Idea: BledDest

**Tutorial**

**Solution (Ne0n25)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
int n;
int a[N];
void solve() {
cin >> n;
for (int i = 0; i < n; ++i)
cin >> a[i];
for (int i = 1; i < n - 1; ++i) {
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
cout << "YES" << endl;
cout << i << ' ' << i + 1 << ' ' << i + 2 << endl;
return;
}
}
cout << "NO" << endl;
}
int main() {
int T;
cin >> T;
while (T--)
solve();
}
```

Idea: adedalic

**Tutorial**

**Solution (adedalic)**

```
fun main() {
val winBy = mapOf('R' to 'P', 'S' to 'R', 'P' to 'S')
repeat(readLine()!!.toInt()) {
val s = readLine()!!
val maxCnt = s.groupingBy { it }.eachCount().maxBy { it.value }!!.key
println("${winBy[maxCnt]}".repeat(s.length))
}
}
```

Idea: Roms

**Tutorial**

**Solution (Roms)**

```
for _ in range(int(input())):
n, x = map(int, input().split())
a = sorted(list(map(int, input().split())), reverse=True)
res, cur = 0, 1
for s in a:
if s * cur >= x:
res += 1
cur = 0
cur += 1
print(res)
```

Idea: Roms

**Tutorial**

**Solution (Roms)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = int(2e5) + 9;
int n, m;
long long x, k, y;
int a[N];
int b[N];
bool upd(int l, int r, long long &res) {
if (l > r) return true;
bool canDel = false;
int mx = *max_element(a + l, a + r + 1);
int len = r - l + 1;
if (l - 1 >= 0 && a[l - 1] > mx) canDel = true;
if (r + 1 < n && a[r + 1] > mx) canDel = true;
if (len < k && !canDel) return false;
int need = len % k;
res += need * y;
len -= need;
if (y * k >= x) {
res += len / k * x;
} else if(canDel) {
res += len * y;
} else {
res += (len - k) * y + x;
}
return true;
}
int main(){
scanf("%d %d", &n, &m);
scanf("%lld %lld %lld", &x, &k, &y);
for (int i = 0; i < n; ++i) scanf("%d", a + i);
for (int i = 0; i < m; ++i) scanf("%d", b + i);
long long res = 0;
int lst = -1, posa = 0, posb = 0;
while (posb < m) {
while(posa < n && a[posa] != b[posb]) ++posa;
if (posa == n) {
puts("-1");
return 0;
}
if (!upd(lst + 1, posa - 1, res)) {
puts("-1");
return 0;
}
lst = posa;
++posb;
}
if (!upd(lst + 1, n - 1, res)) {
puts("-1");
return 0;
}
printf("%lld\n", res);
return 0;
}
```

**Tutorial**

**Solution 1 (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
const int N = 200043;
const int L = 20;
vector<int> g[2 * N];
int p[2 * N];
int up[2 * N][L];
int idx[N];
int psum[N];
int cur[N];
int tin[2 * N];
int tout[2 * N];
int T = 0;
void dfs(int x, int y)
{
tin[x] = T++;
p[x] = y;
up[x][0] = y;
for(int i = 1; i < L; i++)
up[x][i] = up[up[x][i - 1]][i - 1];
for(auto z : g[x])
dfs(z, x);
tout[x] = T++;
}
bool is_ancestor(int x, int y)
{
return tin[x] <= tin[y] && tout[x] >= tout[y];
}
int lca(int x, int y)
{
if(is_ancestor(x, y))
return x;
for(int i = L - 1; i >= 0; i--)
if(!is_ancestor(up[x][i], y))
x = up[x][i];
return p[x];
}
int main()
{
int n, m;
scanf("%d %d", &n, &m);
for(int i = 0; i < n; i++)
{
scanf("%d", &idx[i]);
idx[i]--;
}
for(int i = 0; i < m; i++)
cur[i] = i;
for(int i = 0; i < m - 1; i++)
{
int x, y;
scanf("%d %d", &x, &y);
--x;
--y;
int nidx = m + i;
g[nidx].push_back(cur[x]);
g[nidx].push_back(cur[y]);
cur[x] = nidx;
}
int root = m * 2 - 2;
dfs(root, root);
for(int i = 0; i < n - 1; i++)
{
int t = lca(idx[i], idx[i + 1]);
if(t < m)
psum[0]++;
else
psum[t - m + 1]++;
}
for(int i = 0; i < m - 1; i++)
psum[i + 1] += psum[i];
for(int i = 0; i < m; i++)
printf("%d\n", n - 1 - psum[i]);
}
```

**Solution 2 (pikmike)**

```
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
int n, m;
vector<int> p;
vector<vector<int>> val;
vector<int> who;
int getp(int a){
return a == p[a] ? a : p[a] = getp(p[a]);
}
int main(){
scanf("%d%d", &n, &m);
p.resize(m);
val.resize(m);
who.resize(n);
int ans = n - 1;
forn(i, m)
p[i] = i;
forn(i, n){
int x;
scanf("%d", &x);
--x;
who[i] = x;
ans -= (i && who[i] == who[i - 1]);
val[who[i]].push_back(i);
}
printf("%d\n", ans);
forn(i, m - 1){
int v, u;
scanf("%d%d", &v, &u);
v = getp(v - 1), u = getp(u - 1);
if (val[v].size() < val[u].size())
swap(v, u);
for (int x : val[u]){
if (x) ans -= who[x - 1] == v;
if (x < n - 1) ans -= who[x + 1] == v;
}
for (int x : val[u]){
val[v].push_back(x);
who[x] = v;
}
p[u] = v;
printf("%d\n", ans);
}
return 0;
}
```

Idea: Roms

**Tutorial**

**Solution 1 (pikmike)**

```
#include <bits/stdc++.h>
using namespace std;
#define forn(i, n) for(int i = 0; i < int(n); i++)
const int MOD = 998244353;
const int N = 500 * 1000 + 13;
string s;
int add(int a, int b){
a += b;
if (a >= MOD)
a -= MOD;
return a;
}
int mul(int a, int b){
return a * 1ll * b % MOD;
}
struct node{
int val[4], len;
};
node merge(const node &a, const node &b, int l, int r){
node c;
int na = a.len == 1 ? 0 : 1;
int nb = b.len == 1 ? 0 : 2;
c.len = a.len + b.len;
c.val[0] = mul(a.val[na], b.val[nb]);
c.val[1] = mul(a.val[na], b.val[3]);
c.val[2] = mul(a.val[3], b.val[nb]);
c.val[3] = mul(a.val[3], b.val[3]);
if (l == 1){
na = a.len == 1 ? 2 : 0;
nb = b.len == 1 ? 1 : 0;
c.val[na + nb] = add(c.val[na + nb], mul(mul(a.val[0], b.val[0]), 19 - (l * 10 + r)));
c.val[1 + na] = add(c.val[1 + na], mul(mul(a.val[0], b.val[1]), 19 - (l * 10 + r)));
c.val[2 + nb] = add(c.val[2 + nb], mul(mul(a.val[2], b.val[0]), 19 - (l * 10 + r)));
c.val[3] = add(c.val[3], mul(mul(a.val[2], b.val[1]), 19 - (l * 10 + r)));
}
return c;
}
node t[4 * N];
void build(int v, int l, int r){
t[v].len = r - l;
if (l == r - 1){
t[v].val[0] = 1;
t[v].val[3] = s[l] + 1;
return;
}
int m = (l + r) / 2;
build(v * 2, l, m);
build(v * 2 + 1, m, r);
t[v] = merge(t[v * 2], t[v * 2 + 1], s[m - 1], s[m]);
}
void upd(int v, int l, int r, int x, int val){
if (l == r - 1){
s[l] = val;
t[v].val[3] = s[l] + 1;
return;
}
int m = (l + r) / 2;
if (x < m)
upd(v * 2, l, m, x, val);
else
upd(v * 2 + 1, m, r, x, val);
t[v] = merge(t[v * 2], t[v * 2 + 1], s[m - 1], s[m]);
}
int main(){
int n, m;
scanf("%d%d", &n, &m);
static char buf[N];
scanf("%s", buf);
s = buf;
forn(i, n) s[i] -= '0';
memset(t, 0, sizeof(t));
build(1, 0, n);
forn(i, m){
int x, v;
scanf("%d%d", &x, &v);
--x;
upd(1, 0, n, x, v);
printf("%d\n", t[1].val[3]);
}
}
```

**Solution 2 (Roms)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = int(5e5) + 9;
const int MOD = 998244353;
int mul(int a, int b) {
return (a * 1LL * b) % MOD;
}
void add(int &a, int x) {
a += x;
a %= MOD;
}
int bp(int a, int n) {
int res = 1;
for (; n > 0; n >>= 1) {
if (n & 1) res = mul(res, a);
a = mul(a, a);
}
return res;
}
int inv(int a) {
int ia = bp(a, MOD - 2);
assert(mul(a, ia) == 1);
return ia;
}
int n, m;
string s;
int dp[N][10];
int idp[N][10];
set <pair<int, int> > lr;
int res = 1;
void updRes(int l, int r, int c) {
assert(l <= r);
if (c == 1) {
assert(!lr.count(make_pair(l, r)));
lr.insert(make_pair(l, r));
res = mul(res, dp[r - l + (r + 1 != n)][r + 1 == n? 1 : s[r + 1] - '0']);
} else {
assert(lr.count(make_pair(l, r)));
lr.erase(make_pair(l, r));
res = mul(res, inv(dp[r - l + (r + 1 != n)][r + 1 == n? 1 : s[r + 1] - '0']));
}
}
void getLR(int &l, int &r, int pos) {
l = r = -1;
auto it = lr.lower_bound(make_pair(pos, n));
if(it == lr.begin()) return;
--it;
if(!(it->first <= pos && pos <= it->second)) return;
l = it->first, r = it->second;
}
char buf[N];
int main(){
scanf("%d %d\n", &n, &m);
scanf("%s", buf);
s = string(buf);
for (int i = 0; i <= 9; ++i) {
dp[0][i] = i + 1; //idp[0][i] = inv(dp[0][i]);
dp[1][i] = 2 * (i + 1) + (9 - i); //idp[1][i] = inv(dp[1][i]);
}
for (int i = 2; i < N; ++i)
for (int j = 0; j <= 9; ++j) {
dp[i][j] = (2LL * dp[i - 1][j] + 8LL * dp[i - 2][j]) % MOD;
//idp[i][j] = inv(dp[i][j]);
}
for (int i = 0; i < N; ++i) for(int j = 0; j < 10; ++j) assert(dp[i][j] != 0);
for (int l = 0; l < n; ) {
int r = l;
while(r < n && s[r] == '1') ++r;
res = mul(res, dp[r - l - (r == n)][r == n? 1 : s[r] - '0']);
if (l != r) lr.insert(make_pair(l, r - 1));
l = r + 1;
}
for (int i = 0; i < m; ++i) {
int pos;
char x;
scanf("%d %c\n", &pos, &x);
--pos;
if (x == '1') {
if (s[pos] != '1'){
int l1, r1, l2, r2;
getLR(l1, r1, pos - 1);
getLR(l2, r2, pos + 1);
int l = pos, r = pos;
if (l1 != -1) {
assert(r1 == pos - 1);
updRes(l1, r1, -1);
l = l1;
} else {
res = mul(res, inv(dp[0][s[pos] - '0']));
}
if (l2 != -1) {
assert(l2 == pos + 1);
updRes(l2, r2, -1);
r = r2;
} else {
if (pos + 1 != n) res = mul(res, inv(dp[0][s[pos + 1] - '0']));
}
s[pos] = x;
updRes(l, r, 1);
}
} else {
if (s[pos] != '1') {
if (pos - 1 >= 0 && s[pos - 1] == '1') {
int l, r;
getLR(l, r, pos - 1);
updRes(l, r, -1);
s[pos] = x;
updRes(l, r, 1);
} else {
res = mul(res, dp[0][x - '0']);
res = mul(res, inv(dp[0][s[pos] - '0']));
s[pos] = x;
}
} else {
int l, r;
getLR(l, r, pos);
assert(l != -1 && r != -1);
updRes(l, r, -1);
s[pos] = x;
if (l == r) {
res = mul(res, dp[0][s[pos] - '0']);
if (pos + 1 < n) res = mul(res, dp[0][s[pos + 1] - '0']);
} else if (pos == l) {
res = mul(res, dp[0][s[pos] - '0']);
updRes(l + 1, r, 1);
} else if (pos == r) {
if (pos + 1 != n) res = mul(res, dp[0][s[pos + 1] - '0']);
updRes(l, r - 1, 1);
} else {
updRes(l, pos - 1, 1);
updRes(pos + 1, r, 1);
}
}
}
printf("%d\n", res);
}
}
```

Idea: BledDest

**Tutorial**

At first, let's say that the expected value is equal to the average of total earnings over all positions and is equal to the sum of earnings over all positions divided by $$$n$$$. So we can trasition to minimizing the sum.

Let's learn how to solve the task for some fixed $$$k$$$. Fix some arrangement and rotate the rooms so that the last room contains a mimic. So now you have $$$cnt_1$$$ regular chests, then a single mimic, $$$cnt_2$$$ regular chests, single mimic, $$$\dots$$$, $$$cnt_k$$$ regular chests, single mimic. All $$$cnt_i \ge 0$$$ and $$$\sum \limits_{i=1}^{k} cnt_i = n - k$$$.

Take a look at some of these intervals of length $$$cnt_i$$$. The last chest in the interval is taken from $$$cnt_i$$$ starting positions, the second-to-last is taken $$$cnt_i - 1$$$ times and so on.

Now let's find the optimal way to choose $$$cnt_i$$$. Fix some values of $$$cnt_i$$$. Take a look at the smallest of these values and the largest of them. Let the values be $$$x$$$ and $$$y$$$. If they differ by at least $$$2$$$ ($$$x \le y - 2$$$), then the smaller result can always be achieved by moving a regular chest from the larger one to the smaller one.

**Proof**

Consider two sequences of coefficients for both intervals: $$$[1, 2, \dots, x - 1, x]$$$ and $$$[1, 2, \dots, y - 1, y]$$$.

However, if you remove one chest, then they will be equal to $$$[1, 2, \dots, x - 1, x, x + 1]$$$ and $$$[1, 2, \dots, y - 1]$$$.

If you only consider the difference between the numbers of both sequences, then you can see that only coefficient $$$y$$$ got removed and coefficient $$$x + 1$$$ was added. So you can rearrange the chests in such a way that all chests are assigned to the same value and only the chest that was assigned to $$$y$$$ becomes assigned to $$$x+1$$$, thus decreasing the total value.

Now we have all $$$cnt_i$$$ set now. The only thing left is to assign chests optimally. Write down the union of all the coefficient sequences from all the intervals $$$\bigcup \limits_{i=1}^{n-k} [1, \dots, cnt_i-1, cnt_i]$$$ and sort them in the non-decreasing order. It's easy to show that the chests should be sorted in the non-increasing order (really classical thing, you can try proving that by showing that any other arrangement can easily be improved once again).

That allows us to write a solution in $$$O(n^2)$$$. Sort all the chests in the beginning, after that for some $$$k$$$ multiply the value of the $$$i$$$-th chest by $$$\lfloor \frac{i}{k} \rfloor$$$ and sum up the results.

Finally, let's speed this up with prefix sums. Notice that the first $$$k$$$ values are multiplied by $$$0$$$, the second $$$k$$$ values — by $$$1$$$ and so on. If $$$n$$$ is not divisible by $$$k$$$, then the last block just has length smaller than $$$k$$$. Thus, we can calculate the answer for some $$$k$$$ in $$$O(\frac{n}{k})$$$. And that's equal to $$$O(\sum \limits_{k=1}^{n} \frac{n}{k})$$$ = $$$O(n \log n)$$$.

Overall complexity: $$$O(n \log n)$$$.

**Solution (pikmike)**

```
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
const int MOD = 998244353;
int add(int a, int b){
a += b;
if (a >= MOD)
a -= MOD;
if (a < 0)
a += MOD;
return a;
}
int mul(int a, int b){
return a * 1ll * b % MOD;
}
int binpow(int a, int b){
int res = 1;
while (b){
if (b & 1)
res = mul(res, a);
a = mul(a, a);
b >>= 1;
}
return res;
}
vector<int> pr;
int main() {
int n;
scanf("%d", &n);
vector<int> c(n);
forn(i, n)
scanf("%d", &c[i]);
sort(c.begin(), c.end(), greater<int>());
pr.push_back(0);
forn(i, n)
pr.push_back(add(pr.back(), c[i]));
int invn = binpow(n, MOD - 2);
for (int k = 1; k <= n; ++k){
int ans = 0;
for (int i = 0, j = 0; i < n; i += k, ++j)
ans = add(ans, mul(j, add(pr[min(n, i + k)], -pr[i])));
printf("%d ", mul(ans, invn));
}
puts("");
return 0;
}
```

Fastest editorial for an educational round I think.

Btw, in E merging without “smallest to biggest” optimization somehow passes the tests :)

Nothing surprising. There is even such a heuristic: randomly connecting sets. It works for O(nlog(n)).

Oh, that makes sense, thanks

No it's not $$$O(n \log n)$$$. Let's say we have $$$n$$$ one element sets and in $$$i$$$-th round we merge $$$i+1$$$-st set to su of first $$$i$$$. Then on $$$i$$$-th turn we have $$$\frac 12$$$ probability to do one insert, and $$$\frac 12$$$ to do $$$i$$$ inserts, which means on avearge $$$O(i)$$$ inserts on $$$i$$$-th turn, which adds to aveeage of $$$O(n^2)$$$ inserts.

That's one way to look at the probabilistic formulation. If you have broader probability space, for example, if operations are also chosen randomly, then the probability of big set being in the operation can be small, and I think it won't be O(n^2). I'm not sure how to show it, but my solution passed in the beginning, until they added a counterexample as yours, so if tests were generated using random at first, I think it shows that average complexity is not so bad.

First time solving a div 2D in a contest and this happens :(

Haha!

In the solution of problem E, visualising the towers and its merged states as tree is quite impressive. Thanks for the LCA solution.

To me, Small-to-large merge sounds like a greedy approach. It would be helpful if you can elaborate the alternate solution.

If you are looking for dsu solution.... If you always merge smaller set to larger set then it is o(n log(n))... Because a single element will not change it set more than log(n) time..... Let say we have a element 'a' which is in set 1 of size x we merge set 1 to set 2 of size greter than x suppose y.... Than we have atleast 2*x element which get in same set as y>x..whenever we are merging we are atleast doubling the size of. set

Statement of problem B was difficult to understand until watching the image at last as didn't know about the game.

Seriously! You don't know stone paper scissor! LOL!

r u patrick living under a rock?

Can anyone explain to me why greedy works in C? It was kind of intuitive to me, I coded it up and it passed but I still can't figure out that why's it optimal?

It's always optimal to take more skilled programmers over less skilled ones, and there is no advantage to expanding the team size once you've reached the minimum requirement of $$$x$$$ (since you risk adding a low-skill programmer that decreases the overall team skill below $$$x$$$, and you take programmers who could have otherwise formed more teams and improved the answer).

https://codeforces.com/contest/1380/submission/86713545

Can you please explain why i am getting memory limit exceeded on this solution of problem A?

Sorry to bother, I found the problem.

I am trying some different approach,can anyone please tell me what is wrong in this approach. https://codeforces.com/contest/1380/submission/86678593

you are sorting it in non-decreasing order, sort it in non-increasing order

We can solve with greedy in O(n) also. If we start from last index and keep the number of unused programmers and whenever for a particular index if no of programmers needed == no of unused we increment count and make unused 0 again.

I don't understand problem-B.Anyone help me to understand..And I don't know about this game.

Seriously bro? There is an image explaining it at the end of the problem. If you still don't understand : wiki page

Got it, now..

Why are submissions low on problems? Were they hard, imo first 3 were pretty easy?

You didn't even gave the round......Better first participate in the live round and then give your wishful verdict on whether the problems were easy or tough and why there are more or less submissions on a particular problemset

Do you even realize that this is an alt/troll handle

Codeforces was down for the first half-hour of the round; by the time it came back up, the round was declared unrated and many competitors left.

problem G. 'Let the values be x and y. If they differ by at least 2 (x≤y−2), then the smaller result can always be achieved by moving a regular chest from the larger one to the smaller one.' I could not understand the proof below. I indeed turn a coefficient y into a smaller one (x + 1), but how can I assign every regular chest to the old value, with the changes of this two interval lengths?

Sorry for the misread of the statement. Every chest's position can be changed.

https://codeforces.com/contest/1380/submission/86713545

Can anyone explain why i am getting memory limit exceeded error? It is the solution to problem A and is not much different from what explained in editorial

UPD: I have found the problem. Sorry to bother.

Can someone explain why the second solution for problem A works?

@spectre_gd I had the same doubt but I have figured it out while I was trying to contradict second solution approach. Go ahead post a test case(if you can find one) where this approach fails.

imagine array as a mountain and we have two situations when traversing 2nd to 2nd last element: 1. we find element lower on left side and element lower on right side (we got answer) 2. we dont get element lower on left side (means our current element is lowest and array is descending upto current element so it cant be the peak), do the same thing on right side

Second solution is based on negation of the problem. We are exploring those cases where answer does not exists and it will be when array is either non-decreasing or non-increasing. So, to break this condition, there must be local maxima i.e. an index j where both (j-1) & (j+1) elements are less. And this index is our answer!

Can someone explain in problem B why picking the greedy ci for the most frequent will somehow work for the rest of string?

How does that make sense? Can't there be a value in the string where ci fails?

Your choice string is played agains the given string on every position. This means, all positions of the two Strings are paired once. Or, in other words, it does not matter in which order you put the symbols into the choice string.

Since the order does not matter, every single position in your string conributes independend of any other position. Finally, for a single position it is obvious that the best choice is based on the frequency of the symbols.

you have to understand that we are comparing each of element of s with c1 BUT we are doing the same thing with c2, c3, c4.... after just rotating the s

for eg s = RSPRR R is most frequent so we choose P for c1 for most wins Now for c2 we are doing the same comparisons and P will give best result overall again (cause R is most frequest) so its basically the same

My Video Solution of B and C where i have tried to explain my thought process and why the solution worked .

n = 5, x = 10,

7 11 2 9 5

how can we choose 2 teams from this? shouldn't it be just 1 team cause min = 2 multiplied by 5 members = 10 so only one team?

choose 1 member for one team....this team is formed only 11...bcz 11*1=11 > x... then choose 9 & 7 ..then you get min(9,7)*2=14>x... another way to choose is (11,9) & (7,5)

It is given that elements are distinct

If the elements are distinct just suppose, then how we can solve it?

I suppose it can be done in many ways. Store the index of each element occurring in

`a`

. Let's call it`apos`

. Also make a copy of array`a`

. Let's call it`astrip`

(I know the naming is bad). Now while iterating with`b`

, mark all positions in`strip`

where this number occurred.`astrip[apos[b[i]]] = some_identifier`

. Because all elements are unique, this'll work. Now just iterate through array`astrip`

and figure out the segments. Along with this you also need to make sure`apos[b[i]] > apos[b[i - 1]]`

because you want b to be a subsequence of`a`

.86693530

Elements are pairwise distinct. Doesn't that mean something like 1,2,1,2,1,2 could be a possible test case?

My bad :(

my bad.... thanks all for responding.....

in question E merge towers: for the given test case [[5,1],[2],[7,4,3],[6]] are the towers before any query on merging tower 1 and 3 according to me it should be done like: from tower [5,1] ,1 goes to tower 3-> [5],[7,4,3,1] -> 1st operation from tower [7,4,3,1] , 1,3 and 4 goes to tower 1 -> [5,4,3,1],[7] -> 2nd operation from tower [5,4,3,1], 5,4,1 and 3 goes to 3rd tower -> [],[7,5,4,3,1] -> 3rd operation it can be done in three operation but it shows 5 as a result . Where am i wrong?

Put some clothes on and format the question ;)

The result must be one tower [7,6,5,4,3,2,1]. Somehow you are loosing the 2 and 6.

The rule for merging is fairly simple: Take the top segment from the tower where the 1 is, and put the whole segment onto the tower where it fits. The moved segments in the example would be:

The last move creates the complete tower.

We can also solve F by matrix-DP .. First we calculate the matrix for every dp transition .. then we just need the product of these matrices .. (To simplify append 0 at the begining and 9 at the end ) and build segTree in [0,n] and then remains only two point updates at index(x) and index(x-1) .. (Make sure to have dp[n+1]=1 (even if its nine ))

Can someone explain C more elaborately?

.

In D problem, if the elements are not distinct then how we can solve it??

according to me.i think complexity will be >=qudratic

There's another solution O(n) for A.

`n`

will definitely satisfy this condition as middle element and the ends of the array as first and last element. (Because`n`

will be bigger than anything and that's what we want here).`n`

is at the ends of the array, then`n`

can't be the answer. We ignore that element, update ends of the array and do the same thing again. The highest element in the array is now`n - 1`

and if it's somewhere in the middle, that should be the answer.86668426

Can also be done with prefix and suffix max in 0(n)

can anyone please explain what is this line doing in problem D solution

`res += (len - k) * y + x;`

. this line will be executed when we cannot perform berserk because one of the elements in segment is greater than both`l`

and`r`

and cost of berserk is lesser than fireball to delete everything. i thought this should have been`res+=len/k*x`

so that we delete everything remaining using fireball. Thank you!!The tutorial does the operations in irritating order. However, if the biggest monster is bigger than l and r we need to use Fire at least once. And independend of anything else, we need to use Berserk on at least len%k elements. The order of these two operations or the others operations does not matter.

thank you so much !! understood now

In problem C, I calculated minimum number of programmers required starting from programmer at position 'i'. This is will basically give me n (or less) segments. I then run an algorithm same as activity selection problem on this array of segements, to find the maximum number of teams. https://codeforces.com/contest/1380/submission/86686449 This comes out as wrong on test case 8. My answer is higher than the correct answer. I am not able to figure out why is this method wrong. Some help?

In problem G why can we ignore the initial order of chests and sort? I was thinking on a case like 4 40 400 400 600 and k=1. Where can I put the mimic so that the answer gives 330 (as it does with the editorial solution)?

You misread the problem: "Print n integers — the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics."

"the chests are placed into the rooms in some order and exactly k of the chests are mimics." so you can choose any order.

Oh. Now It makes sense. Ty

In B ,what if there is an extra constraint that "You will lose point if the bot's choice is superior",So doing the same thing which we actually did for the original problem,would it be still optimal ?

It is still the case that every position is played against any other position. So the order of the symbols in the choosen string does not matter. Since the order does not matter it is still optimal to use one symbol for the whole string. As a pragmatic solution, just try all three possibilities and choose the best.

Anyone please explain the update part in problem F .

how we are doing updates ?

Just curious:

Does there exist some kind of dp solution for D?

Constraints are too high for a DP solution for problem C(assuming you mean by DP is to brute force and use dynamic programming to decrease number of total operations made).

Disclaimer: It is a little messy because I tried to code as fast I could and submit it during contest.Here you go! 87810909

ExplanationThe idea is simple. We precompute minimum cost of deleting subarrays of size 1 to n such that either only single or zero element remains at last.

When we are computing the answer we keep in mind that cost of deleting all elements but one with added cost of $$$y$$$ is lesser or the cost of deleting all elements is lesser i.e.

$$$ans = ans + min(dp[size][1] + y, dp[size][0])$$$

problem C is enjoyable

86766761 Can someone help me identify what did I do wrongly here on problem C?

I didn't understand your code completely...

You can see this one might be helpful 86676269

I first consider all those with skill >= x because they can form independent group of size 1.

I then have an array, say A, for the rest of those whose skillsets are strictly less than x, which means they need to team up with others to use the length of their team to compensante for x. I sorted this array A and loop it from the smallest value to the greatest.

(continue from above) Within each iteration in my array A, I calculate the corresponding length that I need such that I can form a team by taking the ceiling quotient num = x/a (where a represents the element of A and num represents the required team length)

# Example: Array A = [1,1,2,3,4], given x = 5

In my first loop, num = ceil(5/1) = 5 (which is reasonable, given that I would need a team of length 5 to fit the team formation requirement)

# So, if in my iteration I find that my array A has the length greater than or equal to "num", then my group count will += 1 and I will delete the partition of my array from the original.

If in my iteration I realize that given the element, it is impossible to form such a team from my original array A, then I remove the element from my original array A.

Now that I think of this logic again, I feel like in certain cases it might undercount. But for now the failing case that I have is that the solution is 0 and my code gives 1, which is an overcount

I have applied part of your logic here 86781167

But without using the

`ceil`

part, which is not needed.And if you think of it carefully you will realize that there is no need for deleting elements from array A, since you are iterating throw elements...

Note that

`ceil`

won't be useful if it is used with integers, and it can sometimes cause some troubles if not used properly, so avoid using it as you can.Thank you, but I would be more enlightened if you could help me find what went wrong with my implementation or my thinking process

If you start small, you are guilty of waste. M = 8, for example,,2,7,7,7,7,7,7 [1]; Instead of [7.7][7.7][7.7] [7.7] As for the answer greater than correct, if m=10,[4,4]; You only need 10/4=2, but (10/4)*2=8 does not equal 10

Thank you!

In C I miss read the question and was thinking that each team should have a product at least x. can anyone tell how to solve if this was the question ??

In this case also, just sort the array and keep two variables. One containing the product of the present array and one containg the count of number of arrays. The code will look something like this: https://pastebin.com/zL2pxcfZ

I don't think it's that easy. here is the testcase

answer : 3 your code output: 2

Ya, it's not easy. Sorry for the wrong explanation. Need a better approach for this.

I think the correct answer would be: 1. Because only the minimum element of the group and the size of the group matters. if I'm wrong I would be happy if u correct me.

I think i am not asking problem C. how about you read my comment again.

Can anybody help me with Problem D Berserk and Fireball I am facing some implementation issues? My submission — 86799781 Thanks

Do you really expect a meaningful answer to such a vague question?

Can you help me with my submission?

You noticed the diagnostics at the end of testcase 7?

Haa But I couldn't make out the error like what does it mean?

I cannot spot the bug either. But I see some ways to greatly simplify your code.

Do not use a set for index. The values you are inserting are distinct and sorted, just use vector.

Additionally put a -1 as first element into that vector, and a n as last. So you can delete the copy of the main loops body.

The first loop where you collect index is unnice. Still not sure if there is an off by one somehow. You better loop over a[i] and check every index if a[i]==b[j]. If yes insert i and increment j.

how's solution of 1380A-Three Indices correct w O(n), as we don't know that those three will be the consecutive no.s.

If there exists a solution then it is guaranteed that three consecutive numbers are definitely a solution otherwise, the solution doesn't exist. e.g.-> Consider array --> [3,1,4,2,5] one solution is i=1, j=3, k=4 but instead, to make the algorithm O(n) we are more interested in i=2, j=3, k=4

upd:done

pikmike can you explain what 0,1,2,3 represent in the solution of problem F. Thank you.

0 = 00 — neither is taken

1 = 01 — leftmost isn't taken, rightmost is taken

2 = 10 — leftmost is taken, rightmost isn't taken

3 = 11 — both taken

@pikmike Can you please explain me the problem -e test case when we merge the 3 and 1 tower how the ans coming out to be 4 instead of 3 . Ok I can tell you how i got 3.

1st tower contain — 5 1 and 3rd tower contain — 7 4 3

1st move — you shift (1) from tower 1 to tower 3 now 3rd tower contain — 7 4 3 1 and 1st tower conatin — 5

2nd move — you shift (4 3 1) from tower 3 to tower 1 now tower 3 contain -7 and tower 1 conatin — 5 4 3 1

3rd move — now you shift (5 4 3 1) from tower 1 to tower 3 now tower 3 contain — 7 5 4 3 1 and tower 1 conatin nothing .So my answer coming out to be 3 instead of 4 . I know i am missing something could you please point out.Thankyou.

Reread the problem statement carefully.

We are not calculating the number of operations needed to merge the towers in the query. Instead, after each query, we want to calculate the number of operations required to merge all current towers into one.

why I'm getting Time Limit Error at test set 6 ? https://codeforces.com/contest/1380/submission/86946265 can anybody explain? thanks in advance

I think, there is a mistake in editorial of the problem 'C'. There should be 'Non-increasing' instead of 'non-decreasing'.

In problem E it is not at all clear how to achieve these answers.

I have tried to make editorial for questions A-E . please have a look. Language :- Hindi

https://www.youtube.com/playlist?list=PLrT256hfFv5UUIBTJBUL7ZlZVHaWWEmGB

Was F inspired by this meme?

I still don't understand the tutorial for F.

Specifically, I don't understand how a segment tree could optimize this dp. What will the nodes store and what the merging looks like. I get the part about the dp but become totally confused when they start talking about segment tree.

Can anyone explain it for me? Maybe some visual example would help. Thanks in advance.

4 6 1 2 5 3 This is the second testcase for problem A.. Why doesn't 4 < 6 > 1 give 1 2 3 as output ????