2019 Multi-University Training Contest 4 - K-th Closest Distance

Chairman of the tree + half answers

For the range of issues within this range is generally solved by the Chairman of the tree.

Because the data range only 1e6, so do not discrete, direct contribution can be.

Title required number of k from p to find near interval, it is contemplated that this problem is monotonic (with a length greater than an interval k values, the longer such k values ​​are also greater than the ratio of the section length)

So we can answer half this distance, the distance p from the enumeration of the mid range (max (1, p - mid), min (p + mid, 1e6)

So every time we asked whether the President of the number of trees within the range of greater than or equal k determine the scope of the enumeration mid feasibility, and finally stopped mid minimum is the answer.

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 200005;
const int M = 1000005;
int _, n, m, tot, a[N], lc[M*20], rc[M*20], tree[M*20], root[N];

int buildTree(int l, int r){
    int cur = ++ tot;
    if(l == r) return cur;
    int mid = (l + r) >> 1;
    buildTree(l, mid);
    buildTree(mid + 1, r);
    return cur;
}

int insert(int rt, int l, int r, int val){
    int cur = ++ tot;
    tree[cur] = tree[rt] + 1, lc[cur] = lc[rt], rc[cur] = rc[rt];
    if(l == r) return cur;
    int mid = (l + r) >> 1;
    if(val <= mid) lc[cur] = insert(lc[rt], l, mid, val);
    else rc[cur] = insert(rc[rt], mid + 1, r, val);
    return cur;
}

int query(int a, int b, int l, int r, int ql, int qr){
    if(l == ql && r == qr){
        return tree[b] - tree[a];
    }
    int mid = (l + r) >> 1;
    if(qr <= mid) return query(lc[a], lc[b], l, mid, ql, qr);
    else if(ql > mid) return query(rc[a], rc[b], mid + 1, r, ql, qr);
    return query(lc[a], lc[b], l, mid, ql, mid) + query(rc[a], rc[b], mid + 1, r, mid + 1, qr);
}

bool calc(int l, int r, int p, int k, int mid){
    int ret = query(root[l - 1], root[r], 1, M, max(1, p - mid), min(p + mid, M));
    return ret >= k;
}

int main(){

    //freopen("data.txt", "r", stdin);
    for(_ = read(); _; _ --){
        tot = 0;
        n = read(), m = read();
        for(int i = 1; i <= n; i ++) a[i] = read();
        root[0] = buildTree(1, M);
        for(int i = 1; i <= n; i ++){
            root[i] = insert(root[i - 1], 1, M, a[i]);
        }
        int ans = 0;
        while(m --){
            int L = read(), R = read(), p = read(), k = read();
            L ^= ans, R ^= ans, p ^= ans, k ^= ans;
            int l = 0, r = 1e6;
            while(l < r){
                int mid = (l + r) >> 1;
                if(calc(L, R, p, k, mid)) r = mid;
                else l = mid + 1;
            }
            printf("%d\n", l);
            ans = l;
        }
    }
    return 0;
}