20240913

Mr. Wow and Lucky Array

我们可以打一个表找规律，我们会发现，最大的合法的就是 \(0, 1, 0, 1......\)，那么假如这个重合了，我们可以把 \(2, 3\) 调换一下顺序即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e5 + 5;

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

void Solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = 0;
  }
  for (int i = 2; i <= n; i += 2) {
    b[i] = 1;
  }
  bool flag = true;
  for (int i = 1; i <= n; i++) {
    if (b[i] != a[i]) {
      flag = false;
      break;
    }
  }
  if (n == 1) {
    cout << "-1\n";
    return ;
  }
  if (!flag) {
    for (int i = 1; i <= n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  else {
    b[2] = 0, b[3] = 1;
    for (int i = 1; i <= n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
}

signed main() {
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}

Mr. Wow and Dislikes

我们可以发现，数据范围中 \(a > b\)，那么我们可以贪心的维护一个 \(priority_queue\)，每次将最大的，也就是对顶去除，将其减去 \(b\)，闲不住的可以用线段树完成，但是我们可以打一个标记表示对当前这个点减了几次 \(b\) 即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 2e5 + 5;

int t, n, a, b, c[N];

bool check(int x) {
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    int tmp = max(0ll, c[i] - x * b);
    sum += (tmp + a - b - 1) / (a - b);
  }
  return sum <= x;
}

void Solve() {
  cin >> n >> a >> b;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  int l = 0, r = 1e9 + 10;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      r = mid;
    }
    else l = mid + 1;
  }
  cout << l << "\n";
}

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