Javascript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 
 

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.

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：

Javascript

// javascript program to find Bitwise OR of two
// equal halves of an array after performing
// K right circular shifts
    const MAX = 100005;
 
    // Array for storing
    // the segment tree
    var seg = Array(4 * MAX).fill(0);
 
    // Function to build the segment tree
    function build(node , l , r , a) {
        if (l == r)
            seg[node] = a[l];
 
        else {
            var mid = parseInt((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]
    function query(node , l , r , start , end , 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
        var mid = parseInt((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
    function orsum(a , n , q , k) {
 
        // Function to build the segment Tree
        build(1, 0, n - 1, a);
 
        // Loop to handle q queries
        for (j = 0; j < q; j++) {
 
            // Effective number of
            // right circular shifts
            var 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]
            var 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]
            var first = (query(1, 0, n - 1, 0, n / 2 - 1 - i, a) | query(1, 0, n - 1, n - i, n - 1, a));
 
            var temp = sec + first;
 
            // Print final answer to the query
            document.write(temp + "<br/>");
        }
    }
 
    // Driver Code
        var a = [ 7, 44, 19, 86, 65, 39, 75, 101 ];
        var n = a.length;
        var q = 2;
 
        var k = [ 4, 2 ];
 
        orsum(a, n, q, k);
 
// This code is contributed by Rajput-Ji.

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！