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 longconst 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 longconst 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;
}