\(\mathcal{Description}\)

  Link.

  給定序列 \(\{a_n\}\)，\(q\) 組詢問，給定 \(a<b<c<d\)，求 \(l\le[a,b],r\le[c,d]\) 的子序列 \([l,r]\) 的中位數最大值。若長度為偶數，中位數取中間兩數較大的一個。強制線上。

  \(n\le2\times10^4\)，\(q\le2.5\times10^4\)。

\(\mathcal{Solution}\)

crashed：眾所周知，中位數是可以二分的。

  考慮單組詢問，二分中位數 \(mid\)，把序列中大於等於 \(mid\) 的值設為 \(1\)

  多組詢問，發現對於每個 \(mid\)，序列的長相都不盡相同。但序列的變化是極其有限的——當 \(mid\) 遞增，每個位置只會又 \(1\) 變為 \(0\) 一次。

  由此可以想到主席樹。離散化之後，預處理 \(mid=1,2,\dots,n\) 時的序列，建成主席樹。樹上維護區間和，區間最大字首，區間最大字尾。處理詢問時仍二分 \(mid\)，利用以 \(mid\) 為根的這棵權值線段樹的資訊，求出區間最大和。顯然最大和為 \([a,b)\text{最大字尾}+[b,c]\text{之和}+(c,d]\text{最大字首}\)

  複雜度 \(\mathcal O(n\log^2n)\)。

\(\mathcal{Code}\)

#include <cstdio>
#include <vector>
#include <algorithm>

typedef std::pair<int, int> pii;

const int MAXN = 20000;
int n, a[MAXN + 5], tval[MAXN + 5], root[MAXN + 5];
std::vector<int> apr[MAXN + 5];

inline int rint () {
	int x = 0; char s = getchar ();
	for ( ; s < '0' || '9' < s; s = getchar () );
	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
	return x;
}

inline int max_ ( const int a, const int b ) { return a < b ? b : a; }

struct PersistentSegmentTree {
	static const int MAXND = MAXN * 40;
	int cntnd, ch[MAXND + 5][2], sum[MAXND + 5], lmx[MAXND + 5], rmx[MAXND + 5];

	inline void pushup ( const int rt ) {
		sum[rt] = sum[ch[rt][0]] + sum[ch[rt][1]];
		lmx[rt] = max_ ( lmx[ch[rt][0]], sum[ch[rt][0]] + lmx[ch[rt][1]] );
		rmx[rt] = max_ ( rmx[ch[rt][1]], sum[ch[rt][1]] + rmx[ch[rt][0]] );
	}

	inline void build ( int& rt, const int l, const int r ) {
		rt = ++ cntnd;
		if ( l == r ) return sum[rt] = lmx[rt] = rmx[rt] = 1, void ();
		int mid = l + r >> 1;
		build ( ch[rt][0], l, mid ), build ( ch[rt][1], mid + 1, r );
		pushup ( rt );
	}

	inline void update ( int& rt, const int l, const int r, const int x ) {
		int old = rt; rt = ++ cntnd;
		ch[rt][0] = ch[old][0], ch[rt][1] = ch[old][1];
		sum[rt] = sum[old], lmx[rt] = lmx[old], rmx[rt] = rmx[old];
		if ( l == r ) return sum[rt] = -1, lmx[rt] = rmx[rt] = 0, void ();
		int mid = l + r >> 1;
		if ( x <= mid ) update ( ch[rt][0], l, mid, x );
		else update ( ch[rt][1], mid + 1, r, x );
		pushup ( rt );
	}

	inline int qrySum ( const int rt, const int l, const int r, const int ql, const int qr ) {
		if ( ql <= l && r <= qr ) return sum[rt];
		int mid = l + r >> 1, ret = 0;
		if ( ql <= mid ) ret += qrySum ( ch[rt][0], l, mid, ql, qr );
		if ( mid < qr ) ret += qrySum ( ch[rt][1], mid + 1, r, ql, qr );
		return ret;
	}

	inline pii qryLmx ( const int rt, const int l, const int r, const int ql, const int qr ) {
		if ( ql <= l && r <= qr ) return { lmx[rt], sum[rt] };
		int mid = l + r >> 1, ret = 0, s = 0; pii tmp;
		if ( ql <= mid ) {
			tmp = qryLmx ( ch[rt][0], l, mid, ql, qr );
			ret = tmp.first, s = tmp.second;
		}
		if ( mid < qr ) {
			tmp = qryLmx ( ch[rt][1], mid + 1, r, ql, qr );
			ret = max_ ( ret, s + tmp.first ), s += tmp.second;
		}
		return { ret, s };
	}

	inline pii qryRmx ( const int rt, const int l, const int r, const int ql, const int qr ) {
		if ( ql <= l && r <= qr ) return { rmx[rt], sum[rt] };
		int mid = l + r >> 1, ret = 0, s = 0; pii tmp;
		if ( mid < qr ) {
			tmp = qryRmx ( ch[rt][1], mid + 1, r, ql, qr );
			ret = tmp.first, s = tmp.second;
		}
		if ( ql <= mid ) {
			tmp = qryRmx ( ch[rt][0], l, mid, ql, qr );
			ret = max_ ( ret, s + tmp.first ), s += tmp.second;
		}
		return { ret, s };
	}
} pst;

int main () {
	n = rint ();
	for ( int i = 1; i <= n; ++ i ) a[i] = tval[i] = rint ();
	std::sort ( tval + 1, tval + n + 1 );
	int lim = std::unique ( tval + 1, tval + n + 1 ) - tval - 1;
	for ( int i = 1; i <= n; ++ i ) {
		a[i] = std::lower_bound ( tval + 1, tval + lim + 1, a[i] ) - tval;
		apr[a[i]].push_back ( i );
	}
	pst.build ( root[1], 1, n ); // 注意這裡n的意義成為了序列長度，不要與lim搞混。
	for ( int i = 2; i <= lim; ++ i ) {
		root[i] = root[i - 1];
		for ( int p: apr[i - 1] ) pst.update ( root[i], 1, n, p );
	}
	for ( int q = rint (), ans = 0, tmp[4]; q --; ) {
		for ( int i = 0; i < 4; ++ i ) tmp[i] = ( rint () + ans ) % n + 1;
		std::sort ( tmp, tmp + 4 );
		int l = 1, r = lim;
		while ( l <= r ) {
			int mid = l + r >> 1;
			int cnt = pst.qrySum ( root[mid], 1, n, tmp[1], tmp[2] )
					+ pst.qryLmx ( root[mid], 1, n, tmp[2] + 1, tmp[3] ).first
					+ pst.qryRmx ( root[mid], 1, n, tmp[0], tmp[1] - 1 ).first;
			if ( cnt >= 0 ) l = ( ans = mid ) + 1;
			else r = mid - 1;
		}
		printf ( "%d\n", ans = tval[ans] );
	}
	return 0;
}