Java程序在K循环移位后将给定数组分成两半后使用按位或查找数组和

给定一个长度为N的数组 A[] ，其中 N 是偶数，任务是回答Q个独立查询，其中每个查询由一个正整数K组成，表示对数组执行的循环移位的次数，并找到元素的总和通过对分割后的数组执行按位或运算。
注意：每个查询都从原始数组开始。
例子：

Input: A[] = {12, 23, 4, 21, 22, 76}, Q = 1, K = 2
Output: 117
Explanation:
Since K is 2, modified array A[]={22, 76, 12, 23, 4, 21}.
Bitwise OR of first half of array = (22 | 76 | 12) = 94
Bitwise OR of second half of array = (21 | 23 | 4) = 23
Sum of OR values is 94 + 23 = 117
Input: A[] = {7, 44, 19, 86, 65, 39, 75, 101}, Q = 1, K = 4
Output: 238
Since K is 4, modified array A[]={65, 39, 75, 101, 7, 44, 19, 86}.
Bitwise OR of first half of array = 111
Bitwise OR of second half of array = 127
Sum of OR values is 111 + 127 = 238

编程需要懂一点英语

天真的方法：
要解决上面提到的问题，最简单的方法是将数组的每个元素移动K % (N / 2) ，然后遍历数组来计算每个查询的两半的 OR。但是这种方法效率不高，因此可以进一步优化。
有效的方法：
为了优化上述方法，我们可以借助 Segment Tree 数据结构。

Observation:

We can observe that after exactly N / 2 right circular shifts the two halves of the array become the same as in the original array. This effectively reduces the number of rotations to K % (N / 2).
Performing a right circular shift is basically shifting the last element of the array to the front. So for any positive integer X performing X right circular shifts is equal to shifting the last X elements of the array to the front.

编程需要懂一点英语

以下是解决问题的步骤：

为原始数组A[]构造一个段树并分配一个变量，比如说i = K % (N / 2) 。
然后对于每个查询，我们使用查找按位或的段树；即从末尾开始的 i 个元素的按位或或第一个(N / 2) – i – 1 个元素的按位或。
然后计算[(N / 2) – i, N – i – 1]范围内元素的按位或。
将这两个结果相加以获得第 i 个查询的答案。

下面是上述方法的实现：

Java

// Java program to find Bitwise OR of two
// equal halves of an array after performing
// K right circular shifts
import java.util.*;
 
class GFG{
     
static int MAX = 100005;
 
// Array for storing
// the segment tree
static int []seg = new int[4 * MAX];
 
// Function to build the segment tree
static void build(int node, int l,
                  int r, int a[])
{
    if (l == r)
        seg[node] = a[l];
 
    else
    {
        int mid = (l + r) / 2;
 
        build(2 * node, l, mid, a);
        build(2 * node + 1, mid + 1, r, a);
 
        seg[node] = (seg[2 * node] |
                     seg[2 * node + 1]);
    }
}
 
// Function to return the OR
// of elements in the range [l, r]
static int query(int node, int l, int r,
                 int start, int end, int a[])
{
     
    // Check for out of bound condition
    if (l > end || r < start)
        return 0;
 
    if (start <= l && r <= end)
        return seg[node];
 
    // Find middle of the range
    int mid = (l + r) / 2;
 
    // Recurse for all the elements in array
    return ((query(2 * node, l, mid,
                   start, end, a)) |
            (query(2 * node + 1, mid + 1,
                   r, start, end, a)));
}
 
// Function to find the OR sum
static void orsum(int a[], int n,
                  int q, int k[])
{
     
    // Function to build the segment Tree
    build(1, 0, n - 1, a);
 
    // Loop to handle q queries
    for(int j = 0; j < q; j++)
    {
         
        // Effective number of
        // right circular shifts
        int i = k[j] % (n / 2);
 
        // Calculating the OR of
        // the two halves of the
        // array from the segment tree
 
        // OR of second half of the
        // array [n/2-i, n-1-i]
        int sec = query(1, 0, n - 1,
                        n / 2 - i,
                        n - i - 1, a);
 
        // OR of first half of the array
        // [n-i, n-1]OR[0, n/2-1-i]
        int first = (query(1, 0, n - 1, 0,
                           n / 2 - 1 - i, a) |
                     query(1, 0, n - 1,
                           n - i, n - 1, a));
 
        int temp = sec + first;
 
        // Print final answer to the query
        System.out.print(temp + "
");
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    int a[] = { 7, 44, 19, 86, 65, 39, 75, 101 };
    int n = a.length;
    int q = 2;
 
    int k[] = { 4, 2 };
 
    orsum(a, n, q, k);
}
}
 
// This code is contributed by 29AjayKumar

输出：

238
230

时间复杂度： O(N + Q*logN)

辅助空间： O(4*MAX)

有关详细信息，请参阅完整的文章在 K 循环移位后将给定数组分成两半后使用按位或查找数组和！