题面
Sol
给一个数列，有m个询问，每次问数列[l,r]区间中所有数的第一次出现的位置的中位数是多少，强制在线
主席树
询问区间内不同的数的个数
树上二分找到那个中位数
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 5);
const int __(8e6);
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, rt[_], tot, o[_], len, a[_], vis[_];
struct HJT{
int ls, rs, sz;
} T[__];
IL void Modify(RG int &x, RG int l, RG int r, RG int p, RG int v){
T[++tot] = T[x], T[x = tot].sz += v;
if(l == r) return;
RG int mid = (l + r) >> 1;
if(p <= mid) Modify(T[x].ls, l, mid, p, v);
else Modify(T[x].rs, mid + 1, r, p, v);
}
IL int Query(RG int x, RG int l, RG int r, RG int L, RG int R){
if(!x) return 0;
if(L <= l && R >= r) return T[x].sz;
RG int mid = (l + r) >> 1, ret = 0;
if(L <= mid) ret = Query(T[x].ls, l, mid, L, R);
if(R > mid) ret += Query(T[x].rs, mid + 1, r, L, R);
return ret;
}
IL int Calc(RG int x, RG int l, RG int r, RG int k){
if(l == r) return l;
RG int mid = (l + r) >> 1, s = T[T[x].ls].sz;
if(k <= s) return Calc(T[x].ls, l, mid, k);
return Calc(T[x].rs, mid + 1, r, k - s);
}
int main(RG int argc, RG char* argv[]){
for(RG int t = Input(), Case = 1; Case <= t; ++Case){
Fill(T, 0), Fill(rt, 0), Fill(vis, 0), tot = 0;
n = Input(), m = Input();
for(RG int i = 1; i <= n; ++i) o[i] = a[i] = Input();
sort(o + 1, o + n + 1), len = unique(o + 1, o + n + 1) - o - 1;
for(RG int i = n; i; --i){
a[i] = lower_bound(o + 1, o + len + 1, a[i]) - o;
rt[i] = rt[i + 1];
if(!vis[a[i]]) vis[a[i]] = i, Modify(rt[i], 1, n, i, 1);
else{
Modify(rt[i], 1, n, vis[a[i]], -1);
vis[a[i]] = i;
Modify(rt[i], 1, n, i, 1);
}
}
printf("Case #%d:", Case);
for(RG int i = 1, ans = 0; i <= m; ++i){
RG int l = Input(), r = Input(), ql, qr, num;
ql = min((l + ans) % n + 1, (r + ans) % n + 1);
qr = max((l + ans) % n + 1, (r + ans) % n + 1);
num = Query(rt[ql], 1, n, ql, qr);
printf(" %d", ans = Calc(rt[ql], 1, n, (num + 1) >> 1));
}
puts("");
}
return 0;
}