求表达式的范围值

原文:https://www . geesforgeks . org/find-range-value-of-the-expression/

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:

输入: L = 6，R = 12 输出: 0.09 输入: L = 5，R = 6 输出: 0.06

进场: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}) 于是，答案会是(1/L)–(1/(R+1))。 以下是上述方法的实施:

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 语言(一种计算机语言，尤用于创建网站)

// 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

Python 3

# 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

java 描述语言

<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

时间复杂度: O(1)

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