Python3 程序查找范围和查询，以按 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.

编程需要懂一点英语

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：

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

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！