function calling makePolygon(a) # This code is contributed # by ANKITRAI1 |
C#
// C# implementation of // above approach using System; class GFG { // Function to check whether // it is possible to make a // regular polygon with a // given angle. static void makePolygon( double a) { // N denotes the number of // sides of polygons possible double n = 360 / (180 - a); if (n == ( int )n) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); } // Driver code static void Main() { double a = 90; // function to print // the required answer makePolygon(a); } } // This code is contributed by mits |
PHP
<?php // PHP implementation of above approach // Function to check whether it // is possible to make a regular // polygon with a given angle. function makePolygon( $a ) { // N denotes the number of // sides of polygons possible $n = 360 / (180 - $a ); if ( $n == (int) $n ) echo "YES" ; else echo "NO" ; } // Driver code $a = 90; // function to print the // required answer makePolygon( $a ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // JavaScript implementation of above approach // Function to check whether it is possible // to make a regular polygon with a given angle. function makePolygon(a) { // N denotes the number of sides // of polygons possible var n = parseFloat(360 / (180 - a)); if (n === parseInt(n)) document.write( "YES" ); else document.write( "NO" ); } // Driver code var a = 90; // function to print the required answer makePolygon(a); </script> |
输出:
YES
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