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# Find Range Value of the Expression

• Last Updated : 17 Mar, 2021

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 using namespace std; // Function to return the value// of the given expressiondouble 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 codeint 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 approachimport java.util.*; class GFG{ // Function to return the value// of the given expressionstatic 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 codepublic 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 expressiondef 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 codeif __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 approachusing System; public class GFG{  // Function to return the value// of the given expressionstatic 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 codepublic 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

 
Output:
0.09

Time Complexity: O(1)

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