查找表达式的范围值

给定两个整数L和R ，任务是计算表达式的值：

    $$$F = \sum_{i=L}^{R} \frac{1}{i^2 + i} $$$

例子：

Input: L = 6, R = 12 
Output: 0.09
Input: L = 5, R = 6 
Output: 0.06 
 

方法：可以观察到$\frac{1}{i^2 + i} = \frac{1}{i} - \frac{1}{i + 1}$ .
所以， $F = \sum_{i=L}^{R} \frac{1}{i^2 + i} = (\frac{1}{L} - \frac{1}{L + 1}) + (\frac{1}{L + 1} - \frac{1}{L + 2}) + .... + (\frac{1}{R} - \frac{1}{R + 1}) = (\frac{1}{L} - \frac{1}{R + 1})$
因此，答案将是(1 / L) – (1 / (R + 1)) 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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.

输出：

0.09

时间复杂度： O(1)

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