The optimal sequence of operations is very simple.

The optimal sequence of operations is adding $$$k-1$$$ 1-s into the set each time, at the same time decreasing $$$n$$$ by $$$k-1$$$. This implies that the answer is $$$\lceil \frac{n-1}{k-1}\rceil$$$.

t = (int)(input())
for _ in range(t):
    n, k = map(int, input().split())
    print((n - 1 + k - 2) // (k - 1))

"Most sequences" can be transformed into $$$[1]$$$. Conditions for a sequence to be un-transformable is stringent.

Find several simple substrings that make the string transformable.

We list some simple conditions for a string to be transformable:

• If 111 exists somewhere (as a substring) in the string, the string is always transformable.
• If 11 appears at least twice in the string, the string is always transformable.
• If the string both begins and ends with 1, it is always transformable.
• If the string begins or ends with 1 and 11 exists in the string, it is always transformable.

These can be found by simulating the operation for short strings on paper.

Contrarily, if a string does not meet any of the four items, it is always not transformable. This can be proved using induction (as an exercise).

Observe that the number of 1 never increases in an operation. We can change a continuous segment of zeros into one zero. After that, it suffices to check that the number of occurences of 1 is larger than the number of occurences of 0.

t = (int)(input())
for _ in range(t):
    n = (int)(input())
    a = input()
    ok = 0
    if a.count("111") >= 1:
        ok = 1
    if a.count("11") >= 2:
        ok = 1
    if a.count("11") >= 1 and (a[0] == "1" or a[-1] == "1"):
        ok = 1
    if a[0] == "1" and a[-1] == "1":
        ok = 1
    if ok:
        print("Yes")
    else:
        print("No")

The answer is a simple construction.

The maximum length for $$$2^k-1\ (k \gt 1)$$$ is $$$k+1$$$.

It's obvious that the answer only depends on the popcount of $$$n$$$. Below, we assume $$$n=2^k-1$$$. If $$$k=1$$$, it is shown in the samples that the length is $$$1$$$.

Otherwise, the maximum sequence length for $$$2^k-1$$$ is $$$k+1$$$. This can be achived by $$$a_i=n-2^{i-1}\ (1\le i\le k),a_{k+1}=n$$$.

t = (int)(input())
for _ in range(t):
    n = (int)(input())
    a = []
    for i in range(62, -1, -1):
        x = 1 << i
        if ((x & n) == x) and (x != n):
            a.append(n - x)
    a.append(n)
    print(len(a))
    for i in a:
        print(i, end=" ")
    print("")

Formulate the problem.

Some variables can't be large.

Suppose monster $$$i$$$ is killed in round $$$b_i$$$. Then, the total health decrement is the sum of $$$a_i\times b_i$$$. The "independent set" constraint means that for adjacent vertices, their $$$b$$$-s must be different.

Observation: $$$b_i$$$ does not exceed $$$\lfloor \log_2 n\rfloor+1$$$. In this problem, $$$b_i\le 19$$$ always holds for at least one optimal $$$a_i$$$.

By dp, we can find the answer in $$$O(n\log n)$$$ or $$$O(n\log^2 n)$$$, depending on whether you use prefix/suffix maximums to optimize the taking max part.

Bonus: Find a counterexample for $$$b_i\le 18$$$ when $$$n=300000$$$. (Pretest 2 is one case)

Note that $$$b_x\le \deg x$$$. By carefully handing the dp transitions, we can achieve a $$$O(\sum \deg_x)=O(n)$$$ solution.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
ll a[300005], f[300005][24], smn[30];
vector<int> g[300005];
void dfs(int x, int fa) {
	for (int i = 1; i <= 22; i++) f[x][i] = i * a[x];
	for (int y : g[x]) {
		if (y == fa) continue;
		dfs(y, x);
		ll tt = 8e18;
		smn[23] = 8e18;
		for (int i = 22; i >= 1; i--) {
			smn[i] = min(smn[i + 1], f[y][i]);
		}
		for (int i = 1; i <= 22; i++) {
			f[x][i] += min(tt, smn[i + 1]);
			tt = min(tt, f[y][i]);
		}
	}
}
void Solve() {
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	for (int i = 1, x, y; i < n; i++) {
		cin >> x >> y;
		g[x].push_back(y), g[y].push_back(x);
	}
	dfs(1, 0);
	cout << *min_element(f[1] + 1, f[1] + 23) << '\n';
	for (int i = 1; i <= n; i++) {
		g[i].clear();
		memset(f[i], 0, sizeof(f[i]));
	}
}
int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	int t;
	cin >> t;
	while (t--) Solve();
}

Formulate the problem on a cartesian tree.

Find a clever bruteforce. Calculate the time complexity carefully.

Build the cartesian tree of $$$a$$$. Let $$$lc_x,rc_x$$$ respectively be $$$x$$$'s left and right children.

Consider a vertex $$$x$$$. If we delete it, what would happen to the tree? Vertices that are on the outside of $$$x$$$'s subtree will not be affected. Vertices inside the subtree of $$$x$$$ will "merge". Actually, we can see that, only the right chain of $$$x$$$'s left child ($$$lc_x\to rc_{lc_x}\to rc_{rc_{lc_x}}\to \dots$$$) and the left chain of $$$x$$$'s right child will merge in a way like what we do in mergesort.

Now, if we merge the chains by bruteforce (use sorting or std::merge), the time complexity is $$$O(n)$$$! It's easy to see that each vertex will only be considered $$$O(1)$$$ times.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pr;
int n, a[500005], st[500005], c[500005][2], top, sz[500005];
ll ans[500005], atv[500005], sum;
void dfs1(int x) {
	sz[x] = 1;
	for (int i = 0; i < 2; i++)
		if (c[x][i]) dfs1(c[x][i]), sz[x] += sz[c[x][i]];
	sum += (atv[x] = (sz[c[x][0]] + 1ll) * (sz[c[x][1]] + 1ll) * a[x]);
}
void dfs2(int x, ll dlt) {
	ll val = sum - dlt - atv[x];
	vector<int> lr, rl;
	vector<pr> ve;
	int y = c[x][0];
	while (y) lr.push_back(y), val -= atv[y], y = c[y][1];
	y = c[x][1];
	while (y) rl.push_back(y), val -= atv[y], y = c[y][0];
	{
		int i = 0, j = 0;
		while (i < lr.size() || j < rl.size()) {
			if (j >= rl.size() || (i < lr.size() && j < rl.size() && a[lr[i]] < a[rl[j]])) {
				ve.push_back(pr(lr[i], 0));
				i++;
			} else {
				ve.push_back(pr(rl[j], 1));
				j++;
			}
		}
	}
	// cout << x << ' ' << val << '\n';
	int cursz = 0;
	for (int i = (int)ve.size() - 1; i >= 0; i--) {
		int p = ve[i].first, q = ve[i].second;
		// cout << "I:" << i << ' ' << cursz << '\n';
		val += (cursz + 1ll) * (sz[c[p][q]] + 1ll) * a[p];
		cursz += sz[c[p][q]] + 1;
	}
	ans[x] = val;
	for (int i = 0; i < 2; i++)
		if (c[x][i]) dfs2(c[x][i], dlt + 1ll * (sz[c[x][i ^ 1]] + 1) * a[x]);
}
void Solve() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	st[top = 0] = 0;
	for (int i = 1; i <= n; i++) {
		while (top && a[st[top]] > a[i]) top--;
		c[i][0] = c[st[top]][1], c[st[top]][1] = i;
		st[++top] = i;
	}
	int rt = st[1];
	sum = 0, dfs1(rt), dfs2(rt, 0);
	for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
	cout << '\n';
	for (int i = 0; i <= n; i++) c[i][0] = c[i][1] = 0;
}
int main() {
	int t;
	scanf("%d", &t);
	while (t--) Solve();
}

We can use dp to calculate the number of permutations of $$$1\sim n$$$ with $$$i$$$ prefix maximums and $$$j$$$ ascents, $$$f(n,i,j)$$$: consider where 1 is inserted, we will have a $$$O(n^3)$$$ dp that finds $$$f(n,i,j)$$$ for all suitable $$$(n,i,j)$$$-s.

For suffix maximums, (the number of permutations of $$$1\sim n$$$ with $$$i$$$ prefix maximums and $$$j$$$ ascents, $$$g(n,i,j)$$$, we can just reverse some dimension of $$$f$$$).

To calculate the answer, consider the position of $$$n$$$. Suppose it's $$$p$$$. The the answer is

Let $$$u(x,y)=\sum_{i}f(x,i,y)a_{i+1}$$$, $$$v(x,y)=\sum_{z}g(x,i,y)b_{i+1}$$$ (both of these are calculated in $$$O(n^3)$$$), then the answer is

By seeing $$$u$$$ and $$$v$$$ as 2D polynomials, this can be calculated with 2D FFT in $$$O(n^2\log n)$$$.

#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int Power(int x, int y) {
	int r = 1;
	while (y) {
		if (y & 1) r = 1ll * r * x % mod;
		x = 1ll * x * x % mod, y >>= 1;
	}
	return r;
}
namespace Conv_998244353 {

const int g = 3, invg = ((mod + 1) % 3 == 0 ? (mod + 1) / 3 : (2 * mod + 1) / 3);
int wk[1050005], ta[1050005], tb[1050005];
void DFT(int *a, int n) {
	for (int i = n >> 1; i; i >>= 1) {
		int w = Power(g, (mod - 1) / (i << 1));
		wk[0] = 1;
		for (int j = 1; j < i; j++) wk[j] = 1ll * wk[j - 1] * w % mod;
		for (int j = 0; j < n; j += (i << 1)) {
			for (int k = 0; k < i; k++) {
				int x = a[j + k], y = a[i + j + k], z = x;
				x += y, (x >= mod && (x -= mod)), a[j + k] = x;
				z -= y, (z < 0 && (z += mod)), a[i + j + k] = 1ll * z * wk[k] % mod;
			}
		}
	}
}
void IDFT(int *a, int n) {
	for (int i = 1; i < n; i <<= 1) {
		int w = Power(invg, (mod - 1) / (i << 1));
		wk[0] = 1;
		for (int j = 1; j < i; j++) wk[j] = 1ll * wk[j - 1] * w % mod;
		for (int j = 0; j < n; j += (i << 1)) {
			for (int k = 0; k < i; k++) {
				int x = a[j + k], y = 1ll * a[i + j + k] * wk[k] % mod, z = x;
				x += y, (x >= mod && (x -= mod)), a[j + k] = x;
				z -= y, (z < 0 && (z += mod)), a[i + j + k] = z;
			}
		}
	}
	for (int i = 0, inv = Power(n, mod - 2); i < n; i++) a[i] = 1ll * a[i] * inv % mod;
}
vector<int> conv(vector<int> A, vector<int> B) {
	for (auto &i : A) i %= mod;
	for (auto &i : B) i %= mod;
	int sa = A.size(), sb = B.size();
	vector<int> ret(sa + sb - 1);
	int len = 1;
	while (len < ret.size()) len <<= 1;
	for (int i = 0; i < len; i++) ta[i] = tb[i] = 0;
	for (int i = 0; i < sa; i++) ta[i] = A[i];
	for (int i = 0; i < sb; i++) tb[i] = B[i];
	DFT(ta, len), DFT(tb, len);
	for (int i = 0; i < len; i++) ta[i] = 1ll * ta[i] * tb[i] % mod;
	IDFT(ta, len);
	for (int i = 0; i < ret.size(); i++) ret[i] = ta[i];
	return ret;
}

} // namespace Conv_998244353
vector<int> conv(vector<int> A, vector<int> B) {
	return Conv_998244353::conv(A, B);
}
int n, a[705], b[705], c[705], f[705][705], u[705][705], v[705][705];
int ny[705], jc[705];
void upd(int &x, int y) { x += y, (x >= mod && (x -= mod)); }
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	for (int i = 1; i <= n; i++) cin >> b[i];
	for (int i = 0; i < n; i++) cin >> c[i];
	f[0][0] = jc[0] = ny[0] = 1;
	for (int i = 1; i <= n; i++) {
		jc[i] = 1ll * jc[i - 1] * i % mod, ny[i] = Power(jc[i], mod - 2);
	}
	vector<int> A(500000), B(500000);
	for (int i = 0; i <= n; i++) {
		for (int j = i; j >= 0; j--) {
			for (int k = i; k >= 0; k--) {
				if (!f[j][k]) continue;
				// cout << i << ' ' << j << ' ' << k << ' ' << f[j][k] << '\n';
				upd(u[i][k], 1ll * f[j][k] * a[j + 1] % mod);
				upd(v[i][k], 1ll * f[j][k] * b[j + 1] % mod);
				upd(f[j + 1][k + (i != 0)], f[j][k]);
				upd(f[j][k + 1], 1ll * f[j][k] * (i - (i != 0) - k) % mod);
				f[j][k] = 1ll * f[j][k] * (k + (i != 0)) % mod;
			}
		}
		if (i > 0) reverse(v[i], v[i] + i);
		for (int j = 0; j <= i; j++) {
			if (i) A[703 * i + j] = 1ll * u[i][j] * ny[i] % mod;
			B[703 * i + j] = 1ll * v[i][j] * ny[i] % mod;
		}
	}
	A = conv(A, B);
	for (int i = 1; i <= n; i++) {
		int ans = 0;
		for (int y = 0; y <= i - 1; y++) {
			ans = (ans + 1ll * u[0][0] * v[i - 1][y] % mod * c[y]) % mod;
		}
		for (int j = 0; j < i; j++) {
			ans = (ans + 1ll * A[703 * (i - 1) + j] * jc[i - 1] % mod * c[j + 1]) % mod;
		}
		cout << ans << ' ';
	}
}

#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353, N = 705;
int n, a[N + 5], b[N + 5], c[N + 5], f[N + 5][N + 5], u[N + 5][N + 5], v[N + 5][N + 5],
    C[N + 5][N + 5];
int g[N + 5], h[N + 5], t[N + 5][N + 5], p[N + 5][N + 5], ny[N + 5], o[N + 5];
void upd(int &x, int y) { x += y, (x >= mod && (x -= mod)); }
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	for (int i = 1; i <= n; i++) cin >> b[i];
	for (int i = 0; i < n; i++) cin >> c[i];
	f[0][0] = ny[1] = ny[0] = 1;
	for (int i = 0; i <= n; i++) {
		C[i][0] = 1;
		for (int j = 1; j <= i; j++) C[i][j] = (C[i &mdash; 1][j] + C[i &mdash; 1][j &mdash; 1]) % mod;
	}
	for (int i = 2; i <= n; i++) ny[i] = 1ll * ny[mod % i] * (mod &mdash; mod / i) % mod;
	for (int i = 2; i <= n; i++) ny[i] = 1ll * ny[i &mdash; 1] * ny[i] % mod;
	for (int i = 0; i < n; i++) {
		for (int j = i; j >= 0; j--) {
			for (int k = i; k >= 0; k--) {
				if (!f[j][k]) continue;
				// cout << i << ' ' << j << ' ' << k << ' ' << f[j][k] << '\n';
				upd(u[i][k], 1ll * f[j][k] * a[j + 1] % mod);
				upd(v[i][k], 1ll * f[j][k] * b[j + 1] % mod);
				upd(f[j + 1][k + (i != 0)], f[j][k]);
				upd(f[j][k + 1], 1ll * f[j][k] * (i &mdash; (i != 0) &mdash; k) % mod);
				f[j][k] = 1ll * f[j][k] * (k + (i != 0)) % mod;
			}
		}
		if (i > 0) reverse(v[i], v[i] + i);
	}
	for (int x = 1; x <= n; x++) {
		for (int i = 0; i < n; i++) {
			g[i] = h[i] = 0;
			for (int j = i; j >= 0; j--) {
				g[i] = (1ll * g[i] * x + u[i][j]) % mod;
				h[i] = (1ll * h[i] * x + v[i][j]) % mod;
			}
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= i; j++) {
				upd(t[i][x], 1ll * g[j &mdash; 1] * h[i &mdash; j] % mod * C[i &mdash; 1][j &mdash; 1] % mod *
				                 (j > 1 ? x : 1) % mod);
			}
		}
	}
	p[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < i; j++) p[i][j] = p[0][j], o[j] = 0;
		for (int j = 0; j <= i; j++) {
			if (j != i) {
				for (int k = i &mdash; 1; k >= 0; k--) {
					upd(p[j][k + 1], p[j][k]);
					p[j][k] = 1ll * p[j][k] * (mod &mdash; i) % mod;
				}
			}
			if (!j) continue;
			int s = 1ll * t[i][j] * ny[j &mdash; 1] % mod * ny[i &mdash; j] % mod;
			if ((i &mdash; j) & 1) s = mod &mdash; s;
			for (int k = 0; k < i; k++) {
				upd(o[k], 1ll * s * p[j][k] % mod);
			}
		}
		int ans = 0;
		for (int k = 0; k < i; k++) {
			ans = (ans + 1ll * o[k] * c[k]) % mod;
		}
		cout << ans << ' ';
	}
}