817E. Choosing The Commander trie字典树

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题意：现有3种操作 加入一个值，删除一个值，询问pi^x<k的个数

思路：很像以前lightoj上写过的01异或的字典树，用字典树维护数求异或值即可

/** @Date    : 2017-07-02 18:58:02
  * @FileName: 817E trie.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e6+20;
const double eps = 1e-8;

int a[N*4][3];
int sum[N*4];
int cnt = 0;

void ope(int x, int f)
{
	int bit[32];
	MMF(bit);
	int len = 0;
	while(x)
	{
		bit[len++] = x % 2;
		x >>= 1;
	}
	int np = 0;
	for(int i = 31; i >= 0; i--)
	{
		if(!a[np][bit[i]])
			a[np][bit[i]] = ++cnt;
		np = a[np][bit[i]];
		sum[np] += f;
	}
}

int query(int x, int k)
{
	int bit[32], tmp[32];
	MMF(bit);
	MMF(tmp);
	int len = 0;
	while(x)
	{
		bit[len++] = x % 2;
		x >>= 1;
	}
	len = 0;
	while(k)
	{
		tmp[len++] = k % 2;
		k >>= 1;
	}

	int np = 0;
	int res = 0;
	for(int i = 31; i >= 0; i--)
	{
		if(!tmp[i] && !a[np][bit[i]])
			break;
		if(tmp[i]==0)
			np = a[np][bit[i]];
		else
		{
			if(a[np][bit[i]])
				res += sum[a[np][bit[i]]];
			if(a[np][!bit[i]] == 0)
				break;
			np = a[np][!bit[i]];
		}
	}
	return res;
}

int main()
{
	int q;
	while(cin >> q)
	{
		MMF(sum);
		while(q--)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			if(x == 1)
				ope(y, 1);
			else if(x == 2)
				ope(y, -1);
			else if(x == 3)
			{
				int k;
				scanf("%d", &k);
				printf("%d\n", query(y, k));
			}
		}
	}
    return 0;
}
//要求有异或操作 为了快速运算 使用trie字典树保存每个人对应二进制位上的数字
//并存入sum[]中
//注意空间1e5*32位以上