计算从 L 到 R 范围内的数字之和

给定两个数字L和R 。任务是找到L到R范围内的数字的总和。

例子：

Input: L = 3, R = 6
Output: 40
Explanation: 3 + 3+4 + 3+4+5 + 3+4+5+6 = 40

Input: L = 5, R = 6
Output: 16

方法：这个问题是基于公式的。对于下面给出的说明，观察每个数字在总和中重复的次数，并根据该次数计算最终总和。

插图： L = 3， R = 6

总和 = 3 + 3+4 + 3+4+5 + 3+4+5+6 = 3+3+3+3 + 4+4+4 + 5+5 + 6（分组时）
即等于 3*4 + 4*3 + 5*2 + 6*1

因此，对于任何范围 L 到 R，总和可以计算为：

L*D + (L+1)*(D-1) + (L+2)*(D-2) + … + (R-1)*(2) + R*1

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
  
// Function to return sum
int findSum(int L, int R)
{
    // Initializing the variables
    int sum = 0, d = R - L + 1;
  
    for (int i = L; i <= R; i++) {
        sum += (i * d);
        d--;
    }
  
    // Return Sum as the final result.
    return sum;
}
  
// Driver Code
int main()
{
    int L = 3, R = 6;
  
    // Function call
    cout << findSum(L, R);
  
    return 0;
}

Java

// Java code to implement above approach
import java.util.*;
public class GFG {
  
// Function to return sum
static int findSum(int L, int R)
{
    
    // Initializing the variables
    int sum = 0, d = R - L + 1;
  
    for (int i = L; i <= R; i++) {
        sum += (i * d);
        d--;
    }
  
    // Return Sum as the final result.
    return sum;
}
  
// Driver code
public static void main(String args[])
{
    int L = 3, R = 6;
  
    // Function call
    System.out.println(findSum(L, R));
  
}
}
  
// This code is contributed by Samim Hossain Mondal.

Python

# Pyhton program for above approach
  
# Function to return sum
def findSum(L, R):
      
    # Initializing the variables
    sum = 0
    d = R - L + 1
  
    for i in range(L, R + 1):
        sum += (i * d)
        d = d - 1
  
    # Return Sum as the final result.
    return sum
  
# Driver Code
L = 3
R = 6
  
# Function call
print(findSum(L, R))
  
# This code is contributed by Samim Hossain Mondal.

C#

// C# code to implement above approach
using System;
public class GFG {
  
  // Function to return sum
  static int findSum(int L, int R)
  {
  
    // Initializing the variables
    int sum = 0, d = R - L + 1;
  
    for (int i = L; i <= R; i++) {
      sum += (i * d);
      d--;
    }
  
    // Return Sum as the final result.
    return sum;
  }
  
  // Driver code
  public static void Main()
  {
    int L = 3, R = 6;
  
    // Function call
    Console.WriteLine(findSum(L, R));
  }
}
  
// This code is contributed by ukasp.

Javascript

   // JavaScript code for the above approach
 
 
   // Function to return sum
   function findSum(L, R) {
     // Initializing the variables
     let sum = 0, d = R - L + 1;
 
     for (let i = L; i <= R; i++) {
       sum += (i * d);
       d--;
     }
 
     // Return Sum as the final result.
     return sum;
   }
 
   // Driver Code
 
   let L = 3, R = 6;
 
   // Function call
   document.write(findSum(L, R));
 // This code is contributed by Potta Lokesh
 

输出

40

时间复杂度： O(R-L+1)

辅助空间： O(1)