用于范围和查询的 Javascript 程序，用于按 K 索引对 Array 进行逆时针旋转

给定一个包含N个元素的数组arr和以下两种类型的Q个查询：

• 1 K ：对于这种类型的查询，数组需要从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：

• 创建一个大小为arr两倍的前缀数组，并将arr的第i^个索引处的元素复制到[0, N)中所有 i 的第i^个和第N + i^个前缀索引。
• 为该数组的每个索引预先计算前缀和并存储在前缀中。
• 将指针设置为从0开始以表示初始数组的起始索引。
• 对于类型 1 的查询，将start移至

((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或打印的值
  

prefix[start + R]

• 除此以外。

下面的代码是上述方法的实现：

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 pointer 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 pointing 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O(1) 。
辅助空间复杂度： O(N)

有关更多详细信息，请参阅关于数组逆时针旋转 K 索引的范围和查询的完整文章！