找出对角线倾斜并排成一排的正方形的一面

这里给出n个正方形，它们在顶点之间相互倾斜并相互接触，并排成一行，给出第一个和最后一个正方形的中心之间的距离，这些正方形的边长相等，任务是找到每个正方形的侧面。
例子：

Input :d = 42, n = 4 
Output :The side of each square is 9.899
Input :d = 54, n = 7 
Output :The side of each square is 6.364 
 

方法：
有n个正方形，每个正方形的边长为a，第一个和最后一个正方形之间的距离等于d。从图中可以明显看出，它们是通过对角线连接的。每个对角线的长度等于a√2 。
对于第一个和最后一个正方形，对角线的一半仅覆盖长度d;对于其余(n-2)个正方形，对角线的整个覆盖范围为d。因此，a和d之间的关系如下：

a/√2 + a/√2 + (n-2)*a√2 = d 
=> a√2 + √2na – 2a√2 = d 
=> n√2a – a√2 = d 
=> a = d/((n-1)*(√2)) 
Side of the square = distance between centers/((no. of squares-1) * sqrt(2)). 
 

下面是上述方法的实现：

C++

// C++ program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
#include 
using namespace std;
 
void radius(double n, double d)
{
    cout << "The side of each square is "
         << d / ((n - 1) * sqrt(2)) << endl;
}
 
// Driver code
int main()
{
    double d = 42, n = 4;
    radius(n, d);
    return 0;
}

Java

// Java program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
import java.io.*;
 
class GFG
{
 
 
static void radius(double n, double d)
{
    System.out.print( "The side of each square is "+
        d / ((n - 1) * Math.sqrt(2)));
}
 
// Driver code
public static void main (String[] args)
{
    double d = 42, n = 4;
    radius(n, d);
}
}
 
// This code is contributed by anuj_67..

Python3

# Python program to find side of the squares
# inclined and touch each other externally
# at vertices and are lined in a row
# and distance between the
# centers of first and last squares is given
 
def radius(n, d):
 
    print("The side of each square is ",
        d / ((n - 1) * (2**(1/2))));
 
# Driver code
d = 42; n = 4;
radius(n, d);
 
# This code is contributed by Rajput-Ji

C#

// C# program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
using System;
 
class GFG
{
 
 
static void radius(double n, double d)
{
    Console.WriteLine( "The side of each square is "+
        d / ((n - 1) * Math.Sqrt(2)));
}
 
// Driver code
public static void Main ()
{
    double d = 42, n = 4;
    radius(n, d);
}
}
 
// This code is contributed by anuj_67..

Javascript

 
// javascript program to find side of the squares
// inclined and touch each other externally
// at vertices and are lined in a row
// and distance between the
// centers of first and last squares is given
 
function radius(n , d)
{
    document.write( "The side of each square is "+
        (d / ((n - 1) * Math.sqrt(2))).toFixed(5));
}
 
// Driver code
var d = 42, n = 4;
radius(n, d);
 
 
// This code is contributed by Amit Katiyar
 

输出：

The side of each square is 9.89949

时间复杂度： O(1)