[BZOJ 4103] [Thu Summer Camp 2015] 异或运算 【可持久化Trie】

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int MaxN = 1000 + 5, MaxM = 300000 + 5, MaxNode = 10000000 + 5;

int n, m, p, Index, Ans;
int A[MaxN], B[MaxM], Root[MaxM], Son[MaxNode][2], T[MaxNode], F[MaxN], Q[MaxN];

bool OK[MaxN];

void Build(int &x, int y, int Num, int Bit)
{
	if (!x) x = ++Index;
	if (Bit == 0)
	{
		T[x] = T[y] + 1;
		return;
	}
	if (Num & (1 << (Bit - 1)))
	{
		Son[x][0] = Son[y][0];
		Build(Son[x][1], Son[y][1], Num, Bit - 1);
	}
	else
	{
		Son[x][1] = Son[y][1];
		Build(Son[x][0], Son[y][0], Num, Bit - 1);
	}
	T[x] = T[Son[x][0]] + T[Son[x][1]];
}

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) scanf("%d", &A[i]);
	for (int i = 1; i <= m; ++i) 
	{
		scanf("%d", &B[i]);
		Build(Root[i], Root[i - 1], B[i], 31);
	}
	scanf("%d", &p);
	int u, d, l, r, kk;
	for (int i = 1; i <= p; ++i)
	{
		scanf("%d%d%d%d%d", &u, &d, &l, &r, &kk);
		kk = (d - u + 1) * (r - l + 1) - kk + 1; 
		Ans = 0;
		for (int j = u; j <= d; ++j)
		{
			F[j] = Root[r];
			Q[j] = Root[l - 1];
		}
		for (int j = 30; j >= 0; --j)
		{
			int x, Temp = 0;
			for (int k = u; k <= d; ++k)
			{
				x = (A[k] & (1 << j));
				if (x) Temp += T[Son[F[k]][1]] - T[Son[Q[k]][1]];
				else Temp += T[Son[F[k]][0]] - T[Son[Q[k]][0]];
			}
			if (Temp >= kk)
			{
				for (int k = u; k <= d; ++k)
				{
					x = (A[k] & (1 << j));
					if (x) 
					{
						F[k] = Son[F[k]][1];
						Q[k] = Son[Q[k]][1];
					}
					else
					{
						F[k] = Son[F[k]][0];
						Q[k] = Son[Q[k]][0];
					}
				}
			}
			else
			{
				kk -= Temp;
				Ans |= (1 << j);				
				for (int k = u; k <= d; ++k)
				{
					x = (A[k] & (1 << j));
					if (x) 
					{
						F[k] = Son[F[k]][0];
						Q[k] = Son[Q[k]][0];
					}
					else
					{
						F[k] = Son[F[k]][1];
						Q[k] = Son[Q[k]][1];
					}
				}
			}
		}
		printf("%d\n", Ans);
	}
	return 0;
}