Given here are n squares which touch each other externally, and are lined up in a row. The distance between the centers of the first and last square is given. The squares have equal side length. The task is to find the side of each square.
Examples:
Input: d = 42, n = 4 Output: The side of each square is 14 Input: d = 36, n = 5 Output: The side of each square is 9
Approach:
Suppose there are n squares each having side of length a.
Let, the distance between the first and last squares = d
From the figure, it is clear,
a/2 + a/2 + (n-2)*a = d
a + na – 2a = d
na – a = d
so, a = d/(n-1)
C++
// C++ program to find side of the squares // which are lined in a row and distance between the // centers of first and last squares is given #include <bits/stdc++.h> using namespace std;
void radius( int n, int d)
{ cout << "The side of each square is "
<< d / (n - 1) << endl;
} // Driver code int main()
{ int d = 42, n = 4;
radius(n, d);
return 0;
} |
Java
// Java program to find side of the squares // which 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( int n, int d)
{ System.out.print( "The side of each square is "
+ d / (n - 1 ));
} // Driver code public static void main (String[] args)
{ int d = 42 , n = 4 ;
radius(n, d);
} } // This code is contributed by vt_m. |
Python3
# Python program to find side of the squares # which 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 ));
d = 42 ; n = 4 ;
radius(n, d); # This code contributed by PrinciRaj1992 |
C#
// C# program to find side of the squares // which are lined in a row and distance between the // centers of first and last squares is given using System;
class GFG
{ static void radius( int n, int d)
{ Console.Write( "The side of each square is "
+ d / (n - 1));
} // Driver code public static void Main ()
{ int d = 42, n = 4;
radius(n, d);
} } // This code is contributed by anuj_67.. |
Javascript
<script> // javascript program to find side of the squares // which 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));
} // Driver code var d = 42, n = 4;
radius(n, d); // This code is contributed by 29AjayKumar </script> |
Output:
The side of each square is 14
Time Complexity: O(1)
Auxiliary Space: O(1)
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