Given two integers L and R, the task is to calculate the value of the expression:
Examples:
Input: L = 6, R = 12
Output: 0.09
Input: L = 5, R = 6
Output: 0.06
Approach: It can be observed that
So,
Hence, the answer will be (1 / L) – (1 / (R + 1)).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the value // of the given expression double get( double L, double R)
{ // Value of the first term
double x = 1.0 / L;
// Value of the last term
double y = 1.0 / (R + 1.0);
return (x - y);
} // Driver code int main()
{ int L = 6, R = 12;
// Get the result
double ans = get(L, R);
cout << fixed << setprecision(2) << ans;
return 0;
} |
Java
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to return the value // of the given expression static double get( double L, double R)
{ // Value of the first term
double x = 1.0 / L;
// Value of the last term
double y = 1.0 / (R + 1.0 );
return (x - y);
} // Driver code public static void main(String []args)
{ int L = 6 , R = 12 ;
// Get the result
double ans = get(L, R);
System.out.printf( "%.2f" , ans);
} } // This code is contributed by Surendra_Gangwar |
Python3
# Python3 implementation of the approach # Function to return the value # of the given expression def get(L, R) :
# Value of the first term
x = 1.0 / L;
# Value of the last term
y = 1.0 / (R + 1.0 );
return (x - y);
# Driver code if __name__ = = "__main__" :
L = 6 ; R = 12 ;
# Get the result
ans = get(L, R);
print ( round (ans, 2 ));
# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;
public class GFG
{ // Function to return the value // of the given expression static double get ( double L, double R)
{ // Value of the first term
double x = 1.0 / L;
// Value of the last term
double y = 1.0 / (R + 1.0);
return (x - y);
} // Driver code public static void Main(String []args)
{ int L = 6, R = 12;
// Get the result
double ans = get (L, R);
Console.Write( "{0:F2}" , ans);
} } // This code contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript implementation of the approach // Function to return the value // of the given expression function get(L, R)
{ // Value of the first term
let x = 1.0 / L;
// Value of the last term
let y = 1.0 / (R + 1.0);
return (x - y);
} // Driver code let L = 6, R = 12;
// Get the result
let ans = get(L, R);
document.write(Math.round(ans * 100) / 100);
// This code is contributed by Surbhi Tyagi. </script> |
Output:
0.09
Time Complexity: O(1)
Auxiliary Space: O(1)
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