用c++编译下面内容Input The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array. The second line contains the space-separated integers from the array. All numbers are non-negative integers strictly less than 230. Output Print a single integer — the required maximal xor of a segment of consecutive elements.

以下是C++代码实现：

#include <iostream>
#include <vector>
using namespace std;

struct TrieNode {
    TrieNode *left;
    TrieNode *right;
};

void insert(TrieNode *root, int num) {
    TrieNode *curr = root;

    for (int i = 31; i >= 0; i--) {
        int bit = (num >> i) & 1;

        if (bit == 0) {
            if (!curr->left) curr->left = new TrieNode();
            curr = curr->left;
        }
        else {
            if (!curr->right) curr->right = new TrieNode();
            curr = curr->right;
        }
    }
}

int findMaxXor(TrieNode *root, vector<int> nums) {
    int maxXor = 0;

    for (int num : nums) {
        int currXor = 0;
        TrieNode *curr = root;

        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;

            if (bit == 0) {
                if (curr->right) {
                    currXor += (1 << i);
                    curr = curr->right;
                }
                else {
                    curr = curr->left;
                }
            }
            else {
                if (curr->left) {
                    currXor += (1 << i);
                    curr = curr->left;
                }
                else {
                    curr = curr->right;
                }
            }
        }

        maxXor = max(maxXor, currXor);
    }

    return maxXor;
}

int main() {
    int n;
    cin >> n;

    vector<int> nums(n);
    for (int i = 0; i < n; i++) {
        cin >> nums[i];
    }

    TrieNode *root = new TrieNode();
    insert(root, nums[0]);
    int maxXor = nums[0];

    for (int i = 1; i < n; i++) {
        insert(root, nums[i]);
        maxXor = max(maxXor, findMaxXor(root, nums));
    }

    cout << maxXor << endl;

    return 0;
}

该算法的时间复杂度为 O(nlogn)。