假设四边形的所有角度都在AP中且具有共同的差’d’,则任务是找到所有角度。
例子:
Input: d = 10
Output: 75, 85, 95, 105
Input: d = 20
Output: 60, 80, 100, 120
方法:
We know that the angles of the quadrilateral are in AP and having the common difference ‘d’.
So, if we assume the first angle to be ‘a’ then the other angles can be calculated as,
‘a+d’, ‘a+2d’ and ‘a+3d’
And, from the properties of quadrilaterals, the sum of all the angles of a quadrilateral is 360. So,
(a) + (a + d) + (a + 2*d) + (a + 3*d) = 360
4*a + 6*d = 360
a = (360 – (6*d)) / 4
where ‘a’ is the angle assumed in the beginning and ‘d’ is the common difference.
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
#define ll long long int
using namespace std;
// Driver code
int main()
{
int d = 10;
double a;
// according to formula derived above
a = (double)(360 - (6 * d)) / 4;
// print all the angles
cout << a << ", " << a + d << ", " << a + (2 * d)
<< ", " << a + (3 * d) << endl;
return 0;
}
Java
// java implementation of the approach
import java.io.*;
class GFG {
// Driver code
public static void main (String[] args) {
int d = 10;
double a;
// according to formula derived above
a = (double)(360 - (6 * d)) / 4;
// print all the angles
System.out.print( a + ", " + (a + d) + ", " + (a + (2 * d))
+ ", " + (a + (3 * d)));
}
}
//This code is contributed
//by inder_verma
Python
# Python implementation
# of the approach
d = 10
a = 0.0
# according to formula
# derived above
a=(360 - (6 * d)) / 4
# print all the angles
print(a,",", a + d, ",", a + 2 * d,
",", a + 3 * d, sep = ' ')
# This code is contributed
# by sahilshelangia
C#
// C# implementation of the approach
using System;
class GFG
{
// Driver code
public static void Main ()
{
int d = 10;
double a;
// according to formula derived above
a = (double)(360 - (6 * d)) / 4;
// print all the angles
Console.WriteLine( a + ", " + (a + d) +
", " + (a + (2 * d)) +
", " + (a + (3 * d)));
}
}
// This code is contributed
// by anuj_67
PHP
Javascript
输出:
75, 85, 95, 105