范围和查询以K索引逆时针旋转Array

给定一个数组arr ，该数组arr由N个元素和以下两种类型的Q个查询组成：

• 1 K ：对于这种类型的查询，阵列需要从其当前状态逆时针旋转K个索引。
• 2 LR ：对于此查询，需要计算索引[L，R]中存在的数组元素的总和。

例子：

Input: arr = { 1, 2, 3, 4, 5, 6 }, query = { {2, 1, 3}, {1, 3}, {2, 0, 3}, {1, 4}, {2, 3, 5} } 
Output: 
9 
16 
12 
Explanation: 
For the 1st query {2, 1, 3} -> Sum of the elements in the indices [1, 3] = 2 + 3 + 4 = 9. 
For the 2nd query {1, 3} -> Modified array after anti-clockwise rotation by 3 places is { 4, 5, 6, 1, 2, 3 } 
For the 3rd query {2, 0, 3} -> Sum of the elements in the indices [0, 3] = 4 + 5 + 6 + 1 = 16. 
For the 4th query {1, 4} -> Modified array after anti-clockwise rotation by 4 places is { 2, 3, 4, 5, 6, 1 } 
For the 5th query {2, 3, 5} -> Sum of the elements in the indices [3, 5] = 5 + 6 + 1 = 12. 
 

方法：

• 创建前缀阵列是ARR的两倍大小和ARR的^个索引复制元件在i至i^个和^第N + I前缀^个索引对于所有的i在[0，N)。
• 为该数组的每个索引预先计算前缀和，并将其存储在prefix中。
• 将指针起始位置设置为0，以表示初始数组的起始索引。
• 对于类型1的查询时，移位开始

((start + K) % N)th position

• 对于类型2的查询，计算

prefix[start + R]
 - prefix[start + L- 1 ]

• 如果start + L> = 1或打印

prefix[start + R]

• 除此以外。

下面的代码是上述方法的实现：

C++

// C++ Program to calculate range sum
// queries for anticlockwise
// rotations of array by K
 
#include 
using namespace std;
 
// Function to execute the queries
void rotatedSumQuery(
    int arr[], int n,
    vector >& query,
    int Q)
{
    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    int prefix[2 * n];
 
    // Copy elements to the new array
    for (int i = 0; i < n; i++) {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }
 
    // Calculate the prefix sum
    // for every index
    for (int i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];
 
    // Set start pointer as 0
    int start = 0;
 
    for (int q = 0; q < Q; q++) {
 
        // Query to perform
        // anticlockwise rotation
        if (query[q][0] == 1) {
            int k = query[q][1];
            start = (start + k) % n;
        }
 
        // Query to answer range sum
        else if (query[q][0] == 2) {
 
            int L, R;
            L = query[q][1];
            R = query[q][2];
 
            // If pointing to 1st index
            if (start + L == 0)
 
                // Display the sum upto start + R
                cout << prefix[start + R] << endl;
 
            else
 
                // Subtract sum upto start + L - 1
                // from sum upto start + R
                cout << prefix[start + R]
                            - prefix[start + L - 1]
                     << endl;
        }
    }
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5, 6 };
 
    // Number of query
    int Q = 5;
 
    // Store all the queries
    vector > query
        = { { 2, 1, 3 },
            { 1, 3 },
            { 2, 0, 3 },
            { 1, 4 },
            { 2, 3, 5 } };
 
    int n = sizeof(arr) / sizeof(arr[0]);
    rotatedSumQuery(arr, n, query, Q);
 
    return 0;
}

Java

// Java program to calculate range sum
// queries for anticlockwise
// rotations of array by K
class GFG{
 
// Function to execute the queries
static void rotatedSumQuery(int arr[], int n,
                            int [][]query, int Q)
{
     
    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    int []prefix = new int[2 * n];
 
    // Copy elements to the new array
    for(int i = 0; i < n; i++)
    {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }
 
    // Calculate the prefix sum
    // for every index
    for(int i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];
 
    // Set start pointer as 0
    int start = 0;
 
    for(int q = 0; q < Q; q++)
    {
 
        // Query to perform
        // anticlockwise rotation
        if (query[q][0] == 1)
        {
            int k = query[q][1];
            start = (start + k) % n;
        }
 
        // Query to answer range sum
        else if (query[q][0] == 2)
        {
            int L, R;
            L = query[q][1];
            R = query[q][2];
 
            // If pointing to 1st index
            if (start + L == 0)
 
                // Display the sum upto start + R
                System.out.print(prefix[start + R] + "\n");
 
            else
 
                // Subtract sum upto start + L - 1
                // from sum upto start + R
                System.out.print(prefix[start + R] -
                                 prefix[start + L - 1] +
                                 "\n");
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
 
    // Number of query
    int Q = 5;
 
    // Store all the queries
    int [][]query = { { 2, 1, 3 },
                      { 1, 3 },
                      { 2, 0, 3 },
                      { 1, 4 },
                      { 2, 3, 5 } };
 
    int n = arr.length;
    rotatedSumQuery(arr, n, query, Q);
}
}
 
// This code is contributed by Rohit_ranjan

Python3

# Python3 program to calculate range sum
# queries for anticlockwise
# rotations of the array by K
 
# Function to execute the queries
def rotatedSumQuery(arr, n, query, Q):
 
    # Construct a new array
    # of size 2*N to store
    # prefix sum of every index
    prefix = [0] * (2 * n)
 
    # Copy elements to the new array
    for i in range(n):
        prefix[i] = arr[i]
        prefix[i + n] = arr[i]
 
    # Calculate the prefix sum
    # for every index
    for i in range(1, 2 * n):
        prefix[i] += prefix[i - 1];
 
    # Set start pointer as 0
    start = 0;
 
    for q in range(Q):
 
        # Query to perform
        # anticlockwise rotation
        if (query[q][0] == 1):
            k = query[q][1]
            start = (start + k) % n;
 
        # Query to answer range sum
        elif (query[q][0] == 2):
            L = query[q][1]
            R = query[q][2]
 
            # If pointing to 1st index
            if (start + L == 0):
 
                # Display the sum upto start + R
                print(prefix[start + R])
 
            else:
 
                # Subtract sum upto start + L - 1
                # from sum upto start + R
                print(prefix[start + R]-
                      prefix[start + L - 1])
         
# Driver code
arr = [ 1, 2, 3, 4, 5, 6 ];
 
# Number of query
Q = 5
 
# Store all the queries
query= [ [ 2, 1, 3 ],
         [ 1, 3 ],
         [ 2, 0, 3 ],
         [ 1, 4 ],
         [ 2, 3, 5 ] ]
 
n = len(arr);
rotatedSumQuery(arr, n, query, Q);
 
# This code is contributed by ankitkumar34

C#

// C# program to calculate range sum
// queries for anticlockwise
// rotations of array by K
using System;
 
class GFG{
 
// Function to execute the queries
static void rotatedSumQuery(int[] arr, int n,
                            int[,] query, int Q)
{
     
    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    int[] prefix = new int[2 * n];
 
    // Copy elements to the new array
    for(int i = 0; i < n; i++)
    {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }
 
    // Calculate the prefix sum
    // for every index
    for(int i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];
 
    // Set start pointer as 0
    int start = 0;
 
    for(int q = 0; q < Q; q++)
    {
 
        // Query to perform
        // anticlockwise rotation
        if (query[q, 0] == 1)
        {
            int k = query[q, 1];
            start = (start + k) % n;
        }
 
        // Query to answer range sum
        else if (query[q, 0] == 2)
        {
            int L, R;
            L = query[q, 1];
            R = query[q, 2];
 
            // If pointing to 1st index
            if (start + L == 0)
 
                // Display the sum upto start + R
                Console.Write(prefix[start + R] + "\n");
 
            else
 
                // Subtract sum upto start + L - 1
                // from sum upto start + R
                Console.Write(prefix[start + R] -
                              prefix[start + L - 1] +
                              "\n");
        }
    }
}
 
// Driver code
public static void Main()
{
    int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };
 
    // Number of query
    int Q = 5;
 
    // Store all the queries
    int[,] query = new int[,] { { 2, 1, 3 },
                                { 1, 3, 0 },
                                { 2, 0, 3 },
                                { 1, 4, 0 },
                                { 2, 3, 5 } };
 
    int n = arr.Length;
    rotatedSumQuery(arr, n, query, Q);
}
}
 
// This code is contributed by sanjoy_62

Javascript

     
// Javascript program to calculate range sum
// queries for anticlockwise
// rotations of array by K
 
// Function to execute the queries
function rotatedSumQuery(arr, n,
                            query, Q)
{
       
    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    let prefix = [];
   
    // Copy elements to the new array
    for(let i = 0; i < n; i++)
    {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }
   
    // Calculate the prefix sum
    // for every index
    for(let i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];
   
    // Set start poleter as 0
    let start = 0;
   
    for(let q = 0; q < Q; q++)
    {
   
        // Query to perform
        // anticlockwise rotation
        if (query[q][0] == 1)
        {
            let k = query[q][1];
            start = (start + k) % n;
        }
   
        // Query to answer range sum
        else if (query[q][0] == 2)
        {
            let L, R;
            L = query[q][1];
            R = query[q][2];
   
            // If poleting to 1st index
            if (start + L == 0)
   
                // Display the sum upto start + R
                document.write(prefix[start + R] + "<br/>");
   
            else
   
                // Subtract sum upto start + L - 1
                // from sum upto start + R
                document.write(prefix[start + R] -
                                 prefix[start + L - 1] +
                                 "<br/>");
        }
    }
}
     
// Driver code
 
    let arr = [ 1, 2, 3, 4, 5, 6 ];
   
    // Number of query
    let Q = 5;
   
    // Store all the queries
    let query = [[ 2, 1, 3 ],
                      [ 1, 3 ],
                      [ 2, 0, 3 ],
                      [ 1, 4 ],
                      [ 2, 3, 5 ]];
   
    let n = arr.length;
    rotatedSumQuery(arr, n, query, Q);
     
    // This code is contributed by susmitakundugoaldanga.

输出：

9
16
12

时间复杂度：每个查询为O(1) 。
辅助空间复杂度： O(N)