在给定范围内所有数字的总和可被6整除

在给定范围LR的情况下，找到范围LR内可被6整除的所有数字的总和
L和R非常大。
例子：

Input : 1 20
Output : 36 
Explanation: 6 + 12 + 18 = 36 

Input : 5 7  
Output : 6
Explanation: 6 is the only divisible number 
in range 5-7 

一个简单的方法是从L到R循环运行，并对所有可被6整除的数字求和。
一种有效的方法是将所有可被6整除的数字相加，直到R，再将所有可被6整除的数字相加，直到L-1。然后减法将是答案。
总和= 6 + 12 + 8 + …….(R / 6)项。
总和= 6(1 + 2 + 3…R / 6项)

sumR = 3 * (R/6) * (R/6+1) 

同样，我们得到

sumL as 3 * ((L-1)/6) * ((L-1/6)+1) 

最终答案为sumR – sumL。

C++

// CPP program to find sum of numbers divisible
// by 6 in a given range.
#include 
using namespace std;
 
// function to calculate the sum of
// all numbers divisible by 6 in range L-R..
int sum(int L, int R)
{
    // no of multiples of 6 upto r
    int p = R / 6;
 
    // no of multiples of 6 upto l-1
    int q = (L - 1) / 6;
 
    // summation of all multiples of 6 upto r
    int sumR = 3 * (p * (p + 1));
 
    // summation of all multiples of 6 upto l-1
    int sumL = (q * (q + 1)) * 3;
 
    // returns the answer
    return sumR - sumL;
}
 
// driver program to test the above function
int main()
{
    int L = 1, R = 20;
    cout << sum(L, R);
    return 0;
}

Java

// Java program to find sum of numbers
// divisible by 6 in a given range.
import java.io.*;
class GFG {
 
// function to calculate the sum
// of all numbers divisible by 6
// in range L-R..
static int sum(int L, int R)
{
    // no of multiples of 6 upto r
    int p = R / 6;
 
    // no of multiples of 6 upto l-1
    int q = (L - 1) / 6;
 
    // summation of all multiples of
    // 6 upto r
    int sumR = 3 * (p * (p + 1));
 
    // summation of all multiples of
    // 6 upto l-1
    int sumL = (q * (q + 1)) * 3;
 
    // returns the answer
    return sumR - sumL;
}
 
// driver program
public static void main(String[] args)
{
    int L = 1, R = 20;
    System.out.println(sum(L, R));
}
}
 
// This code is contributed by Prerna Saini

Python

# Python3 program to find sum of numbers divisible
# by 6 in a given range.
 
def sumDivisible(L, R):
   
    # no of multiples of 6 upto r
    p = int(R/6)
 
    # no of multiples of 6 upto l-1
    q = int((L-1)/6)
 
    # summation of all multiples of 6 upto r
    sumR = 3 * (p * (p + 1))
 
    # summation of all multiples of 6 upto l-1
    sumL = (q * (q + 1)) * 3
 
    # returns the answer
    return sumR - sumL
 
# driver code
L = 1
R = 20
print(sumDivisible(L,R))
 
# This code is contributed by 'Abhishek Sharma 44'.

C#

// C# program to find sum of numbers
// divisible by 6 in a given range.
using System;
 
class GFG {
     
    // function to calculate the sum
    // of all numbers divisible by 6
    // in range L-R..
    static int sum(int L, int R)
    {
         
        // no of multiples of 6 upto r
        int p = R / 6;
     
        // no of multiples of 6 upto l-1
        int q = (L - 1) / 6;
     
        // summation of all multiples of
        // 6 upto r
        int sumR = 3 * (p * (p + 1));
     
        // summation of all multiples of
        // 6 upto l-1
        int sumL = (q * (q + 1)) * 3;
     
        // returns the answer
        return sumR - sumL;
    }
     
    // driver program
    public static void Main()
    {
        int L = 1, R = 20;
         
        Console.WriteLine(sum(L, R));
    }
}
 
// This code is contributed by Anant Agarwal.

Javascript

// Javascript program to find sum of numbers divisible
// by 6 in a given range.

     
    // function to calculate the sum of
    // all numbers divisible by 6 in range L-R..
    function sum(L, R)
    {
        // no of multiples of 6 upto r
        let p = Math.floor(R / 6);
 
        // no of multiples of 6 upto l-1
        let q = Math.floor((L - 1) / 6);
 
        // summation of all multiples of 6 upto r
        let sumR = Math.floor(3 * (p * (p + 1)));
 
        // summation of all multiples of 6 upto l-1
        let sumL = Math.floor((q * (q + 1)) * 3);
 
        // returns the answer
        return sumR - sumL;
    }
 
    // Driver Code
    let L = 1, R = 20;
    document.write(sum(L, R));
     
    // This code is contributed by ajaykrsharma132.

输出：

36

时间复杂度：O(1)