Thanks for participation!↵
↵
**Update**: added alternative solutions/proofs for A,B,D,F↵
↵
[problem:1988A]↵
↵
<spoiler summary="Hint 1">↵
The optimal sequence of operations is very simple.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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 $\lceil \frac{n-1}{k-1}\rceil$.↵
↵
<spoiler summary="Why?">↵
I failed to find a Div2-A level proof. If you have a simpler proof please share it in the comments.↵
↵
Consider the number of elements that is $\equiv 1\pmod{(k-1)}$ in the set. The number of such elements increase by at most $k-1$ in each operation, and the aforementioned sequence of operation achieves the maximum increment.↵
↵
---↵
↵
A simpler proof in the comments: consider the number of elements in the set. It increases by at most $k-1$ each time, and our construction reaches the maximum increment.↵
</spoiler>↵
↵
</spoiler>↵
↵
<spoiler summary="Code (python)">↵
```python↵
t = (int)(input())↵
for _ in range(t):↵
n, k = map(int, input().split())↵
print((n - 1 + k - 2) // (k - 1))↵
```↵
</spoiler>↵
↵
[problem:1988B]↵
↵
<spoiler summary="Hint 1">↵
"Most sequences" can be transformed into $[1]$. Conditions for a sequence to be un-transformable is stringent.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
Find several simple substrings that make the string transformable.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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).↵
</spoiler>↵
↵
<spoiler summary="Another Solution">↵
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.↵
</spoiler>↵
↵
<spoiler summary="Code (python)">↵
```python↵
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")↵
```↵
</spoiler>↵
↵
[problem:1988C]↵
↵
<spoiler summary="Hint 1">↵
The answer is a simple construction.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
The maximum length for $2^k-1\ (k>1)$ is $k+1$.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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 $a_i=n-2^{i-1}\ (1\le i\le k),a_{k+1}=n$.↵
↵
<spoiler summary="Why?">↵
Consider the value of $a_1$. Note that its $k$-th bit (indexed from $2^0$ to $2^{k-1}$) should be 0, otherwise we would not make use of the largest bit.↵
↵
Since $a_1$ and $a_2$'s OR is $n$, $a_2$'s $k$-th bit should be 1, and thus the construction of $a_2\sim a_{k+1}$ boils down to the subproblem when $n=2^{k-1}-1$. This shows that $f(k)\le f(k-1)+1$ where $k$ is the popcount of $n$ and $f(k)$ is the maximum sequence length. We know that $f(2)=3$, so $f(k)\le k+1$ for all $k\ge 2$.↵
</spoiler>↵
↵
</spoiler>↵
↵
<spoiler summary="Code (python)">↵
```python↵
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("")↵
```↵
</spoiler>↵
↵
[problem:1988D]↵
↵
<spoiler summary="Hint 1">↵
Formulate the problem.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
Some variables can't be large.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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 $a_i$.↵
↵
<spoiler summary="Proof">↵
Let the mex of a set be the smallest positive integer that does not appear in it. Note that in an optimal arrangement, $b_i=\mathrm{mex}_{(j,i)\in E} b_j$.↵
↵
Consider an vertex with the maximum $b$, equal to $u$. Root the tree at this vertex $x$. The vertices connected with $x$ should take all $b$-s from $1$ to $u-1$. Denote $f(u)$ as the minimum number of vertices to make this happen, we have↵
↵
$$↵
f(1)=1,\ f(u)\ge 1+\sum_{1\le i<u}f(i)\to f(u)\ge 2^{u-1}↵
$$↵
↵
This ends the proof.↵
</spoiler>↵
↵
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)↵
</spoiler>↵
↵
<spoiler summary="Another Solution">↵
Note that $b_x\le \deg x$. By carefully handing the dp transitions, we can achieve a $O(\sum \deg_x)=O(n)$ solution.↵
</spoiler>↵
↵
<spoiler summary="Code (C++)">↵
```cpp↵
#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();↵
}↵
```↵
</spoiler>↵
↵
[problem:1988E]↵
↵
<spoiler summary="Hint 1">↵
Formulate the problem on a cartesian tree.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
Find a clever bruteforce. Calculate the time complexity carefully.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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.↵
</spoiler>↵
↵
<spoiler summary="Code (C++)">↵
```cpp↵
#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();↵
}↵
```↵
</spoiler>↵
↵
[problem:1988F]↵
↵
<spoiler summary="Solution">↵
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↵
$$↵
\sum_{p=1}^n\sum_{i=0}^{p-1}\sum_{j=0}^{n-p}\sum_{x=0}^{p-1}\sum_{y=0}^{n-p}f(p-1,i,x)g(n-p,j,y)\binom{n-1}{p-1}a_{i+1}b_{j+1}c_{x+y+[p>1]}↵
$$↵
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↵
$$↵
\sum_{p=1}^n\sum_{x=0}^{p-1}\sum_{y=0}^{n-p}u(p-1,x)v(n-p,y)\binom{n-1}{p-1}c_{x+y+[p>1]}↵
$$↵
↵
By seeing $u$ and $v$ as 2D polynomials, this can be calculated with 2D FFT in $O(n^2\log n)$.↵
↵
<spoiler summary="I don't know what FFT is, can I also solve the problem?">↵
Of course! This problem can also be solved with modulo $10^9+7$ or on an algebraic structure where FFT is not easy, but interpolation can be successfully carried out.↵
↵
We still need to consider $u(p-1,*)$ and $v(n-p,*)$ as polynomials $\sum_{i}u(p-1,i)x^i,\sum_{i}v(n-p,i)x^i$. Instead of doing FFT, consider substitute $x$ with $0,1,\dots,n+1$, and use interpolation to recover the final coefficients.↵
↵
Suppose we have a value of $x$. Then, calculating $u(p-1)$ and $v(n-p)$ takes $O(n^2)$ in total. For fixed $x$ and $n$, the answer for $n$ is (here, note how $x$ is multiplied)↵
$$↵
\sum_{p=1}^nu(p-1)v(n-p)\binom{n-1}{p-1}x^{p>1}↵
$$↵
This also takes $O(n^2)$ in total. So, for each $x$, the calculation is $O(n^2)$.↵
↵
For each $n$, the naive implementation of interpolation runs in $O(n^2)$. After we recover the coefficients, we multiply it with $c$ and update the answer.↵
↵
So, the time complexity is $O(n^3)$.↵
</spoiler>↵
↵
<spoiler summary="An even easier solution, found by jiangly">↵
We are trying to iterate over $i,x,j,y$ and do $ans_{i+j}\leftarrow f(i,x)g(j,y)h_{x+y}$. We can achieve this with two steps: first, iterate over $i,x,y$ and do $v_{i,xy}\leftarrow f(i,x)h_{x+y}$. Then, iterate over $i,j,xy$ and do $ans_{i+j}\leftarrow g(j,y)v_{i,xy}$.↵
↵
This runs in $O(n^3)$, and does not need to calculate inverses, and thus can be carried out on any ring.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="Code (C++, FFT)">↵
```cpp↵
#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 << ' ';↵
}↵
}↵
```↵
</spoiler>↵
↵
↵
<spoiler summary="Code (C++, Interpolation)">↵
```cpp↵
#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 — 1][j] + C[i — 1][j — 1]) % mod;↵
}↵
for (int i = 2; i <= n; i++) ny[i] = 1ll * ny[mod % i] * (mod — mod / i) % mod;↵
for (int i = 2; i <= n; i++) ny[i] = 1ll * ny[i — 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 — (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 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 — 1] * h[i — j] % mod * C[i — 1][j — 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 — 1; k >= 0; k--) {↵
upd(p[j][k + 1], p[j][k]);↵
p[j][k] = 1ll * p[j][k] * (mod — i) % mod;↵
}↵
}↵
if (!j) continue;↵
int s = 1ll * t[i][j] * ny[j — 1] % mod * ny[i — j] % mod;↵
if ((i — j) & 1) s = mod — 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 << ' ';↵
}↵
}↵
```↵
</spoiler>↵
↵
↵
**Update**: added alternative solutions/proofs for A,B,D,F↵
↵
[problem:1988A]↵
↵
<spoiler summary="Hint 1">↵
The optimal sequence of operations is very simple.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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 $\lceil \frac{n-1}{k-1}\rceil$.↵
↵
<spoiler summary="Why?">↵
I failed to find a Div2-A level proof. If you have a simpler proof please share it in the comments.↵
↵
Consider the number of elements that is $\equiv 1\pmod{(k-1)}$ in the set. The number of such elements increase by at most $k-1$ in each operation, and the aforementioned sequence of operation achieves the maximum increment.↵
↵
---↵
↵
A simpler proof in the comments: consider the number of elements in the set. It increases by at most $k-1$ each time, and our construction reaches the maximum increment.↵
</spoiler>↵
↵
</spoiler>↵
↵
<spoiler summary="Code (python)">↵
```python↵
t = (int)(input())↵
for _ in range(t):↵
n, k = map(int, input().split())↵
print((n - 1 + k - 2) // (k - 1))↵
```↵
</spoiler>↵
↵
[problem:1988B]↵
↵
<spoiler summary="Hint 1">↵
"Most sequences" can be transformed into $[1]$. Conditions for a sequence to be un-transformable is stringent.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
Find several simple substrings that make the string transformable.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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).↵
</spoiler>↵
↵
<spoiler summary="Another Solution">↵
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.↵
</spoiler>↵
↵
<spoiler summary="Code (python)">↵
```python↵
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")↵
```↵
</spoiler>↵
↵
[problem:1988C]↵
↵
<spoiler summary="Hint 1">↵
The answer is a simple construction.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
The maximum length for $2^k-1\ (k>1)$ is $k+1$.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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 $a_i=n-2^{i-1}\ (1\le i\le k),a_{k+1}=n$.↵
↵
<spoiler summary="Why?">↵
Consider the value of $a_1$. Note that its $k$-th bit (indexed from $2^0$ to $2^{k-1}$) should be 0, otherwise we would not make use of the largest bit.↵
↵
Since $a_1$ and $a_2$'s OR is $n$, $a_2$'s $k$-th bit should be 1, and thus the construction of $a_2\sim a_{k+1}$ boils down to the subproblem when $n=2^{k-1}-1$. This shows that $f(k)\le f(k-1)+1$ where $k$ is the popcount of $n$ and $f(k)$ is the maximum sequence length. We know that $f(2)=3$, so $f(k)\le k+1$ for all $k\ge 2$.↵
</spoiler>↵
↵
</spoiler>↵
↵
<spoiler summary="Code (python)">↵
```python↵
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("")↵
```↵
</spoiler>↵
↵
[problem:1988D]↵
↵
<spoiler summary="Hint 1">↵
Formulate the problem.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
Some variables can't be large.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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 $a_i$.↵
↵
<spoiler summary="Proof">↵
Let the mex of a set be the smallest positive integer that does not appear in it. Note that in an optimal arrangement, $b_i=\mathrm{mex}_{(j,i)\in E} b_j$.↵
↵
Consider an vertex with the maximum $b$, equal to $u$. Root the tree at this vertex $x$. The vertices connected with $x$ should take all $b$-s from $1$ to $u-1$. Denote $f(u)$ as the minimum number of vertices to make this happen, we have↵
↵
$$↵
f(1)=1,\ f(u)\ge 1+\sum_{1\le i<u}f(i)\to f(u)\ge 2^{u-1}↵
$$↵
↵
This ends the proof.↵
</spoiler>↵
↵
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)↵
</spoiler>↵
↵
<spoiler summary="Another Solution">↵
Note that $b_x\le \deg x$. By carefully handing the dp transitions, we can achieve a $O(\sum \deg_x)=O(n)$ solution.↵
</spoiler>↵
↵
<spoiler summary="Code (C++)">↵
```cpp↵
#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();↵
}↵
```↵
</spoiler>↵
↵
[problem:1988E]↵
↵
<spoiler summary="Hint 1">↵
Formulate the problem on a cartesian tree.↵
</spoiler>↵
↵
<spoiler summary="Hint 2">↵
Find a clever bruteforce. Calculate the time complexity carefully.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
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.↵
</spoiler>↵
↵
<spoiler summary="Code (C++)">↵
```cpp↵
#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();↵
}↵
```↵
</spoiler>↵
↵
[problem:1988F]↵
↵
<spoiler summary="Solution">↵
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↵
$$↵
\sum_{p=1}^n\sum_{i=0}^{p-1}\sum_{j=0}^{n-p}\sum_{x=0}^{p-1}\sum_{y=0}^{n-p}f(p-1,i,x)g(n-p,j,y)\binom{n-1}{p-1}a_{i+1}b_{j+1}c_{x+y+[p>1]}↵
$$↵
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↵
$$↵
\sum_{p=1}^n\sum_{x=0}^{p-1}\sum_{y=0}^{n-p}u(p-1,x)v(n-p,y)\binom{n-1}{p-1}c_{x+y+[p>1]}↵
$$↵
↵
By seeing $u$ and $v$ as 2D polynomials, this can be calculated with 2D FFT in $O(n^2\log n)$.↵
↵
<spoiler summary="I don't know what FFT is, can I also solve the problem?">↵
Of course! This problem can also be solved with modulo $10^9+7$ or on an algebraic structure where FFT is not easy, but interpolation can be successfully carried out.↵
↵
We still need to consider $u(p-1,*)$ and $v(n-p,*)$ as polynomials $\sum_{i}u(p-1,i)x^i,\sum_{i}v(n-p,i)x^i$. Instead of doing FFT, consider substitute $x$ with $0,1,\dots,n+1$, and use interpolation to recover the final coefficients.↵
↵
Suppose we have a value of $x$. Then, calculating $u(p-1)$ and $v(n-p)$ takes $O(n^2)$ in total. For fixed $x$ and $n$, the answer for $n$ is (here, note how $x$ is multiplied)↵
$$↵
\sum_{p=1}^nu(p-1)v(n-p)\binom{n-1}{p-1}x^{p>1}↵
$$↵
This also takes $O(n^2)$ in total. So, for each $x$, the calculation is $O(n^2)$.↵
↵
For each $n$, the naive implementation of interpolation runs in $O(n^2)$. After we recover the coefficients, we multiply it with $c$ and update the answer.↵
↵
So, the time complexity is $O(n^3)$.↵
</spoiler>↵
↵
<spoiler summary="An even easier solution, found by jiangly">↵
We are trying to iterate over $i,x,j,y$ and do $ans_{i+j}\leftarrow f(i,x)g(j,y)h_{x+y}$. We can achieve this with two steps: first, iterate over $i,x,y$ and do $v_{i,
↵
This runs in $O(n^3)$, and does not need to calculate inverses, and thus can be carried out on any ring.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="Code (C++, FFT)">↵
```cpp↵
#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 << ' ';↵
}↵
}↵
```↵
</spoiler>↵
↵
↵
<spoiler summary="Code (C++, Interpolation)">↵
```cpp↵
#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 — 1][j] + C[i — 1][j — 1]) % mod;↵
}↵
for (int i = 2; i <= n; i++) ny[i] = 1ll * ny[mod % i] * (mod — mod / i) % mod;↵
for (int i = 2; i <= n; i++) ny[i] = 1ll * ny[i — 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 — (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 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 — 1] * h[i — j] % mod * C[i — 1][j — 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 — 1; k >= 0; k--) {↵
upd(p[j][k + 1], p[j][k]);↵
p[j][k] = 1ll * p[j][k] * (mod — i) % mod;↵
}↵
}↵
if (!j) continue;↵
int s = 1ll * t[i][j] * ny[j — 1] % mod * ny[i — j] % mod;↵
if ((i — j) & 1) s = mod — 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 << ' ';↵
}↵
}↵
```↵
</spoiler>↵
↵
