给定范围内的几何级数(GP)元素的总和

在arr []和Q查询中以[L，R]的形式给定几何级数级数，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“几何级数”元素的总和。

注意：范围为1-索引和1≤L，R≤N，其中N是ARR的大小。
例子：

Input: arr[] = {2, 4, 8, 16, 32, 64, 128, 256}, Q = [[2, 4], [2, 6], [5, 8]] 
Output: 
28 
124 
480
Explanation: 
Range 1: arr = {4, 8, 16}. Therefore sum = 28 
Range 2: arr = {4, 8, 16, 32, 64}. Therefore sum = 124 
Range 3: arr = {32, 64, 128, 256}. Therefore sum = 480

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

方法：由于给定的序列是几何级数，因此可以很容易地通过两步轻松地求出和：

1. 获取范围的第一个元素。
2. 如果d = 1，则将d * k乘以它，否则将(d ^k – 1)/(d – 1)乘以它，其中d是GP的公共比率， k是该范围内的元素数。

例如：
假设a [i]是该范围的第一个元素， d是GP的公共比率， k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-1] 
= a[i] + (a[i] * d) + (a[i] * d * d) + …. + (a[i] *  d^k) 
= a[i] *  (1 + d + … + d^k) 
= a[i] * (d^k – 1)/(d – 1)

下面是上述方法的实现：

C++

// C++ program to find the sum
// of elements of an GP 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 + 1;
 
    // Find the common difference
    int d = arr[1] / arr[0];
 
    // Find the sum
    int ans = arr[left - 1];
     
    if (d == 1)
        ans = ans * d * k;
    else
        ans = ans * ((int)pow(d, k) - 1 /
                                 (d - 1));
         
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 4, 8, 16, 32,
                  64, 128, 256 };
    int queries = 3;
    int q[][2] = { { 2, 4 }, { 2, 6 },
                   { 5, 8 } };
     
    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;
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program to find the sum
// of elements of an GP in the
// given range
import java.io.*;
import java.util.*;
 
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 + 1;
 
    // Find the common difference
    int d = arr[1] / arr[0];
 
    // Find the sum
    int ans = arr[left - 1];
     
    if (d == 1)
        ans = ans * d * k;
    else
        ans = ans * ((int)Math.pow(d, k) - 1 /
                                (d - 1));
         
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int[] arr = { 2, 4, 8, 16, 32,
                64, 128, 256 };
    int queries = 3;
    int[][] q = { { 2, 4 }, { 2, 6 }, { 5, 8 } };
     
    int n = arr.length;
     
    for(int i = 0; i < queries; i++)
        System.out.println(findSum(arr, n, q[i][0],
                                        q[i][1]));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to
# find the sum of elements
# of an GP 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 + 1
 
    # Find the common difference
    d = arr[1] // arr[0]
 
    # Find the sum
    ans = arr[left - 1]
    if d == 1:
        ans = ans * d * k
    else:
        ans = ans * (d ** k - 1) // (d -1)
    return ans
 
# Driver code
if __name__ == '__main__':
    arr = [ 2, 4, 8, 16, 32, 64, 128, 256 ]
    queries = 3
    q = [[ 2, 4 ], [ 2, 6 ], [ 5, 8 ]]
    n = len(arr)
 
    for i in range(queries):
        print(findSum(arr, n, q[i][0], q[i][1]))

C#

// C# program to find the sum
// of elements of an GP 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 + 1;
 
    // Find the common difference
    int d = arr[1] / arr[0];
 
    // Find the sum
    int ans = arr[left - 1];
     
    if (d == 1)
        ans = ans * d * k;
    else
        ans = ans * ((int)Math.Pow(d, k) - 1 /
                                      (d - 1));
         
    return ans;
}
 
// Driver Code
public static void Main(string []args)
{
    int[] arr = { 2, 4, 8, 16, 32,
                  64, 128, 256 };
                   
    int queries = 3;
    int[,] q = { { 2, 4 }, { 2, 6 }, { 5, 8 } };
     
    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 rutvik_56

Javascript

// JavaScript program to find the sum
// of elements of an GP in the
// given range
 
// Function to find sum in the given range
function findSum(arr, n, left, right) {
 
    // Find the value of k
    let k = right - left + 1;
 
    // Find the common difference
    let d = arr[1] / arr[0];
 
    // Find the sum
    let ans = arr[left - 1];
 
    if (d == 1)
        ans = ans * d * k;
    else
        ans = ans * (Math.pow(d, k) - 1 / (d - 1));
 
    return ans;
}
 
// Driver Code
let arr = [2, 4, 8, 16, 32,
    64, 128, 256];
 
let queries = 3;
 
let q = [[2, 4], [2, 6],
[5, 8]];
 
let n = arr.length;
 
for (let i = 0; i < queries; i++)
    document.write(findSum(arr, n, q[i][0], q[i][1]));
 
// This code is contributed by blalverma92
 

输出：

28
124
480

• 时间复杂度： O(Q)
• 空间复杂度： O(1)