数组中两两异或大于m的个数_求n个数两两组合大于m的数有几个-优快云博客

本文介绍了一种使用前缀树（Trie树）实现的高效算法，该算法用于解决给定两个整数数组A和整数m的情况下，计算数组A中所有两两异或结果大于m的数量的问题。通过构建Trie树并进行查询，可以有效地统计出满足条件的异或结果数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

给定整数m以及n各数字A1,A2,..An，将数列A中所有元素两两异或，共能得到n(n-1)/2个结果，请求出这些结果中大于m的有多少个

1. aDigit=1， mDigit=1时，字典中第k位为0，异或结果为1，需要继续搜索更低位，第k位为1，异或结果为0，小于mDigit，不用理会；

2. aDigit=0， mDigit=1时，字典中第k位为1，异或结果为1，需要继续搜索更低位，第k位为0，异或结果为0，小于mDigit，不用理会；

3. aDigit=1， mDigit=0时，字典中第k位为0，异或结果为1，与对应分支所有数异或，结果都会大于m，第k位为1，异或结果为0，递归获得结果；

4. aDigit=0， mDigit=0时，字典中第k位为1，异或结果为1，与对应分支所有数异或，结果都会大于m，第k位为0，异或结果为0，递归获得结果；

public class Main 
{
    private static class TrieTree 
    {
        TrieTree[] next = new TrieTree[2];
        int count = 1;
    }
    private static long solve(int[] a, int m) 
    {
        TrieTree trieTree = buildTrieTree(a);
        long result = 0;
        for (int i = 0; i < a.length; i++) 
        {
            result += queryTrieTree(trieTree, a[i], m, 31);
        }
        return result / 2;
    }
    private static long queryTrieTree(TrieTree trieTree, int a, int m, int index) 
    {
        if(trieTree == null)
            return 0;
        TrieTree current = trieTree;
        for (int i = index; i >= 0; i--) 
        {
            int aDigit = (a >> i) & 1;
            int mDigit = (m >> i) & 1;
            if(aDigit == 1 && mDigit == 1) 
            {
                if(current.next[0] == null)
                    return 0;
                current = current.next[0];
            } 
            else if (aDigit == 0 && mDigit == 1) 
            {
                if(current.next[1] == null)
                    return 0;
                current = current.next[1];
            } 
            else if (aDigit == 1 && mDigit == 0) 
            {
                long p = queryTrieTree(current.next[1], a, m, i - 1);
                long q = current.next[0] == null ? 0 : current.next[0].count;
                return p + q;
            } 
            else if (aDigit == 0 && mDigit == 0) 
            {
                long p = queryTrieTree(current.next[0], a, m, i - 1);
                long q = current.next[1] == null ? 0 : current.next[1].count;
                return p + q;
            }
        }
        return 0;
    }
    private static TrieTree buildTrieTree(int[] a) 
    {
        TrieTree trieTree = new TrieTree();
        for (int i = 0; i < a.length; i++) 
        {
            TrieTree current = trieTree;
            for (int j = 31; j >= 0; j--) 
            {
                int digit = (a[i] >> j) & 1;
                if(current.next[digit] == null) 
                {
                    current.next[digit] = new TrieTree();
                } 
                else 
                {
                    current.next[digit].count ++;
                }
                current = current.next[digit];
            }
        }
        return trieTree;
    }
}