20241006

Back to School '24 P1 - Kicking

按照題意模擬即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e5 + 5;

int n, m, k;

char c;

vector<int> sum[2][N];

vector<int> a[N];

signed main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m + 1; j++) {
      sum[0][i].push_back(0);
      sum[1][i].push_back(0);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> c;
      a[i].push_back(c);
      sum[0][i][j] = sum[0][i][j - 1] + (c == 'A');
      sum[1][i][j] = sum[1][i][j - 1] + (c == 'B');
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (a[i][j - 1] == '.') {
        cout << ".";
      }
      else if (a[i][j - 1] == 'A') {
        if (sum[1][i][min(j + k, m)] - sum[1][i][j - 1] > 0) {
          cout << "N";
        }
        else cout << "Y";
      }
      else {
        if (sum[0][i][j] - sum[0][i][max(0ll, j - k - 1)] > 0) {
          cout << "N";
        }
        else cout << "Y";
      }
    }
    cout << "\n";
  }
  return 0;
}

Back to School '24 P2 - Cheating

注意到一個重要的性質,就是 \(a_i\) 為遞增序列,那麼我們會發現,他改的區間一定是上一個區間後面的,模擬即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e6 + 5;

int n, m, a[N], b[N], ans[N];

signed main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
  }
  int last = 1;
  for (int i = 1; i <= n; i++) {
    if (last + b[i] - 1 <= m) {
      ans[last] += a[i];
      ans[last + b[i]] -= a[i];
      last += b[i];
      if (last == m + 1) {
        last = 1;
      }
    }
    else {
      ans[last] += a[i];
      ans[m + 1] -= a[i];
      b[i] -= (m - last + 1);
      ans[1] += a[i];
      ans[min(b[i] + 1, last)] -= a[i];
      last = min(b[i] + 1, last);
    }
  }
  for (int i = 1; i <= m; i++) {
    ans[i] += ans[i - 1];
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

Back to School '24 P3 - Tournament

我們可以發現,他合併答案是簡直和線段樹一摸一樣,所以直接線段樹維護即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = (1 << 18) + 5;

struct node {
  int maxi, ans;
  void clean() {
    maxi = ans = 0;
  }
}tr[N * 4];

int n, p, a[N];

bool vis[N];

void build(int i, int l, int r) {
  if (l == r) {
    if (l == 1) {
      tr[i].maxi = p, tr[i].ans = 0;
    }
    else tr[i].maxi = a[l - 1], tr[i].ans = 0;
    return ;
  }
  int mid = (l + r) >> 1;
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  tr[i].ans = tr[i * 2].ans + tr[i * 2 + 1].ans + (tr[i * 2].maxi - tr[i * 2 + 1].maxi) * (tr[i * 2].maxi - tr[i * 2 + 1].maxi);
  tr[i].maxi = max(tr[i * 2].maxi, tr[i * 2 + 1].maxi);
}

void modify(int i, int l, int r, int p, int x) {
  if (l == r) {
    tr[i].maxi = x;
    tr[i].ans = 0;
    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].ans = tr[i * 2].ans + tr[i * 2 + 1].ans + (tr[i * 2].maxi - tr[i * 2 + 1].maxi) * (tr[i * 2].maxi - tr[i * 2 + 1].maxi);
  tr[i].maxi = max(tr[i * 2].maxi, tr[i * 2 + 1].maxi);
}

signed main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> a[i];
    vis[a[i]] = true;
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      p = i;
    }
  }
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    cout << tr[1].ans << " ";
    if (i < n) {
      modify(1, 1, n, i, a[i]);
      modify(1, 1, n, i + 1, p);
    }
  }
  return 0;
}

Back to School '24 P4 - Candidates

我們假如考慮每一種因素,他一定要滿足在每個區間內是遞增的,用 \(set\) 維護,做一個類似於拓撲排序的即可

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;

int n, k, a[N], cnt[N];

vector<int> r[N], v[N], ans;

bool vis[N];

priority_queue<int, vector<int>, greater<int> > q;

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    r[i].push_back(0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1, c; j <= k; j++) {
      cin >> c;
      r[i].push_back(c);
    }
  }
  for (int j = 1; j <= k; j++) {
    for (int i = 1; i < n; i++) {
      if (r[a[i]][j] < r[a[i + 1]][j]) {
        cnt[j]++;
        vis[i] = true;
        v[i].push_back(j);
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!cnt[i]) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int cur = q.top();
    q.pop();
    ans.push_back(cur);
    vector<int> tmp;
    for (int i = 1; i < n; i++) {
      if (r[a[i]][cur] > r[a[i + 1]][cur] && vis[i]) {
        tmp.push_back(i);
      }
    }
    for (auto cur : tmp) {
      for (auto u : v[cur]) {
        cnt[u]--;
        if (!cnt[u]) {
          q.push(u);
        }
      }
      vis[cur] = false;
      v[cur].clear();
    }
  }
  for (int i = 1; i <= k; i++) {
    if (cnt[i]) {
      cout << "-1";
      return 0;
    }
  }
  for (auto cur : ans) {
    cout << cur << " ";
  }
  return 0;
}