Enjoy!

**A. Balanced Rating Changes**

Tutorial is loading...

**Solution by tourist**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
int flag = 1;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
if (x % 2 == 0) {
cout << x / 2 << '\n';
} else {
cout << (x + flag) / 2 << '\n';
flag *= -1;
}
}
return 0;
}
```

**B. Balanced Tunnel**

Tutorial is loading...

**Solution by tourist**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<int> a(n), b(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
--a[i];
}
for (int i = 0; i < n; i++) {
cin >> b[i];
--b[i];
}
vector<int> pos(n);
for (int i = 0; i < n; i++) {
pos[b[i]] = i;
}
vector<int> c(n);
for (int i = 0; i < n; i++) {
c[i] = pos[a[i]];
}
int mx = -1, ans = 0;
for (int i = 0; i < n; i++) {
if (c[i] > mx) {
mx = c[i];
} else {
++ans;
}
}
cout << ans << '\n';
return 0;
}
```

**C1. Balanced Removals (Easier)**

Tutorial is loading...

**Solution by arsijo**

```
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<pair<int, int>, pair<int, int> > ii;
int main(){
int n;
cin >> n;
vector<ii> v(n);
for (int i = 0; i < n; i++){
cin >> v[i].first.first >> v[i].first.second >> v[i].second.first;
v[i].second.second = i + 1;
}
while(!v.empty()){
sort(v.begin(), v.end());
int x = v.back().first.first;
int y = v.back().first.second;
int z = v.back().second.first;
int pos = v.back().second.second;
v.pop_back();
int ind = 0;
ll d = 1e10;
for (int i = 0; i < (int) v.size(); i++){
ll dist = abs(v[i].first.first - x);
dist += abs(v[i].first.second - y);
dist += abs(v[i].second.first - z);
if (dist < d){
d = dist;
ind = i;
}
}
cout << pos << " " << v[ind].second.second << endl;
v.erase(v.begin() + ind);
}
}
```

**C2. Balanced Removals (Harder)**

Tutorial is loading...

**Solution by tourist**

```
#include <bits/stdc++.h>
using namespace std;
const int D = 3;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<vector<int>> p(n, vector<int>(D));
for (int i = 0; i < n; i++) {
for (int j = 0; j < D; j++) {
cin >> p[i][j];
}
}
auto Solve = [&](auto& Self, vector<int> ids, int k) {
if (k == D) {
return ids[0];
}
map<int, vector<int>> mp;
for (int x : ids) {
mp[p[x][k]].push_back(x);
}
vector<int> a;
for (auto& q : mp) {
int cur = Self(Self, q.second, k + 1);
if (cur != -1) {
a.push_back(cur);
}
}
for (int i = 0; i + 1 < (int) a.size(); i += 2) {
cout << a[i] + 1 << " " << a[i + 1] + 1 << '\n';
}
return (a.size() % 2 == 1 ? a.back() : -1);
};
vector<int> t(n);
iota(t.begin(), t.end(), 0);
Solve(Solve, t, 0);
return 0;
}
```

**D. Balanced Playlist**

Tutorial is loading...

**Solution by tourist**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<int> a(3 * n);
for (int i = 0; i < n; i++) {
cin >> a[i];
a[i + n] = a[i + 2 * n] = a[i];
}
vector<int> ans(3 * n);
vector<int> st_max;
vector<int> st_min;
for (int i = 3 * n - 1; i >= 0; i--) {
while (!st_max.empty() && a[st_max.back()] < a[i]) {
st_max.pop_back();
}
while (!st_min.empty() && a[st_min.back()] > a[i]) {
st_min.pop_back();
}
int low = 0, high = (int) st_min.size();
while (low < high) {
int mid = (low + high) >> 1;
if (a[st_min[mid]] * 2 < a[i]) {
low = mid + 1;
} else {
high = mid;
}
}
int nxt = 3 * n;
if (low > 0) {
nxt = min(nxt, st_min[low - 1]);
}
if (!st_max.empty()) {
nxt = min(nxt, st_max.back());
}
if (nxt < 3 * n && a[nxt] >= a[i]) {
ans[i] = ans[nxt];
} else {
ans[i] = nxt;
}
st_min.push_back(i);
st_max.push_back(i);
}
for (int i = 0; i < n; i++) {
if (i > 0) {
cout << " ";
}
cout << (ans[i] == 3 * n ? -1 : ans[i] - i);
}
cout << '\n';
return 0;
}
```

**E. Balanced Binary Search Trees**

Tutorial is loading...

**Solution by tourist**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
int x = 1;
while (x <= n) {
if (n == x || n == x + 1) {
cout << 1 << '\n';
return 0;
}
if (x % 2 == 0) {
x = x + 1 + x;
} else {
x = (x + 1) + 1 + x;
}
}
cout << 0 << '\n';
return 0;
}
```

**F. Balanced Domino Placements**

Tutorial is loading...

**Solution by arsijo**

```
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 998244353;
const int MAX = 3610;
ll dp1[MAX][MAX], dp2[MAX][MAX];
int has1[MAX], has2[MAX];
ll fac[MAX], c[MAX][MAX];
int n, m, k;
void make(int n){
dp1[0][0] = 1;
for (int i = 1; i <= n; i++){
for (int j = 0; j <= m; j++){
dp1[i][j] = dp1[i - 1][j];
if (j && i >= 2 && has1[i] == 0 && has1[i - 1] == 0){
dp1[i][j] += dp1[i - 2][j - 1];
}
dp1[i][j] %= MOD;
}
}
}
int main(){
cin >> n >> m >> k;
int t = max(n, m) + 1;
vector<ll> f(t);
for (int i = 1; i <= k; i++){
int a, b, c, d;
cin >> a >> b >> c >> d;
if (has1[a] || has1[c] || has2[b] || has2[d]){
cout << 0 << endl;
return 0;
}
has1[a] = has1[c] = has2[b] = has2[d] = 1;
}
int N = n, M = m;
for (int i = 1; i <= n; i++){
N -= has1[i];
}
for (int i = 1; i <= m; i++){
M -= has2[i];
}
make(n);
swap(dp1, dp2);
swap(has1, has2);
make(m);
swap(dp1, dp2);
swap(has1, has2);
c[0][0] = 1;
f[0] = 1;
for (int i = 1; i < t; i++){
f[i] = (f[i - 1] * i) % MOD;
c[i][0] = c[i][i] = 1;
for (int j = 1; j < i; j++){
c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
if (c[i][j] >= MOD){
c[i][j] -= MOD;
}
}
}
ll ans = 0;
for (int i = 0; (i << 1) <= N; i++){
for (int j = 0; (j << 1) <= M; j++){
int have1 = N - 2 * i;
int have2 = M - 2 * j;
ll res = dp1[n][i] * dp2[m][j];
res %= MOD;
res *= c[have1][j];
res %= MOD;
res *= c[have2][i];
res %= MOD;
res *= f[i];
res %= MOD;
res *= f[j];
res %= MOD;
ans += res;
}
}
ans %= MOD;
cout << ans << endl;
}
```

**G. Balanced Distribution**

Tutorial is loading...

**Solution by KAN**

```
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define eprintf(...) 42
#endif
using ll = long long;
using ld = long double;
using D = double;
using uint = unsigned int;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
const int maxn = 200005;
const int maxk = 19;
const int inf = 1e9;
ll a[2 * maxn];
ll psum[2 * maxn];
pair2<int> go[maxk][2 * maxn];
map<ll, int> lst[maxn];
int id[2 * maxn];
int n, k;
ll need;
vector<pair<int, vector<ll>>> answer;
void perform(int l)
{
int fn = min(l + k, n);
assert(psum[fn] >= 0);
ll wassum = psum[fn];
vector<ll> exchange = vector<ll>(k, need);
int start = min(n - k, l);
for (int i = start; i < l; i++) exchange[i - start] = need + a[i];
exchange[l - start] += -psum[l];
exchange.back() += psum[fn];
answer.pb({start, exchange});
for (int i = l; i < fn; i++) a[i] = 0;
a[l] += -psum[l];
a[fn - 1] += psum[fn];
for (int i = start; i < start + k; i++) psum[i + 1] = psum[i] + a[i];
assert(wassum == psum[fn]);
}
void restore(int l)
{
if (l >= n) return;
if (a[l] == 0 && psum[l] == 0)
{
restore(l + 1);
return;
}
int fn = min(l + k, n);
if (psum[fn] >= 0)
{
perform(l);
restore(l + k - (psum[fn] > 0));
return;
} else
{
restore(l + k - 1);
assert(psum[fn] == 0);
perform(l);
}
}
int main()
{
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
need = accumulate(a, a + n, 0LL);
assert(need % n == 0);
need /= n;
for (int i = 0; i < n; i++) a[i] -= need;
for (int i = 0; i < n; i++) a[i + n] = a[i];
partial_sum(a, a + 2 * n, psum + 1);
pair2<int> ans = {inf, -1};
for (int j = 0; j < maxk; j++) go[j][2 * n] = {2 * n, 0};
for (int i = 2 * n - 1; i >= 0; i--)
{
int curid = id[i + 1];
if (lst[curid].count(psum[i]))
{
int to = lst[curid][psum[i]];
go[0][i] = {to, (to - i - 1) / (k - 1)};
} else
{
int to = 2 * n;
go[0][i] = {to, (to - i + k - 2) / (k - 1)};
}
for (int j = 1; j < maxk; j++) go[j][i] = {go[j - 1][go[j - 1][i].fi].fi, go[j - 1][i].se + go[j - 1][go[j - 1][i].fi].se};
if (i < n)
{
int to = n + i;
int cur = i;
int curans = 0;
for (int j = maxk - 1; j >= 0; j--) if (go[j][cur].fi <= to)
{
curans += go[j][cur].se;
cur = go[j][cur].fi;
}
if (cur < to) curans += (to - cur - 1 + k - 2) / (k - 1);
ans = min(ans, {curans, i});
}
id[i] = (2 * n - i) % (k - 1);
lst[id[i]][psum[i]] = i;
}
rotate(a, a + ans.se, a + n);
partial_sum(a, a + n, psum + 1);
restore(0);
assert((int)answer.size() == ans.fi);
printf("%d\n", (int)answer.size());
for (auto &t : answer)
{
printf("%d", (t.fi + ans.se) % n);
for (auto tt : t.se) printf(" %lld", tt);
printf("\n");
}
return 0;
}
```

**H. Balanced Reversals**

Tutorial is loading...

**Solution by KAN**

```
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define eprintf(...) 42
#endif
using ll = long long;
using ld = long double;
using D = double;
using uint = unsigned int;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
const int maxn = 4005;
vector<int> ss, tt;
char s[maxn], t[maxn];
int n;
vector<int> answer;
int cnt[5];
vector<int> prepare(char *s)
{
vector<int> ans;
for (int i = 0; i < 2 * n; i += 2) ans.pb((s[i] - '0') * 2 + (s[i + 1] - '0'));
return ans;
}
inline void addanswer(int x)
{
if (x <= 0) return;
answer.pb(x);
}
void doreverse(vector<int> &v, int l)
{
reverse(v.begin(), v.begin() + l);
for (int j = 0; j < l; j++) if (v[j] >= 1 && v[j] <= 2)
{
v[j] = 3 - v[j];
}
}
int main()
{
int NT;
scanf("%d", &NT);
for (int T = 0; T < NT; T++)
{
scanf("%s%s", s, t);
n = strlen(s) / 2;
ss = prepare(s);
tt = prepare(t);
for (auto &t : tt) if (t == 1 || t == 2) t = 3 - t;
cnt[0] = cnt[1] = cnt[2] = cnt[3] = 0;
for (auto t : ss) cnt[t]++;
for (auto t : tt) cnt[t]--;
if (cnt[0] != 0 || cnt[3] != 0)
{
printf("-1\n");
continue;
}
bool found = cnt[1] == 0;
int before = -1;
int after = -1;
for (int i = 1; i <= n && !found; i++)
{
cnt[ss[i - 1]]--;
cnt[3 - ss[i - 1]]++;
if (cnt[1] == 0)
{
before = 2 * i;
doreverse(ss, i);
found = true;
}
}
if (!found)
{
for (int i = 1; i <= n; i++)
{
cnt[ss[i - 1]]++;
cnt[3 - ss[i - 1]]--;
}
}
for (int i = 1; i <= n && !found; i++)
{
cnt[tt[i - 1]]++;
cnt[3 - tt[i - 1]]--;
if (cnt[1] == 0)
{
after = 2 * i;
doreverse(tt, i);
found = true;
}
}
assert(found);
answer.clear();
addanswer(before);
for (int i = n - 1; i >= 0; i--)
{
int wh = -1;
for (int j = n - 1 - i; j < n; j++) if (ss[j] == tt[i]) wh = j;
addanswer(2 * wh);
doreverse(ss, wh);
addanswer(2 * wh + 2);
doreverse(ss, wh + 1);
}
addanswer(after);
for (auto t : answer) reverse(s, s + t);
assert(strcmp(s, t) == 0);
printf("%d ", (int)answer.size());
for (auto t : answer) printf("%d ", t);
printf("\n");
}
return 0;
}
```

tourist Just curious to know, What is the Checker code logic for C?

If you mean arsijo's code for C1, it picks point $$$j$$$ as the closest point to $$$i$$$ using Manhattan metric.

cool. in future we may see problem A using FFT.

Manhattan distance just means $$$|\Delta x| + |\Delta y|$$$.

manish_joshi i see :) hehehe

tourist No. I mean. There are multiple solutions possible for C. So How are you testing whether the solution is correct or wrong?

I guess it's KD-Tree or bitset. :D

Ah, you mean checker. It can probably be implemented in $$$O(n \log^3 n)$$$ and even $$$O(n \log^2 n)$$$, but I did a straighforward $$$O(n^2)$$$, a bit optimized to make it work in about 2 seconds.

tourist can you please explain the Solve function you used on your solution to c2?

`Solve(ids, k)`

stands for "please match points with indices in`ids`

, given that their coordinates in the first`k`

dimensions are exactly the same, and return the id of the only unmatched point, or -1 if`ids`

was of even length".Thanks!

editorials code was hard for me to understand . but this code

is very easy to understand according to editorial

better solution

baal_dao_pari_nacode referred by you is easy to understand but that has dimension limitations whereas editorials code can be expanded to any dimensions.How do you solve such difficult problems tourist?

plz work hard my friend

Test Cases are still not visible.

UPD: It is visible now.

Here's my solution for E with proof:

Perfectly balanced tree must have the condition that every level, except possibly the last, must be completely filled.

What's the condition for striped BSTs? Let's find the condition for each child to match condition of its parent. So, if it's a left child, it should have key parity different than the parent and if it's a right child, it should have key parity different than the parent. Now, observe that for a left child $$$u$$$,

This tells us that every node which is a left child of its parent must have even number of nodes in its right subtree. Similary for every right child $$$u$$$,

This tells us that every node which is a right child of its parent must have odd number of children in its left subtree. With these two observations, we can find number of possible subtrees of height $$$h$$$ like this,

For height $$$1$$$, there is only one tree, one node with no children. For height $$$2$$$, there is only one tree, one node with one child on its left and no child on its right.

For each height $$$h$$$, let's store all the possible trees as a pair $$$(a,b)$$$ where $$$a$$$ is the number of nodes in the left subtree of the root and $$$b$$$ is the number of children in the right subtree of the root. Now, we observe the following simple pattern:

For height $$$1$$$ possible trees: $$$(0,0)$$$, possible sizes: $$$1$$$

For height $$$2$$$: $$$(1,0)$$$, possible sizes: $$$2$$$

For height $$$3$$$: $$$(1, 2)$$$, $$$(2,2)$$$, possible sizes: $$$4$$$, $$$5$$$

For height $$$4$$$: $$$(4, 4)$$$, $$$(5,4)$$$, possible sizes: $$$9$$$, $$$10$$$

For height $$$5$$$: $$$(9,10)$$$, $$$(10,10)$$$, possible sizes: $$$20$$$, $$$21$$$

Can you spot the pattern and prove it? For every height $$$\geq 3$$$, there are exactly two trees possible.

Proof: If for height $$$i$$$ we have possible trees $$$(a,x)$$$ and $$$(b,x)$$$ where $$$x$$$ is even, $$$a$$$ is even and $$$b$$$ is odd, then for height $$$i+1$$$, we can have the right child only as $$$(b,x)$$$ as only then then number of nodes in its left subtree will be even. But the left child can have any of the two values as both satisfy the condition of right subtree having even size. Thus, the new trees are $$$(a+x+1,b+x+1)$$$ and $$$(b+x+1,b+x+1)$$$, which can then be replaced with new $$$A$$$, $$$B$$$ and $$$X$$$ having $$$A = b + x + 1$$$, $$$B = a + x + 1$$$ and $$$X = b + x + 1$$$. Thus, we can say that the pattern holds by induction.

Base case has $$$i = 3$$$, which is satisfied as $$$a = 2$$$, $$$b=1$$$ and $$$x=2$$$.

So, find all possible trees till height $$$\log(\text{max value of n})$$$ and output $$$1$$$ if $$$n$$$ is one of those values and output $$$0$$$ if it isn't.

AC Code: 62737057

IMO the easiest way to solve this is to come up with the brute force DP:

But notice that the sizes of the left and right subtrees can only be floor and ceil (n-1)/2 to be perfectly balanced, so the summation really only has one or zero terms.

Consider a tree with 11 nodes:

This is perfectly balanced. It has 7 in the left part and 3 in the right. Am I missing something?

It's not a BST.

I did not intend for it to be a BST. Replace the numbers with arbitrary numbers, I just wanted to exhibit a structure of a perfectly balanced binary tree that does not satisfy the floor/ceil condition as mentioned in the parent comment. The point was that, if the parity conditions were absent, the dp does not have a single term.

It's not even BST and even more, it doesn't satisfy parity conditions xD.

So, if a BST is perfectly balanced, then its subtrees can only be floor and ceil (n-1)/2, or it must also satisfy parity condition ? Can you please explain a little bit, I can not get the concept behind though ><

According to the definition, it should have minimum sum of depths. BST with minimum sum of depths should have all leaves on the last level or the last but one level. Now we can deduce that a perfect BST of height $$$h$$$ has two perfectly balanced subtrees of height $$$h-1$$$.

Then apply the parity condition to inductively find all striped perfect BST.

Wow, yeah I have no idea how I thought this was correct. Seems I got lucky that the trees asked about in the question actually do satisfy this

LOL, is that the legendary intuition ?

“Perfectly balanced tree must have the condition that every level, except possibly the last, must be completely filled.“Can you prove this? This statement seems incorrect.

I’m facing a special problem that actually I want to turn off the cf code editor and I don’t know how to do that. Can anybody help me please?

Am I only one,who solved B with BIT in O(nlogn)?

I just solved B with ordered set. I feel really stupid now.

https://codeforces.com/contest/1237/submission/62739056

what is bit can you please explain

It's Binary Indexed Tree aka Fenwick tree.

You're not alone :)62707885

You are not alone too :')

https://codeforces.com/contest/1237/submission/62752373

Can you briefly explain how to do that with BIT tree?

When a car is exitting, if the number of cars in front of it is less than the number when it's entering, that means it overtook some cars.

But, there can be the case that there are same no. of cars in front of it even after exiting even though it has overtaken some cars.

Only consider cars which enter earlier than that car.

here's a similar problem from spoj https://www.spoj.com/problems/INVCNT/

If you use lower_bound to find the minimum , it just can be slove be in O(nlogn) xd

Problem C is great. I just loved it.

In problem C2, I just wonder is Solve is a recursive function ?

it is my first time to see a thing like this

It's a lambda expression.

The intended solution of H is very nice!

Do you know who did this during the contest? It seems the submissions are still invisible.

Thanks! Unfortunately, it seems that everyone who solved the problem during the contest used some kind of randomization.

Submissions must be visible now.

Sorry to say that, but it seems you did not follow the rule "if you can't create good tests then don't use this problem". It was kind of obvious to me that some randomized solutions could be created here, but I thought this would be too naive to code it since they seemed pretty simple. As a contestant I did not think carefully how I would exactly design one and how I would ensure this doesn't pass if I were a problemsetter (I decided it's a better decision for me to fight my today's archenemy — problem C xD), but it's always really sad to see hard problem with no legit solutions accepted where many heuristics passed. At least the problem itself is nice :)

Hi, how can randomization be used to solve this problem? I'm just curious, as it wasn't obvious to me.

I was almost there (62731786), but couldn't quite figure out in time how to make the initial situation 'handy' (fixed after contest: 62742219). Fortunately the fifteenth place was (barely) enough for me to reach nutella :).

I tried to do something similar to this during the contest but my last submission for it was missing 2 characters :(

In problem E, I used DP: $$$dp[i][par_{root}][par_{size}]$$$ means the polynomial where the coefficient of $$$x^j$$$ is the number of ways to build a tree such that the depth is $$$i$$$ and the parity of root value is $$$par_{root}$$$, and the parity of the size value is $$$par_{size}$$$. The answer we want is the sum of the $$$n$$$-th term in $$$dp[d][0/1][n\text{ mod }2]$$$.

Then we have $$$dp[i][r][s] = multpoly(dp[i][\neg r][\neg r], dp[i][0][s \oplus r])$$$. Then I used NTT to do the multiplication and solved it barely fitting in the TL.

Now I realize the base cases are all polynomials with only 0/1 term! Which means we can just store an number for each polynomial, and do the multiplication just by adding those numbers! This will run in $$$O(\log n)$$$, and this offers an different way to prove the model solution of this problem.

Another solution of B could be like this: Let A = ingoing car array and B = outgoing car array. Consider indices i and j running over A and B respectively. If Ai == Bj, the current car is in right order. So we insert Ai to a hash_set and we increment i and j such that Ai and Bj aren't in the hash_set. If Ai != Bj, Bj should be fined. We increase fined_car_counter by one and insert Bj to the hash_set. Then we increase j such that Bj is again not in the hash_set. The loop stops if either of i and j hits the end. Works in O(n) time.

if both the arrays are equal will the solution work??

Yes, that'd mean no one overtook. You never hit the 2nd case. Answer is 0. This is my solution: https://codeforces.com/contest/1237/submission/62717631

Please make video solution for problem D,E,F

tourist Wow, solution to H is actually easier to understand than E and F to me (both needing induction/maths of some kind). Thank you for the great contest! Though your coordinators still have not been announced :O

I overkilled D. I used merge sort tree+ segment tree to solve D. It took me almost 2 hours to implement.

Question C1 : O(n^2) solution was not getting accepted using Python 3 even after using fast input and output.

For Problem D, we can find the monotonicity of the position to stop. That is, let $$$s_i$$$ be the position to stop if song $$$i$$$ is the first played song. Then for $$$i = 2, ..., n$$$, it holds $$$s_i \geq s_{i-1}$$$. Here we assume the list is cyclic.

So we can use two pointers to loop over the stopping songs for each starting song. To determine if a song can be added into the current playlist, we need to keep the maximum element of the sliding window. Using STL set can achieve $$$O(nlogn)$$$ time and we can further apply monotonic queue to achieve $$$O(n)$$$ time.

My submission: 62706780. I used STL set in contest, for simplicity of code.

I had the same idea as you as well! Looking at your code, one thing that you might not know is that you don't actually need a multiset of pairs to delete one instance of a number; if you do s.erase(s.find(x)) it will only erase one instance (I believe the first) of x in the multiset.

My code: 62729086

Note: I used %N to eliminate the need for explicitly extending the input array.

Cheers!

I had realized it after the contest XD

In contest I wanted to implement monotonic queue at first, so the pair existed, but later I changed my mind to use STL set. I wrote another simpler version after contest.

EDIT:If anyone interested with the simplified version, here it is: 62751793In Problem A in C++ i'm getting WA for cout<<floor(arr[i]/2.0) and Ac for cout<<(long long int)floor(arr[i]/2.0) , Why?(floor and ceil returns nearest integer!

it returns -0 for few cases so u need to remove '-'

You may want to try the result of following code:

Thanks a lot

1237B — Balanced Tunnel

"Specifically, can $$$a_i$$$ must be fined if $$$c_i$$$ is bigger than $$$max(c_1,c_2,…,c_{i−1})$$$"

I think, it should be "smaller" — not "bigger". And there is a typo — "can" was meant to be "car".

You're right, thanks.

Another solution for D: split the array into sqrt-size blocks and preprocesses whether the first track of each block can go to all tracks in the block, and preprocess the minimum and maximum of each block. Then iterate over every i from 0 to n-1, and for each i go skipping every block that you can hear all the tracks (keeping the current maximum and checking the minimum of the block).

You can hear the whole block if the first track of the block can hear all the block and

`(current_maximum + 1) / 2 <= min_block`

Is this called

`Square Root Decomposition`

?Yes.

tourist For problem C, what I did was to first use a stable sort on x of all points, then on y and then on z. Now taking central pairs from this sorted array was supposed to give me the solution but it failed on a test case. Do you think the order of coordinates in the stable sort matters? I mean should I do it like first on z, then on y and then on x?

The order of coordinates definitely doesn't matter. Consider a simpler test case:

Your solution removes points 2 and 3 first, but they can't be removed due to point 4.

Now I see. Got your point. Thanks.

The cycle formation part in D is not clear.

Isn't it enough to repeat the sequence 2 times ? Please help !

Somewhat funny, slightly different solution for H.

Probably it's more tempting to try to match the last two characters of $$$a$$$ and the last two characters of $$$b$$$. By doing operations like

`abcdefxyghij -> yxfedcbaghij -> jihgabcdefxy`

on either $$$a$$$ or $$$b$$$, we can usually achieve this. The only undesirable situation is when $$$a$$$ ends with $$$01$$$, $$$b$$$ ends with $$$10$$$, $$$a$$$ contains no $$$10$$$, and $$$b$$$ contains no $$$01$$$ (or vice versa).However, this is rather a handy situation in the intended solution! So, if we switch to the intended solution at this point, everyone will be happy. We can even save one more operation.

please can someone provide a simple solution for C2 ,..... I got the approach for it as in editorial but coudn't understand the implementation... please

Spent 2 hours finding tricky cases for A and it happened to be -0 problem because of double. Don't wanna live anymore)

tourist participated in 153 cf rounds and came first in 77 (=ceil(153/2)).

perfectly balancedIt doesn't sum to zero????? Disappointing

did you count one by one in his rank list? BTW nice observation.

Bonus for D: Use a queue/ Double-ended queue:

For i = 1 to 2*n (you can cycle as many as you want, but I think 2 times is enough):

While a[Q.back()] < a[i]: Q.pop_back() and Unify (Q.back(), i). Here means if you start playing from track Q.back(), you can play track i which has greater coolness, so the result of Q.back() should be the same with the result of i.

Push track i to the back of the Deque

Because tracks in the Deque are sorted non-increasingly, so if you push track i to the back of the Deque then you should pop out track from the front of the queue. Suppose j = Q.front() then j is popped out from the front if and only if a[j] > 2*a[i]. Which means if you play from track Q.front() then you must stop at track i.

After the loop, any tracks left in the Queue is UNSTOPPABLE.

By using Disjoint Set Union, every track in the same connected component has the same result. To maximize, we take the result of the coolest track in that connected component.

Finally, calculate the answer. The overall complexity is O(n) (Or close enough, since I use DSU)

Submission 62720568

yes you should cycle atmost 2 times in the worst case in the first cycle you will go through maximum number in the array and in second cycle you should find its answer if you can't find answer then answer is infinity.

Why so many downvotes hmm?

This solution looks awesome,can you please explain the logic eloborately?

How to prove this code?

This $$$O(1)$$$ code is amazing...

WHAT THE CODE !!!

Is there anyone who wrote FFT in E?

let $$$f(i,j,op)$$$ be the quantity of $$$2^i-1+j$$$-nodes perfectly balanced striped binary search trees, the parity of the key of the root is $$$op$$$. $$$0\le j<2^i$$$.

a perfectly balanced striped binary search tree rooted at $$$u$$$ should satisfy:

(1) the subtree size of $$$lc[u]$$$ and the key of $$$lc[u]$$$ have the same parities.

(2) in the subtree of $$$rc[u]$$$, rank of $$$rc[u]$$$ is even.

Consider array $$$a$$$ and $$$b$$$:

if $$$j$$$ is odd, $$$a_j=0$$$, $$$b_j=f(i-1,j,0)$$$ ;

if $$$j$$$ is even, $$$a_j=f(i-1,j,1)$$$, $$$b_j=0$$$ .

Then we can get:

$$$f(i,j,0)=\sum_{x+y=j}a_xf(i-1,y,0)$$$

$$$f(i,j,1)=\sum_{x+y=j}b_xf(i-1,y,0)$$$

Noticed $$$0\le j<2^i$$$, use NTT, it can be solved in $$$O(i\times 2^i)$$$.

if we let $$$k$$$ be the maxium value that $$$2^k-1\le n$$$, then the answer is $$$f(k,n-2^k+1,0)+f(k,n-2^k+1,1)$$$.

the complexity is $$$\sum_{i\le\log n}i\times2^i=O(n\log n)$$$.

Any simple implementation for C2 with same logic?

I tried 3D solution with 3 dimensional ArrayList. Here's my JAVA code: 62776171

can you explain your code ,how it is working?

I think a better way of solving B would be using queue,as it follows First-in First-out and then all we need to do is to check if the element of b is equal to front element of queue and mark it as visited and pop.If it is not equal then counter++.Also while front element of queue is marked visited,pop it.Here is my solution.

I also solved B using queue. Also used HashSet for tracking already gone cars. Kept popping queue while it is already in the gone set. Just whenever next going car isn't in hashset and isn't equal to queue front then increased the answer. Here is my JAVA code : 62716214

I did B using Binary search https://codeforces.com/contest/1237/submission/62707665

Why is it "better"?

As there is no need to create an an array c and also this is exactly what the problem says...No extra thinking or logic required.

So you create the "visited" array instead.

This is subjective — the validity of your approach is not so obvious that it does not require justification.

You can consider queue as a tunnel,push as an entry,pop as an exit and the if-else part(in my code) as a guard who takes fine.

Can't consider queue as the tunnel — cars exit not in the same order as they enter.

Typo- Pop as Expected exit.

What is "expected exit"?

Your queue is not really a queue, because you effectively delete (mark for deletion) items in the middle. You pop items marked for deletion when they reach the front of the queue.

Implementation using std::list is shorter and cleaner (and faster):

https://codeforces.com/contest/1237/submission/62834786

And it is all starting to make sense now.

Start with sequence $$$a$$$ of cars entering tunnel. Then the first car exiting is fined iff it is not the first in $$$a$$$. Remove this car from $$$a$$$ and repeat the same with the next car exiting.

I solved D in $$$O(n^{1.5})$$$.

Check my code for details. https://codeforces.com/contest/1237/submission/62726962

Here is an alternate solution to E that does not require the solver to initially conjecture that the answer is either $$$0$$$ or $$$1$$$:

Suppose we fix the parity of the root (it will be the parity of $$$n$$$). A perfectly balanced tree is a complete binary tree of depth $$$d - 1 +$$$ some leaves at depth $$$d$$$. So we can find the parities of all the nodes in the complete binary tree (i.e. for all nodes with depth $$$\leq d - 1$$$). Write them down in a sequence, like, for $$$n = 10$$$, the following will be the sequence: $$$0, 1, 1, 0, 1, 0, 0$$$ (we can compute this sequence using a recursion).

Now, since the final inorder traversal will have parities like $$$1, 0, 1, 0, 1, 0 \ldots$$$ (because the inorder will be $$$1, 2, 3, 4 \ldots$$$), we must insert a $$$0$$$ whenever there are two consecutive $$$1$$$s and a $$$1$$$ whenever there are two consecutive $$$0$$$s. Also note that if $$$j$$$ corresponds to a leaf, $$$A_j \neq A_{j + 1}$$$ (proof follows because the next node in the inorder is just the first right turn while moving up from the leaf, and a right turn means a different parity), and if $$$j$$$ is a non-leaf, then $$$j - 1$$$ corresponds to a leaf (because of complete binary tree-ness) and $$$A_{j - 1} \neq A_j$$$ (similar proof). So we have:

If $$$A_1 = 0$$$, insert a leaf before (in the left) because the first node has to be 1.

If $$$A_j = A_{j - 1}$$$, insert a leaf here (in the left).

Otherwise, don't (can't) insert a leaf here.

All this is compulsory for a perfectly balanced striped tree. So the answer exists only if the number of leaves to be inserted is $$$n - 2^{d - 1} + 1$$$ (i.e., the number of leaves you have).

P.S.: Note that proving $$$A_j \neq A_{j + 1}$$$ for leaves etc. was not compulsory. We could just not think about them during the contest, and set answer = 0 if $$$A_j = A_{j + 1}$$$, because if it occurs, you cannot insert anything to fix it (because, same parity on the right). Also if for non-leaves $$$A_j = A_{j - 1}$$$, then that case would be handled by the leaf at position $$$j - 1$$$ (who would give an answer of 0 as just mentioned). But still, the answer will remain $$$0$$$ or $$$1$$$, without deciphering the structure of the sequence. That part could be left to be handled by the code itself.P.P.S:If I'm not wrong, this also proves that all leaves in the last level should be left children.Submission: 62757866

For problem C2 I implemented the 3D solution. Here is the java code : 62760448 I sorted the points using Comparator.

Can there be a greedy solution for C2 involving sorting using Manhatten or Eucledian Distance and removing in pairs ?

In tutorial of F

" It follows that $$$R=f(h,dv)∗\binom{h−2dv}{dh}$$$ "

Shouldn't we only choose from rows marked 0 instead of all the h rows? I think it should be

" $$$R=f(h,dv)∗\binom{h- (no of filled rows) − 2dv}{dh}$$$"

Here is an alternate solution for D: lets say we want to find the answer for track x : find the first track after x that has a coolness greater than x (lets call it y) also find the first track after x that has a coolness less than x/2 (lets call it z). if z lies between x and y then the answer for track x will be the number of tracks between x and z. otherwise the answer for track x would be number of track between x and y plus the answer for track y (we can calculate the answers recursively).

Yeah, but how are you finding which one among z and y is closer to x and also how are you finding the position of z or y after knowing that?

I have a question of problem E. If n=5, the tree can be like this: 3 / \ 2 5 / \ 1 4 Also, it can be like this: 3 / \ 2 5 / / 1 4 Which tree dose not meet the conditions? I would appriciate it if you could answer my question!

I have a question of problem E.

If n=5, the tree can be like this:

1 4

Also, it can be like this:

1 4

Which tree dose not meet the conditions?

I would appriciate it if you could answer my question!

Your 1st Tree is not BST

OK，I see! Thanks a lot！

Your first tree is not a BST. $$$4$$$ cannot be in the left subtree of $$$3$$$.

OK，I see! Thanks a lot！

I got WA on Pretest 5 in A and the checker says

`wrong output format Expected integer, but "-0" found`

. What does this mean?Casting the output to

`int`

gives AC. If the error was in output format, why did the first 4 pretests pass?Link to code

https://en.cppreference.com/w/cpp/numeric/math/floor

This function doesn't return an integer. Your code passed the first 4 pretests by chance.

Yes, the

`floor`

and`ceil`

functions will return`double`

datatype. But still it doesn't make sense.What does the checker output mean? I mean what is

`"-0"`

and why is it not an integer?Also, as

`1.0`

is printed as`1`

. How does printing`double`

values instead of`int`

make any difference in the output text file? How can the first 4 pretests pass by chance?Thanks for replying szakubki.

`-0`

or`-0.00...0`

is a valid representation of a floating point number. It's very small but negative; it would have non-zero digits, but they're rounded off on the output. When you multiply it by a very large positive float, you can get a reasonably large negative float, so it has a purpose.In integers, minus zero is zero, so

`-0`

isn't a valid representation of an integer. There's no way to obtain the output`-0`

by printing an integer.One of the

bestD!In problem D, how can one do binary search over a segment tree?

In problem D, what is the reason why repeat the numbers 3 times is enough?

In 1237F, I am facing difficulty in understanding the equation of number of ways to place exactly dh horizontal domino and dv vertical Domino. Can someone explain how we arrived at this equation: R⋅C⋅dh!⋅dv!.

This took me a while to figure out too.

To getting a perfectly balanced placement, we need to

Author’s solutions are good, but I know more interesting way to make this world better — checking M3139 homework ^_^

tourist Wonderful round, as always, but something about it clearly require further consideration. Maybe it’s homework of group m3139...

Very wonderful and complicated problems. But I know some more. For example, the problem of checking M3139 homework.

Amazing balanced problems. I think it's time to check M3139 homework 4more balance :3

Thanks a lot to MikeMirzayanov for Codeforces and Polygon, and also thanks to tourist for checking M3139 homework

An alternative approach for problem H: (a bit not elegant, but able to solve the problem in n operations)

First, check the number of 00s, 11s and 01/10s. Then, judge the situation where there're no 01/10s, which can be solved easily in n operations.

Consider we can do operations on both binary strings, as doing a sequence of operations on B is equivalent to doing the reverse of the sequence on A. And think about each time we make the last 2 characters of A and B equal and do the process on the prefix of length n — 2 of A and B.

Then, assume that at least one of the strings begins with 01 or 10.

Then, discuss about these cases:

in B.)"Recall that car ai must be fined if there exists a car that entered the tunnel earlier and exited the tunnel later." tourist's words i think it's the opposite of what's true.a car should be fined if it enters tunnel later and exits earlier. please explain if i'm wrong

can anyone explain the logic of j and k in question D?

For each $$$i$$$, let $$$j$$$ be the closest position after $$$i$$$ such that $$$a_j > a_i$$$, also let $$$k$$$ be the closest position after $$$i$$$ such that $$$a_k < a_i/2$$$. There are 2 cases:

When $$$k$$$ is closer than $$$j$$$, it means that for sure we know that we will play $$$k - i$$$ tracks, so $$$answer_i = k - i$$$.

When $$$j$$$ is closer than $$$k$$$, it means that if we start playing at $$$i$$$, for sure we will not stop until we reach $$$j$$$ (because we would have to stop

at mostat position $$$k$$$, and $$$k$$$ is after $$$j$$$). Since $$$a_j > a_i$$$, the answer for position $$$i$$$ is the distance to position $$$j$$$ plus the answer for position $$$j$$$. $$$answer_i = j - i + answer_j$$$And how do we find the answer for j? I didn't understand how to use binary seach on segment trees. Can you explain the idea?

There are a couple of ways to find $$$j$$$ and $$$k$$$. I used coordinate compression (there will be at most $$$2n$$$ different values), and a segment tree that for each value kept its minimum position so far. So I went from $$$i=n$$$ to $$$i=1$$$ and for each position did 2 queries, and then updated the segment tree for the element at that position.

This is a step up from the most basic usages of segment trees, so if you're not familiar with them you should try solving easier problems.

Did you use coordinate compression on the entire input array or what?

I didn't get your

atmost 2n different valuesstatement. Well, I know about Segment trees but unable to understand its usage in the given problem.I mean, what to store in each node of the segment trees and why and how is it helping me here. This usage of Segment trees is somewhat new to me.

Your explanation will help me understand a newer application of them. So, Kindly help.

What am I compressing? Notice that for the problem, for each $$$i$$$, we are concerned with at most 2 values: $$$a_i$$$ and $$$\lfloor a_i/2 \rfloor$$$ (if $$$a_i$$$ is even, the second value is actually $$$\lfloor a_i/2 \rfloor - 1$$$, we need this for the segment tree queries later). So, at most $$$2n$$$ values will be needed for the segment tree. This is how many leaves it should have.

The segment tree stores, if we go from the last to the first element, the minimum position so far of a certain value we've passed.

Here is my submission with comments: https://codeforces.com/contest/1237/submission/63557702

You built your tree with

`inf`

and now updating the compressed value of the`a[i]`

with`i+n`

.What's the logic in updating with

`i+n`

?You're right, I guess there's no need to build it with inf and then do those updates. When I was trying to solve the problem, I tried to be as fast as possible, so redundancies like that happen.

can anybody tell me whats wrong with below code for A?

can anybody explain ques B approach please

I can tell you my solution (which I personally find more intuitive). I push all cars from the first array in order into a queue. I also have a "visited" array which keeps track of all cars that we have already processed. This gives us that for any 2 cars a and b such that a is in front of b in the queue, a entered the tunnel before b.

Now, we process, the second array in order. For every iteration of i we do the following:

While the car at the front of the queue has already been processed (check this using visited array), pop it from the front of the queue.

If car at index i (of the second array) does not equal the front of the queue, this means car i should be fined. We know this because the q.front() is guaranteed to have entered the tunnel before car i, and i is leaving the tunnel before q.front(). Thus, we add one to a running count, and mark car i as visited.

if the car at index i does equal the front of the queue, pop the element at the front of the queue.

Output the counter at the end.

My code: 62699992

Another same idea lol

Mine: 62694197

o_o lol

I was 100% sure that in D we need to use the fact that end of path is x/2. But this solution is general, the problem could be given as x-c, and path stops when you reach x-c. But I am glad I still got solution even though I was skeptical that I didnt use the fact x/2 at any point except that x/2 < x.

Hi,please down vote me

Can you make this solution to E any shorter? Maybe by using some other programming language?

Hi, Can you explain your solution?

So by brute forcing/looking at the cases where the answer is 1, and their bit representation, you can notice that they all look like 10101010.. and a few random bits at the end.

Now when you multiply that by 3, it will look like 111111... and a few random bits at the end.

So you need to check if 3*n is of form 2^k-1,2^k-2,2^k-3,2^k-4 or 2^k-5, where k is some number.

Now to check this we can do if n^(n+5)>n.

tourist In editorial of problem C2 why is 'n' not necessarily even when in the question it clearly states that 'n' is even?

The 'n' for the 3 dimensions is even. but for 2 or 1 it's not necessary to be even (at last in every dimension is only one point alone and we know that the number(not coordinates) of these alone points is even so we match them two by two and then we're ok

tourist in your solution of prob D, by taking st_min and st_max, we are looking back i.e. already processed tracks(left direction of current track). But we to check right side of current track, isn't it?

tourist In problem D, can u plz elaborate little more why 3times repetition necessary to pretend cyclic list linear, to me it seems that 2times repetition is sufficient

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