【主席树】 BZOJ 2653 middle

二分答案，判断可不可行的方法如下：比当前元素小的值标记为-1，反之标记为1，然后a，b，c，d内存在区间使得和大于等于0表示可行。。。。。判断区间最大和可以用线段树区间合并。为了降低复杂度，要预先按值建立主席树。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 20005
#define maxm 1000005
#define eps 1e-7
#define mod 998244353
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
//#define ls o<<1
//#define rs o<<1 | 1
//#define lson o<<1, L, mid 
//#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

// 2653 middle

struct node
{
	int lmax, rmax, sum;
	node *ch[2];	
}pool[maxm], *tail;

node *root[maxn], *init_root, *rt;
pair<int, int> a[maxn];
int q[maxn];
int n, m;

node* newnode(int t)
{
	tail->sum = t;
	tail->ch[0] = tail->ch[1] = NULL;
	if(t == 1) tail->lmax = tail->rmax = 1;
	else tail->lmax = tail->rmax = 0;
	return tail++;
}

void pushup(node *o)
{
	o->sum = o->ch[0]->sum + o->ch[1]->sum;
	o->lmax = max(o->ch[0]->lmax, o->ch[0]->sum + o->ch[1]->lmax);
	o->rmax = max(o->ch[1]->rmax, o->ch[1]->sum + o->ch[0]->rmax);
}

node* build(int L, int R)
{
	if(L == R) return newnode(1);
	int mid = (L + R) >> 1;
	node *p = newnode(0);
	p->ch[0] = build(L, mid);
	p->ch[1] = build(mid+1, R);
	pushup(p);
	return p;
}

node* update(node *o, int L, int R, int id)
{
	if(L == R) return newnode(-1);
	int mid = (L + R) >> 1;
	node *p = newnode(0);
	p->ch[0] = o->ch[0];
	p->ch[1] = o->ch[1];
	if(id <= mid) p->ch[0] = update(o->ch[0], L, mid, id);
	else p->ch[1] = update(o->ch[1], mid+1, R, id);
	pushup(p);
	return p;
}

void init()
{
	tail = pool;
}

int query(node *o, int L, int R, int ql, int qr)
{
	if(ql <= L && qr >= R) return o->sum;
	int mid = (L + R) >> 1, ans = 0;
	if(ql <= mid) ans += query(o->ch[0], L, mid, ql, qr);
	if(qr > mid) ans += query(o->ch[1], mid+1, R, ql, qr);
	return ans;
}

int query2(node *o, int L, int R, int ql, int qr)
{
	if(ql == L && qr == R) return o->rmax;
	int mid = (L + R) >> 1;
	if(qr <= mid) return query2(o->ch[0], L, mid, ql, qr);
	else if(ql > mid) return query2(o->ch[1], mid+1, R, ql, qr);
	else return max(query2(o->ch[1], mid+1, R, mid+1, qr), query(o->ch[1], mid+1, R, mid+1, qr) + query2(o->ch[0], L, mid, ql, mid));
}

int query3(node *o, int L, int R, int ql, int qr)
{
	if(ql == L && qr == R) return o->lmax;
	int mid = (L + R) >> 1;
	if(qr <= mid) return query3(o->ch[0], L, mid, ql, qr);
	else if(ql > mid) return query3(o->ch[1], mid+1, R, ql, qr);
	else return max(query3(o->ch[0], L, mid, ql, mid), query(o->ch[0], L, mid, ql, mid) + query3(o->ch[1], mid+1, R, mid+1, qr));
}

bool check(int x)
{
	int t  = lower_bound(a, a+n, mp(x, 0)) - a;
	if(t) rt = root[t-1];
	else rt = init_root;
	
	int ans = 0;
	ans += query(rt, 0, n-1, q[1], q[2]);
	if(q[0] <= q[1]-1) ans += query2(rt, 0, n-1, q[0], q[1]-1);
	if(q[2]+1 <= q[3]) ans += query3(rt, 0, n-1, q[2]+1, q[3]);
	return ans >= 0;
}

void debug(node *o, int L, int R)
{
	printf("%d %d %d %d %d\n", L, R, o->sum, o->lmax, o->rmax);
	if(L == R) return;
	int mid = (L + R) >> 1;
	debug(o->ch[0], L, mid);
	debug(o->ch[1], mid+1, R);
}

void work()
{
	int x;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) scanf("%d", &x), a[i] = mp(x, i);
	scanf("%d", &m);
	int ans = 0;
	sort(a, a+n);
	init_root = build(0, n-1);
	for(int i = 0; i < n; i++) {
		if(i) root[i] = update(root[i-1], 0, n-1, a[i].second);
		else root[i] = update(init_root, 0, n-1, a[i].second);
	}
	int res = 0;
	while(m--) {
		for(int i = 0; i < 4; i++) scanf("%d", &q[i]);
		for(int i = 0; i < 4; i++) q[i] = (q[i] + res) % n;
		sort(q, q+4);
		int bot = a[0].first, top = a[n-1].first, mid;
		while(top >= bot) {
			mid = (top + bot) >> 1;
			if(check(mid)) bot = mid+1, res = mid;
			else top = mid-1;
		}
		printf("%d\n", res);
	}
}

int main()
{
	init();
	work();
	
	return 0;
}