2019 Multi-University Training Contest 4 - K-th Closest Distance
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主席树 + 二分答案
对于这种区间内的值域问题一般用主席树进行求解。
因为数据范围只有1e6,所以不用离散化,直接建树即可。
题目要求找到区间内离p第k近的数,可以想到,这个问题具有单调性(某个区间长度有大于k个值,那么比这个区间长度更长的比如也有大于k个值)
于是我们可以二分答案这个距离,枚举离p距离为mid的范围(max(1, p - mid), min(p + mid, 1e6)
这样我们每次主席树上询问这个值域范围内的个数是否大于等于k判断枚举的mid是否可行,最后mid停下来的最小值就是答案。
#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;
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