**Before the round starts**

Great efforts have been put over last year. We want to say thanks to everybody who helped us to make this round as it is. Cannot wait to see you guys on the next one!

**Testers' predictions**

Tester | A | B | C | D | E | F |
---|---|---|---|---|---|---|

tfg | 800 | 1100 | 1500 | 1600 | 1900 | 2400 |

neko_nyaaaaaaaaaaaaaaaaa | 800 | 1100 | 1700 | 1900 | 2100 | 2600 |

BucketPotato | 800 | 1000 | 1400 | 1800 | 2400 | - |

LetterC67 | 800 | 1200 | 1400 | 1600 | 2100 | - |

_FireGhost_ | 800 | 1100 | 1500 | 1900 | 2000 | 2400 |

fextivity | 800 | 1000 | 1400 | 1700 | 2000 | 2400 |

generic_placeholder_name | 800 | 1100 | 1600 | 1900 | 2200 | - |

**Some comments from testers and authors**

**From BucketPotato**

**From tfg**

**Shower thoughts from vangtrangtan**

**From Fireghost**

**Also from Fireghost**

**From thenymphsofdelphi**

**Huge help from Vladithur when polishing our statements**

**From hydroshiba**

**From thanhchauns2**

**From me**

**From errorgorn**

1713A - Traveling Salesman Problem

**Hint 1**

Do we actually need to go off the axis?

**Hint 2**

How to avoid visiting an axis more than once?

**Tutorial**

### 1713A - Traveling Salesman Problem

Suppose we only have boxes on the $$$Ox+$$$ axis, then the optimal strategy is going in the following way: $$$(0, 0), (x_{max}, 0), (0, 0)$$$. There is no way to do in less than $$$2 \cdot |x_{max}|$$$ moves.

What if we have boxes on two axis? Let's assume it is $$$Oy+$$$, suppose we have a strategy to go in the following way: $$$(0, 0), (x_{max}, 0),..., (0, y_{max}), (0, 0)$$$. In this case it is optimal to fill the three dots with $$$(0, 0)$$$, which is just solving each axis independently.

Therefore, the number of axis does not matters. For each axis that has at least one box, go from $$$(0, 0)$$$ to the farthest one, then come back to $$$(0, 0)$$$.

Time complexity: $$$O(n)$$$

**Solution**

```
def solve():
n = int(input())
minX, minY, maxX, maxY = 0, 0, 0, 0
for i in range(n):
x, y = list(map(int, input().split()))
minX = min(x, minX)
maxX = max(x, maxX)
minY = min(y, minY)
maxY = max(y, maxY)
print(2 * (maxX + maxY - minX - minY))
test = int(input())
while test > 0:
test -= 1
solve()
```

**Hint 1**

How to calculate $$$f(a)$$$?

**Hint 2**

What if $$$a$$$ is intially sorted?

**Hint 3**

Consider $$$a$$$ has $$$3$$$ elements. What if $$$a_1 > a_2$$$ and $$$a_2 < a_3$$$?

**Tutorial**

### 1713B - Optimal Reduction

Let's define $$$max([l, r])$$$ the maximum integer over $$$a_l, a_{l+1}, \dots, a_r$$$. Note that $$$max([1, r])$$$ and $$$max([l, n])$$$ can be precalculated through prefix max and suffix max array.

Consider an indices triplet $$$(i, j, k)$$$ of the array ($$$1 \le i < j < k \le n$$$), and we want to make $$$a_i$$$, $$$a_j$$$ and $$$a_k$$$ all equal to $$$0$$$ in some operations. There are $$$2$$$ cases:

- $$$a_i > a_j$$$ and $$$a_j < a_k$$$: this cost $$$a_i + a_k - a_j$$$ operations.
- Other cases cost the same number of operations: $$$max(a_i, a_j, a_k)$$$.

In the other hand:

- $$$a_i + a_k - a_j = max(a_i, a_k) + min(a_i, a_k) - a_j = max(a_i, a_j, a_k) + min(a_i, a_k) - a_j$$$
- $$$min(a_i, a_k) - a_j > 0$$$. Therefore $$$max(a_i, a_j, a_k) + min(a_i, a_k) - a_j > max(a_i, a_j, a_k)$$$

So as we can see, the first case requires $$$max(a_i, a_j, a_k) + min(a_i, a_k) - a_j$$$ operations while the second case requires only $$$max(a_i, a_j, a_k)$$$ operations. So if the first case is satisfied for some triplet $$$(i, j, k)$$$, then the answer is not optimal.

As a result, we've found the construction of the optimal array. The answer is *NO* if there exists an index $$$i$$$ ($$$2 \le i \le n - 1$$$) such that $$$max([1, i-1]) > a_i$$$ and $$$a_i < max([i+1, n])$$$. Otherwise, the answer is always *YES*.

**How to prove the correctness?**

By constructing a permutation $$$p$$$ where the array $$$a$$$ is sorted, we can prove that $$$f(p) = p[n]$$$. The operations can be done by finding the left most element $$$p_l$$$ such that $$$p_l \neq 0$$$, then decrease all the elements $$$p_l, p_{l+1}, \dots, p_n$$$ by $$$1$$$.

This strategy costs exactly $$$p[n] = max([1, n])$$$ operations and it can obviously be showed that there is no other permutation of $$$a$$$ which costs less operations than when it is sorted. As a result, because we have proved for every permutation $$$p$$$ of $$$a$$$, $$$f(p) \ge max([1, n])$$$, the answer is always $$$YES$$$ if the condition $$$a_i > a_j$$$ and $$$a_j < a_k$$$ is not satisfied for all triplet $$$(i, j, k)$$$ ($$$1 \le i < j < k \le n$$$).

Time complexity: $$$O(n)$$$

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, a[N], pref[N], suff[N];
int main() {
int tc; cin >> tc;
while (tc--) {
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i];
pref[1] = a[1];
for (int i = 2; i <= n; i++)
pref[i] = max(pref[i-1], a[i]);
suff[n] = a[n];
for (int i = n - 1; i >= 1; i--)
suff[i] = max(suff[i+1], a[i]);
bool ok = true;
for (int i = 2; i <= n - 1; i++) {
if (pref[i-1] > a[i] && a[i] < suff[i+1]) {
ok = false;
}
}
cout << (ok ? "YES\n" : "NO\n");
}
}
```

**Hint 1**

Is there any case that the answer doesn't exist?

**Hint 2**

**What if:** $$$n \le 5$$$

**Hint 3**

Construct the suffix instead of the prefix.

**Hint 4**

With any positive integer $$$x$$$, there is at least one square number in $$$[x, 2x]$$$. Proof.

**Tutorial**

### 1713C - Build Permutation

First, let's prove that the answer always exists. Let's call the smallest square number that is not smaller than $$$k$$$ is $$$h$$$. Therefore $$$h \leq 2 \cdot k$$$, which means $$$h - k \leq k$$$.

So we can fill $$$p_i = h - i$$$ for $$$(h - k \leq i \leq k)$$$. Using this method we can recursively reduce $$$k$$$ to $$$h - k - 1$$$, then all the way down to $$$-1$$$.

We can prove that $$$h - k \geq 0$$$, as $$$h \geq k$$$.

Time complexity: $$$O(n)$$$

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, ans[N];
void recurse(int r) {
if (r < 0) return;
int s = sqrt(2*r); s *= s;
int l = s - r; recurse(l - 1);
for (; l <= r; l++, r--) {
ans[l] = r; ans[r] = l;
}
}
int main() {
int tc; cin >> tc;
while (tc--) {
cin >> n; recurse(n - 1);
for (int i = 0; i < n; i++)
cout << ans[i] << ' ';
cout << '\n';
}
}
```

**Hint 1**

We made sure that almost every $$$2^{n - 1}$$$ solutions cannot pass.

**Hint 2**

Did you use the $$$0$$$ query?

**Hint 3**

$$$\frac{2^{n + 1}}{3} = 2^n \cdot \frac{2}{3}$$$, what is the conclusion?

**Tutorial**

### 1713D - Tournament Coundown

There is a way to erase $$$3$$$ participants in every $$$2$$$ queries. Since there are $$$2^n - 1$$$ participants to be removed, the number of queries will be $$$\left \lceil (2^n - 1) \cdot \frac{2}{3} \right \rceil = \left \lfloor \frac{2^{n + 1}}{3} \right \rfloor$$$. Suppose there are only $$$4$$$ participants. In the first query we will ask the judge to compare the $$$1^{st}$$$ and the $$$3^{rd}$$$ participants. There are three cases:

The $$$1^{st}$$$ participant wins more game than the $$$3^{rd}$$$ one: the $$$2^{nd}$$$ and $$$3^{rd}$$$ cannot be the winner.

The $$$3^{rd}$$$ participant wins more game than the $$$1^{st}$$$ one: the $$$1^{st}$$$ and $$$4^{th}$$$ cannot be the winner.

The $$$1^{st}$$$ and $$$3^{rd}$$$ participants' numbers of winning games are equal: both the $$$1^{st}$$$ and $$$3^{rd}$$$ cannot be the winner.

Ask the remaining two participants, find the winner between them.

If there are more than $$$4$$$ participants, we can continuously divide the number by $$$4$$$ again and again, until there are at most $$$2$$$ participants left. Now we can get the winner in one final query.

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int ask(vector<int> &k)
{
cout << "? " << k[0] << ' ' << k[2] << endl;
int x;
cin >> x;
if (x == 1)
{
cout << "? " << k[0] << ' ' << k[3] << endl;
cin >> x;
if (x == 1) return k[0];
return k[3];
}
else if (x == 2)
{
cout << "? " << k[1] << ' ' << k[2] << endl;
cin >> x;
if (x == 1) return k[1];
return k[2];
}
else
{
cout << "? " << k[1] << ' ' << k[3] << endl;
cin >> x;
if (x == 1) return k[1];
return k[3];
}
}
void solve()
{
int n;
cin >> n;
vector<int> a, b;
for (int i = 1; i <= (1LL << n); i++)
{
a.push_back(i);
}
while (a.size() > 2)
{
while (!a.empty())
{
vector<int> k(4);
k[0] = a.back();
a.pop_back();
k[1] = a.back();
a.pop_back();
k[2] = a.back();
a.pop_back();
k[3] = a.back();
a.pop_back();
int win = ask(k);
b.push_back(win);
}
a = b;
b.clear();
}
if (a.size() == 2)
{
cout << "? " << a[0] << ' ' << a[1] << endl;
int x;
cin >> x;
if (x == 2) a[0] = a[1];
}
cout << "! " << a[0] << endl;
}
int main(int argc, char ** argv)
{
int tests;
cin >> tests;
while(tests--) solve();
}
```

**Hint 1**

Think of the most to the least significant cell of the matrix.

**Hint 2**

How many positions in the matrix can a cell travel to after performing the operations?

**Hint 3**

And for each position that that cell can travel to, how many ways are there we can make it?

**Tutorial**

### 1713E - Cross Swapping

Let's take a look at what the lexicographically smallest matrix is. Let's call $$$(x, y)$$$ a cell that is in the intersection of row $$$x$$$ and column $$$y$$$ of the matrix, and the integer written on that cell is $$$A_{x, y}$$$. A cell $$$(x_p, y_p)$$$ of this matrix is called more significant than the another cell $$$(x_q, y_q)$$$ if and only if $$$x_p < x_q$$$, or $$$x_p = x_q$$$ and $$$y_p < y_q$$$. The problem asks us to find the smallest matrix so the best suitable way to solve this problem is to traverse through the most to the least significant cell of the matrix, then determine if the current cell can be minimized or not.

Suppose the current cell we are looking at is $$$(x, y)$$$. If $$$x = y$$$ then its position will not change after performing the operations. But if $$$x \neq y$$$, there are exactly $$$2$$$ operations that swap $$$(x, y)$$$ with another cell, that is $$$k = x$$$ and $$$k = y$$$. Both of these operations swap $$$(x, y)$$$ with the same cell $$$(y, x)$$$. So the only way we can minimize the value of $$$(x, y)$$$ is to try swapping it with $$$(y, x)$$$ (if $$$x < y$$$ and $$$A_{x, y} > A_{y, x}$$$) in some way.

As a result we have our constructive algorithm. Remind that for each operation $$$k = i$$$ of the matrix ($$$1 \le i \le n$$$), there are $$$2$$$ states: it is being performed and not being performed. Suppose we have traversed to the cell $$$(x, y)$$$. If $$$x \ge y$$$, ignore it. If $$$x < y$$$ then we try to make $$$A_{x, y} = min(A_{x, y}, A_{y, x})$$$ by deciding to swap or not to swap the cells. If $$$A_{x, y} > A_{y, x}$$$, try to swap $$$(x, y)$$$ with $$$(y, x)$$$ by making $$$2$$$ operations $$$k = x$$$ and $$$k = y$$$ having different states. And if $$$A_{x, y} < A_{y, x}$$$ then we should keep their positions unchanged by making $$$2$$$ operations $$$k = x$$$ and $$$k = y$$$ having the same state. Note that if $$$A_{x, y} = A_{y, x}$$$, we do nothing.

Let's implement this algorithm using a simple DSU where the $$$ith$$$ node represents the operation $$$k = i$$$. We define the value $$$par[u]$$$ in such a way that, suppose $$$p$$$ is the root of the $$$u$$$ node's component, $$$par[u] = p$$$ if $$$2$$$ operations $$$k = u$$$ and $$$k = p$$$ should have the same state, or $$$par[u] = -p$$$ if $$$2$$$ operations $$$k = u$$$ and $$$k = p$$$ should have different states. Define another function $$$join(i, j)$$$ to union $$$2$$$ nodes $$$i$$$ and $$$j$$$ to the same component. Note that $$$u$$$ and $$$-u$$$ are always in the same component and $$$par[-u] = -par[u]$$$. Thus, for the current cell $$$(x, y)$$$, we want to swap it with $$$(y, x)$$$ by calling $$$join(x, -y)$$$, or keep its position unchanged by calling $$$join(x, y)$$$.

After constructing the graphs, the last thing to do is to determine which operations should be performed. One way to do so is for each root of the components of the DSU, we perform the operation which this root represents for. Then for other nodes just check $$$par[i] > 0$$$ for the $$$ith$$$ node and if it is true, the $$$k = i$$$ operation should be performed. When we have the list of the operations that need to be performed, we can bruteforcely perform each operation from the list one by one and the final matrix will be the lexicographically smallest matrix.

Time complexity: $$$O(n^2)$$$

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 5;
int n, a[N][N];
int par[N];
int getPar(int u) {
if (u < 0) return -getPar(-u);
if (u == par[u]) return u;
return par[u] = getPar(par[u]);
}
void join(int u, int v) {
u = getPar(u); v = getPar(v);
if (abs(u) != abs(v)) {
if (u > 0) par[u] = v;
else par[-u] = -v;
}
}
int main() {
cin.tie(nullptr) -> sync_with_stdio(false);
int tc; cin >> tc;
while (tc--) {
cin >> n;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
cin >> a[i][j];
}
iota(par + 1, par + n + 1, 1);
// set par[i] == i for i in [1, n]
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
if (a[i][j] < a[j][i]) join(i, j);
if (a[i][j] > a[j][i]) join(i, -j);
}
for (int i = 1; i <= n; i++) {
if (getPar(i) < 0) continue;
// we only perform the operation
// if and only if getPar(i) > 0
for (int j = 1; j <= n; j++)
swap(a[i][j], a[j][i]);
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
cout << a[i][j] << ' ';
}
cout << '\n';
}
}
}
```

**Hint 0**

- Is there any case that the answer doesn't exist?
- If exist, are there multiple?

**Hint 1**

- How many times does $$$a_i$$$ contribute to $$$b_{j, n}$$$?

**Hint 1.1**

$$$\rightarrow$$$ Calculate value that $$$a_i$$$ contribute to $$$b_{j, n}$$$.

**Hint 1.2**

**Hint 2**

Consider the inverse problem: Given array $$$a$$$. Construct $$$b_{j, n}$$$ for all $$$j$$$. How can you solve this problem?

**Hint 3**

Consider easier problem: Let construct matrix $$$g$$$ of size $$$(2n + 1) \times (n + 1)$$$ same way as matrix $$$b$$$. Given $$$g_{i, n}$$$ $$$(1 \le i \le 2n)$$$, please reconstruct $$$a$$$. How can you solve this problem?

**Tutorial**

### 1713F - Lost Array

First, we can see that $$$a_i$$$ contribute $$$\binom{(n - i) + (j - 1)}{j - 1}$$$ times to $$$b_{j, n}$$$, which can calculate similar to Pascal's Triangle. It's easy to see that the value that $$$a_i$$$ contribute to $$$b_{j, n}$$$ equal to $$$a_i$$$ when $$$\binom{(n - i) + (j - 1)}{j - 1}$$$ is odd, $$$0$$$ otherwise.

Let's solve the inverse problem: Given array $$$a$$$. Construct $$$b_{j, n}$$$ for all $$$j$$$. $$$(1 \le j \le n)$$$

By Lucas Theorem, $$$\binom{(n - i) + (j - 1)}{j - 1}$$$ is odd when $$$(n - i)\ AND\ (j - 1) = 0$$$

$$$\rightarrow (n - i)$$$ is a submask of $$$\sim(j - 1)$$$ (with $$$\sim a$$$ is inverse mask of $$$a$$$).

Let define $$$m = 2^k$$$ with smallest $$$m$$$ satisfy $$$m \ge n$$$. Set $$$a'_i = a_i$$$ and $$$b'_j = b_{\sim(j - 1)} = b_{(m - 1) - (j - 1)}$$$ then $$$b'$$$ is the Zeta transform of $$$a'$$$.

So we could apply Mobius transform in $$$b'$$$ to get $$$a'$$$. Since the operation is xor, mobius transform is as same as zeta transform. But unlike the inverse problem, there are some differences. We don't know the value of $$$b'_i$$$ for $$$i$$$ in $$$[0, m - n)$$$.

Let $$$c$$$ be the sum over supermasks array of $$$b'$$$ (with $$$i$$$ is supermasks of $$$j$$$ when $$$i\ AND\ j = j)$$$, then set $$$c_k = 0$$$ for $$$k$$$ in $$$[m - n, m)$$$. After that, do another sum over supermasks on $$$c$$$ to get original value of $$$b'$$$. Now we can find $$$a'$$$ from $$$b'$$$ and $$$a$$$ from $$$a'$$$.

Complexity: $$$O(nlog_2(n))$$$

**Solution**

```
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define endl '\n'
#define fi first
#define se second
#define For(i, l, r) for (auto i = (l); i < (r); i++)
#define ForE(i, l, r) for (auto i = (l); i <= (r); i++)
#define FordE(i, l, r) for (auto i = (l); i >= (r); i--)
#define Fora(v, a) for (auto v: (a))
#define bend(a) (a).begin(), (a).end()
#define isz(a) ((signed)(a).size())
using ll = long long;
using ld = long double;
using pii = pair <int, int>;
using vi = vector <int>;
using vpii = vector <pii>;
using vvi = vector <vi>;
const int N = 1 << 19;
int n;
int a[N], b[N], ta[N], tb[N];
int c[N];
int m, lm, all1;
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n;
ForE(i, 1, n){
cin >> b[i];
}
m = 1 << __lg(n);
if (m < n){
m *= 2;
}
lm = __lg(m);
all1 = m - 1;
memset(ta, -1, sizeof(ta));
memset(tb, -1, sizeof(tb));
ForE(i, 1, n){
tb[all1 ^ (i - 1)] = b[i];
ta[n - i] = -2;
}
For(i, 0, m){
c[all1 ^ i] = max(tb[i], 0);
}
For(bit, 0, lm){
For(msk, 0, m){
if (msk >> bit & 1){
c[msk] ^= c[msk ^ (1 << bit)];
}
}
}
For(i, 0, m){
if (tb[i] == -1){
tb[i] = c[all1 ^ i];
}
}
For(i, 0, m){
ta[i] = tb[i];
}
For(bit, 0, lm){
For(msk, 0, m){
if (msk >> bit & 1){
ta[msk] ^= ta[msk ^ (1 << bit)];
}
}
}
ForE(i, 1, n){
a[i] = ta[n - i];
}
ForE(i, 1, n){
cout << a[i] << ' ';
} cout << endl;
}
```