Minimum MEX

用双指针,如果 \(mex\) 小于了就往大扩展

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e5 + 5;

int t, n, a[N], cnt[N], ans;

void Solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] <= n) {
      cnt[a[i]]++;
    }
  }
  int p;
  for (int i = 0; i <= n + 1; i++) {
    if (!cnt[i]) {
      p = i;
      break;
    }
  }
  ans = 1e9;
  fill(cnt, cnt + n + 2, 0);
  for (int i = 1, j = 1, tmp = 0; i <= n; i++) {
    while (j <= n && tmp < p) {
      tmp += (!cnt[a[j]] && a[j] < p);
      if (a[j] < p) {
        cnt[a[j]]++;
      }
      j++;
    }
    if (tmp == p) {
      ans = min(ans, j - i);
    }
    if (a[i] < p) {
      cnt[a[i]]--;
      if (!cnt[a[i]]) {
        tmp--;
      }
    }
  }
  if (p == 0) {
    cout << "1\n";
    return ;
  }
  cout << ans << '\n';
}

signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}

Range Contradiction

将数组排序,然后我们可以得出数组 \(a\) 与 \(b\) 中一定有一个为连续的,那么即可暴力求解

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e5 + 5;

int t, n, a[N], ans[N];

void Solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int p = 1, mini = (a[n / 2] - a[1]) * (a[n] - a[n / 2 + 1]);
  for (int i = 2; i + n / 2 <= n; i++) {
    if (mini > (a[n / 2 + i - 1] - a[i]) * (a[n] - a[1])) {
      p = i;
      mini = (a[n / 2 + i - 1] - a[i]) * (a[n] - a[1]);
    }
  }
  for (int i = p; i <= p + n / 2 - 1; i++) {
    ans[(i - p) * 2 + 1] = a[i];
  }
  int tot = 2;
  for (int i = 1; i < p; i++) {
    ans[tot] = a[i];
    tot += 2;
  }
  for (int i = p + n / 2; i <= n; i++) {
    ans[tot] = a[i];
    tot += 2;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}

signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}
/*
1
10
5 6 1 5 1 5 8 4 7 1
*/

Not a Segment Tree

考虑将反向操作,将 \(b\) 还原回 \(a\),那么如果 \(b_i\)的值大于\(b_{i - k} +...+ b_{i + k} - b_i\),那么 \(b_i\) 一定将被还原,可以使用优先队列,因为如果最大的值都无法还原,那么就意味着 \(b_i\) 无法还原回 \(a_i\)

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 2e6 + 5;

struct node {
  long long x, id;
  bool operator < (const node _y) const {
    return x < _y.x;
  }
};

int t, n, k;

long long a[N], b[N], tr[N * 4], ans;

void build(int i, int l, int r) {
  if (l == r) {
    tr[i] = b[l];
    return ;
  }
  int mid = (l + r) >> 1;
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  tr[i] = tr[i * 2] + tr[i * 2 + 1];
}

int query(int i, int l, int r, int x, int y) {
  if (r < x || l > y) {
    return 0;
  }
  if (l >= x && r <= y) {
    return tr[i];
  }
  int mid = (l + r) >> 1;
  return query(i * 2, l, mid, x, y) + query(i * 2 + 1, mid + 1, r, x, y);
}

int query(int l, int r) {
  if (l > r) {
    return 0;
  }
  return query(1, 1, n, max(1ll, l), min(n, r));
}

void modify(int i, int l, int r, int p, int x) {
  if (l == r) {
    tr[i] = x;
    return ;
  }
  int mid = (l + r) >> 1;
  if (mid >= p) {
    modify(i * 2, l, mid, p, x);
  }
  else modify(i * 2 + 1, mid + 1, r, p, x);
  tr[i] = tr[i * 2] + tr[i * 2 + 1];
}

void Solve() {
  priority_queue<node> q;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    q.push({b[i], i});
  }
  build(1, 1, n);
  ans = 0;
  while (!q.empty() && k) {
    node cur = q.top();
    q.pop();
    long long sum = 0;
    sum += query(cur.id - k, cur.id + k) - b[cur.id];
    sum += query(1, cur.id + k - n);
    sum += query(n - (k - cur.id + 1) + 1, n);
    if (!sum) {
      continue;
    }
    long long p = (b[cur.id] - a[cur.id]) / sum;
    ans += p;
    b[cur.id] -= p * sum;
    modify(1, 1, n, cur.id, b[cur.id]);
    if (p) {
      q.push({b[cur.id], cur.id});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] != a[i]) {
      cout << "-1\n";
      return ;
    }
  }
  cout << ans << "\n";
}

signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}