hdu 5269 ZYB loves Xor I(字典树)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5269

思路分析：当lowbit(AxorB)=2^p 时，表示A与B的二进制表示的0-p-1位相等，第p位不同；考虑维护一棵字母树，将所有数字

转换为二进制形式并且从第0位开始插入树中，并在每个节点中记录通过该结点的数字数目；最后统计答案，对于每一个数字，

对于在其路径中的每一个结点X，假设其为第K层，统计通过与该结点不同的值的结点的数目count，则结果增加count*2^k;

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int MAX_N =  *  + ;
int num[MAX_N];
struct Node{
    Node *child[];
    int count;
    Node(){
        count = ;
        memset(child, , sizeof(child));
    }
    Node(int value){
        count = value;
        memset(child, , sizeof(child));
    }
};

void Insert(Node *head, int value){
    Node *pre = head, *next = NULL;
    for (int i = ; i < ; ++ i){
        if (pre->child[value & ] == NULL){
            pre->child[value & ] = new Node();
            pre = pre->child[value & ];
        }else{
            next = pre->child[value & ];
            next->count++;
            pre = next;
        }
        value >>= ;
    }
}

void MakeEmpty(Node *node){
    node->count = ;
    if (node->child[])
        MakeEmpty(node->child[]);
    if (node->child[])
        MakeEmpty(node->child[]);
}

long long Query(Node *head, int value){
    Node *pre = head, *next = NULL, *other;
    long long ret = ;

    for (int i = ; i < ; ++ i){
        next = pre->child[value & ];
        other = pre->child[(value & ) ^ ];
        if (other)
            ret = (ret + other->count * ( << i)) % ;
        pre = next;
        value >>= ;
    }
    return ret;
}

int main(){
    int case_times, n;
    int case_id = ;

    scanf("%d", &case_times);
    while (case_times--){
        Node *head = new Node();

        scanf("%d", &n);
        for (int i = ; i < n; ++i){
            scanf("%d", &num[i]);
            Insert(head, num[i]);
        }

        long long ans = ;
        for (int i = ; i < n; ++i)
            ans = (ans + Query(head, num[i])) % ;
        MakeEmpty(head);
        printf("Case #%d: %I64d\n", ++case_id, ans);
    }
    return ;
}