HDU4825 - Xor Sum（Trie 字典树）

题目链接 HDU4825

【分析】

用字典树储0 1的存每个数字的每一位，然后根据异或的原理，先把需要查询的数字取反，则从高位到低位和这个取反之后的数字最近接近的数就是答案；

【AC CODE】187ms

#include 
#include 
#define MAXN 100005
typedef unsigned int LL;
int tree[MAXN*16][2], sum,ch;
LL ans;

LL in()
{
    while((ch = getchar()) < '0' || '9' < ch);
    ans = ch-'0';
    while((ch = getchar()) >= '0' && '9' >= ch) ans = ans*10+ch-'0';
    return ans;
}
void add(LL num)
{
    int r = 0,b;
    for(int i = 31; i >= 0; i--)
    {
        b = (num>>i)&1;
        if(~tree[r][b])
            r = tree[r][b];
        else
            r = tree[r][b] = ++sum;
    }
}
LL find(LL num)
{
    int r = 0,b;
    ans = 0;
    for(int i = 31; i >= 0; i--)
    {
        b = (num>>i)&1;
        if(!(~tree[r][b]))
            b = !b;
        r = tree[r][b];
        ans = (ans<<1)+b;
    }
    return ans;
}
int main()
{
#ifdef SHY
    freopen("e:\\1.txt", "r", stdin);
#endif
    int t = in(), count = 0;
    while(t--)
    {
        int n = in(),m = in();
        memset(tree,-1,sizeof(tree));
        sum = 0;
        for(int i = 0; i < n; i++)
            add(in());
        printf("Case #%d:\n", ++count);
        for(int i = 0; i < m; i++)
            printf("%u\n", find(~in()));
    }
    return 0;
}