给定范围内的AP元素总数

给定arr和Q查询中的算术级数，其形式为[L，R] ，其中L是范围的左边界， R是右边界。任务是在给定范围内找到AP元素的总和。

注意：范围为1分度，且1≤L，R≤N，其中N是arr的大小。

例子：

Input: arr[] = {2, 4, 6, 8, 10, 12, 14, 16}, Q = [[2, 4], [2, 6], [5, 8]]
Output:
18
40
52
Explanation:
Range 1: arr = {4, 6, 8}. Therefore sum = 18
Range 2: arr = {4, 6, 8, 10, 12}. Therefore sum = 40
Range 3: arr = {10, 12, 14, 16}. Therefore sum = 52

Input: arr[] = {7, 14, 21, 28, 35, 42}, Q = [[1, 6], [2, 4], [3, 3]]
Output:
147
63
21
Explanation:
Range 1: arr = {7, 14, 21, 28, 35, 42}. Therefore sum = 147
Range 2: arr = {14, 21, 28}. Therefore sum = 63
Range 3: arr = {21}. Therefore sum = 21

方法：由于给定的序列是算术级数，因此可以很容易地分两步轻松地求出和：

1. 将范围的第一个元素乘以范围中的元素数。
2. 在其上加上(d * k *(k + 1))/ 2 ，其中d是AP的共同差异，而k是(间隙数– 1内的元素数)，它对应于间隙数。

例如：
假设a [i]是该范围的第一个元素， d是AP的公共差，而k + 1是给定范围内的元素数。
则范围的总和为

= a[i] + a[i+1] + a[i+2] + ….. + a[i+k]
= a[i] + a[i] + d + a[i] + 2 * d + …. + a[i] + k * d
= a[i] * (k + 1) + d * (1 + 2 + … + k)
= a[i] * (k + 1) + (d * k * (k+1))/2

下面是上述方法的实现：

CPP

// C++ program to find the sum of elements
// of an AP in the given range
  
#include 
using namespace std;
  
// Function to find sum in the given range
int findSum(int* arr, int n,
            int left, int right)
{
    // Find the value of k
    int k = right - left;
  
    // Find the common difference
    int d = arr[1] - arr[0];
  
    // Find the sum
    int ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) / 2;
  
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 4, 6, 8, 10, 12, 14, 16 };
    int queries = 3;
    int q[queries][2] = { { 2, 4 },
                          { 2, 6 },
                          { 5, 6 } };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    for (int i = 0; i < queries; i++)
        cout << findSum(arr, n, q[i][0], q[i][1])
             << endl;
}

Java

// Java program to find the sum of elements
// of an AP in the given range
class GFG{
   
// Function to find sum in the given range
static int findSum(int []arr, int n,
            int left, int right)
{
    // Find the value of k
    int k = right - left;
   
    // Find the common difference
    int d = arr[1] - arr[0];
   
    // Find the sum
    int ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) / 2;
   
    return ans;
}
   
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 6, 8, 10, 12, 14, 16 };
    int queries = 3;
    int q[][] = { { 2, 4 },
                          { 2, 6 },
                          { 5, 6 } };
    int n = arr.length;
   
    for (int i = 0; i < queries; i++)
        System.out.print(findSum(arr, n, q[i][0], q[i][1])
             +"\n");
}
}
  
// This code is contributed by Princi Singh

Python3

# Python program to find the sum of elements
# of an AP in the given range
  
# Function to find sum in the given range
def findSum(arr, n, left, right):
  
    # Find the value of k
    k = right - left;
  
    # Find the common difference
    d = arr[1] - arr[0];
  
    # Find the sum
    ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) // 2;
  
    return ans;
  
# Driver code
if __name__ == '__main__':
    arr = [ 2, 4, 6, 8, 10, 12, 14, 16 ];
    queries = 3;
    q = [[ 2, 4 ],[ 2, 6 ],[ 5, 6 ]];
    n = len(arr);
  
    for i in range(queries):
        print(findSum(arr, n, q[i][0], q[i][1]));
  
# This code is contributed by sapnasingh4991

C#

// C# program to find the sum of elements
// of an AP in the given range
using System;
  
class GFG{
    
// Function to find sum in the given range
static int findSum(int []arr, int n,
            int left, int right)
{
    // Find the value of k
    int k = right - left;
    
    // Find the common difference
    int d = arr[1] - arr[0];
    
    // Find the sum
    int ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) / 2;
    
    return ans;
}
    
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
    int queries = 3;
    int [,]q = { { 2, 4 },
                          { 2, 6 },
                          { 5, 6 } };
    int n = arr.Length;
    
    for (int i = 0; i < queries; i++)
        Console.Write(findSum(arr, n, q[i,0], q[i,1])
             +"\n");
}
}
  
// This code is contributed by PrinciRaj1992

输出：

18
40
22

时间复杂度： O(1)
空间复杂度： O(1)