主席树(BZOJ2653)

考虑二分答案，设为k，将大于等于k的元素设为1，小于的设为-1，如果某一段的和>=0，说明这段的中位数>=k.

对于每组询问，二分完后查询新序列的最大子段和即可。

但是不能开n棵线段树，观察到如果将原序列从小到大排序后，每加一个元素只会修改一个位置的值，所以用个主席树维护最大子段和即可。

#include <cstdio>
#include <algorithm>
using namespace std;
#define lx t[x].l
#define rx t[x].r
#define m ((l+r)>>1)
#define lc lx,l,m
#define rc rx,m+1,r

const int N = ;
int n,q,tt,la,l1,r1,l2,r2,c[],a[N],b[N],rt[N];
struct nd {int l,r,s,ls,rs;}t[];
bool cmp(int x, int y) {return a[x] < a[y];}

void bd(int &x, int l, int r) {
    x = ++tt, t[x].s = t[x].ls = t[x].rs = r-l+;
    if(l == r) return;
    bd(lc), bd(rc);
}

void upd(int lt, int x, int l, int r, int v) {
    if(l == r) {t[x].s = t[x].ls = t[x].rs = -; return;}
    if(v <= m) rx = t[lt].r, t[x].l = ++tt, upd(t[lt].l, lc, v);
    else lx = t[lt].l, t[x].r = ++tt, upd(t[lt].r, rc, v);
    t[x].s=t[lx].s+t[rx].s;
    t[x].ls=max(t[lx].ls,t[lx].s+t[rx].ls),t[x].rs=max(t[rx].rs,t[rx].s+t[lx].rs);
}
int qry2(int x, int l, int r, int L, int R) {
    if(l >= L && r <= R) return t[x].s;
    if(L > m) return qry2(rc, L, R);
    if(R <= m) return qry2(lc, L, R);
    return qry2(lc, L, R)+qry2(rc, L, R);
}
int qry1(int x, int l, int r, int L, int R) {
    if(l >= L && r <= R) return t[x].rs;
    if(L > m) return qry1(rc, L, R);
    if(R <= m) return qry1(lc, L, R);
    return max(qry1(rc,L,R),qry1(lc,L,R)+qry2(rc,L,R));
}
int qry3(int x, int l, int r, int L, int R) {
    if(l >= L && r <= R) return t[x].ls;
    if(L > m) return qry3(rc, L, R);
    if(R <= m) return qry3(lc, L, R);
    return max(qry3(lc,L,R),qry3(rc,L,R)+qry2(lc,L,R));
}

int main() {
    scanf("%d", &n);
    for(int i = ; i <= n; i++) scanf("%d", &a[i]), b[i] = i;
    sort(b+, b++n, cmp), bd(rt[], , n);
    for(int i = ; i <= n; i++) rt[i] = ++tt, upd(rt[i-], rt[i], , n, b[i]);
    scanf("%d", &q);
    while(q--) {
        scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
        c[] = (l1+la)%n+, c[] = (r1+la)%n+, c[] = (l2+la)%n+, c[] = (r2+la)%n+;
        sort(c, c+), l1 = c[], r1 = c[], l2 = c[], r2 = c[];
        int l = , r = n;
        while(l < r) {
            int t1 = qry1(rt[m],,n,l1,r1)+qry3(rt[m],,n,l2,r2),t2;
            if(r1+ < l2) t2 = qry2(rt[m],,n,r1+,l2-); else t2 = ;
            if(t1+t2 >= ) l = m+; else r = m;
        }
        printf("%d\n", la=a[b[l]]);
    }
    return ;
}