Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
t = int(input())
for i in range(t):
l, r = map(int, input().split())
if l * 2 > r:
print(-1, -1)
else:
print(l, l * 2)
```

Idea: Roms

**Tutorial**

Tutorial is loading...

**Solution 1 (pikmike)**

```
for _ in range(int(input())):
n, k, z = map(int, input().split())
a = [int(x) for x in input().split()]
ans = 0
s = 0
mx = 0
for t in range(z + 1):
pos = k - 2 * t
if pos < 0:
continue
mx = 0
s = 0
for i in range(pos + 1):
if i < n - 1:
mx = max(mx, a[i] + a[i + 1])
s += a[i]
ans = max(ans, s + mx * t)
print(ans)
```

**Solution 2 (pikmike)**

```
for _ in range(int(input())):
n, k, z = map(int, input().split())
a = [int(x) for x in input().split()]
ans = 0
s = 0
mx = 0
for i in range(k + 1):
if i < n - 1:
mx = max(mx, a[i] + a[i + 1])
s += a[i]
if i % 2 == k % 2:
tmp = (k - i) // 2
if tmp <= z:
ans = max(ans, s + mx * tmp)
print(ans)
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (Ne0n25)**

```
#include <bits/stdc++.h>
using namespace std;
#define sz(a) int((a).size())
#define forn(i, n) for (int i = 0; i < int(n); ++i)
int solve(const string& s, int x, int y) {
int res = 0;
for (auto c : s) if (c - '0' == x) {
++res;
swap(x, y);
}
if (x != y && res % 2 == 1)
--res;
return res;
}
void solve() {
string s;
cin >> s;
int ans = 0;
forn(x, 10) forn(y, 10)
ans = max(ans, solve(s, x, y));
cout << sz(s) - ans << endl;
}
int main() {
int T;
cin >> T;
while (T--) solve();
}
```

Idea: adedalic

**Tutorial**

Tutorial is loading...

**Solution (adedalic)**

```
import kotlin.math.*
fun main() {
repeat(readLine()!!.toInt()) {
val (n, k) = readLine()!!.split(' ').map { it.toLong() }
val (l1, r1) = readLine()!!.split(' ').map { it.toLong() }
val (l2, r2) = readLine()!!.split(' ').map { it.toLong() }
var ans = 1e18.toLong()
if (max(l1, l2) <= min(r1, r2)) {
val rem = max(0L, k - n * (min(r1, r2) - max(l1, l2)))
val maxPossible = n * (abs(l1 - l2) + abs(r1 - r2))
ans = min(rem, maxPossible) + max(0L, rem - maxPossible) * 2
} else {
val invest = max(l1, l2) - min(r1, r2)
for (cntInv in 1..n) {
var curAns = invest * cntInv
val maxPossible = (max(r1, r2) - min(l1, l2)) * cntInv
curAns += min(k, maxPossible) + max(0L, k - maxPossible) * 2
ans = min(ans, curAns)
}
}
println(ans)
}
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (pikmike)**

```
fun gcd(a: Long, b: Long): Long {
if (a == 0L) return b
return gcd(b % a, a)
}
fun main() {
repeat(readLine()!!.toInt()) {
val (m, d, w) = readLine()!!.split(' ').map { it.toLong() }
val w2 = w / gcd(d - 1, w)
val mn = minOf(m, d)
var cnt = mn / w2
println((2 * (mn - w2) - w2 * (cnt - 1)) * cnt / 2)
}
}
```

Idea: Roms

**Tutorial**

Tutorial is loading...

**Solution 1 (Ne0n25)**

```
#include <bits/stdc++.h>
using namespace std;
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define forn(i, n) for (int i = 0; i < int(n); ++i)
typedef pair<int, int> pt;
const int N = 200 * 1000;
int n;
int l[N], r[N], t[N];
set<pt> st[2];
int main() {
scanf("%d", &n);
forn(i, n) scanf("%d%d%d", &l[i], &r[i], &t[i]), --t[i];
vector<pair<int, pt>> ev;
forn(i, n) {
ev.pb(mp(l[i], mp(0, i)));
ev.pb(mp(r[i], mp(1, i)));
}
sort(all(ev));
int ans = 0;
for (auto it : ev) {
int i = it.y.y;
if (it.y.x) {
int j = t[i];
int k = j ^ 1;
if (st[j].count(mp(r[i], i)) && !st[k].empty()) {
++ans;
st[k].erase(st[k].begin());
}
if (st[j].count(mp(r[i], i))) st[t[i]].erase(mp(r[i], i));
} else {
st[t[i]].insert(mp(r[i], i));
}
}
printf("%d\n", n - ans);
}
```

**Solution 2 (Ne0n25)**

```
#include <bits/stdc++.h>
using namespace std;
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define sz(a) int((a).size())
#define all(a) (a).begin(), (a).end()
#define forn(i, n) for (int i = 0; i < int(n); ++i)
typedef pair<int, int> pt;
const int INF = 1e9;
const int N = 200 * 1000;
int n;
vector<pt> a[2];
struct segtree {
int n;
vector<int> t, ps;
segtree(int n) : n(n) {
t.resize(4 * n, -INF);
ps.resize(4 * n, 0);
}
void push(int v, int l, int r) {
if (l + 1 != r) {
ps[v * 2 + 1] += ps[v];
ps[v * 2 + 2] += ps[v];
}
t[v] += ps[v];
ps[v] = 0;
}
void upd(int v, int l, int r, int pos, int val) {
push(v, l, r);
if (l + 1 == r) {
t[v] = val;
return;
}
int m = (l + r) >> 1;
if (pos < m) upd(v * 2 + 1, l, m, pos, val);
else upd(v * 2 + 2, m, r, pos, val);
t[v] = max(t[v * 2 + 1], t[v * 2 + 2]);
}
void add(int v, int l, int r, int L, int R, int val) {
push(v, l, r);
if (L >= R) return;
if (l == L && r == R) {
ps[v] += val;
push(v, l, r);
return;
}
int m = (l + r) >> 1;
add(v * 2 + 1, l, m, L, min(m, R), val);
add(v * 2 + 2, m, r, max(m, L), R, val);
t[v] = max(t[v * 2 + 1], t[v * 2 + 2]);
}
int get(int v, int l, int r, int L, int R) {
push(v, l, r);
if (L >= R) return -INF;
if (l == L && r == R)
return t[v];
int m = (l + r) >> 1;
int r1 = get(v * 2 + 1, l, m, L, min(m, R));
int r2 = get(v * 2 + 2, m, r, max(m, L), R);
t[v] = max(t[v * 2 + 1], t[v * 2 + 2]);
return max(r1, r2);
}
void upd(int pos, int val) {
return upd(0, 0, n, pos, val);
}
void add(int l, int r, int val) {
return add(0, 0, n, l, r, val);
}
int get(int l, int r) {
return get(0, 0, n, l, r);
}
};
int main() {
scanf("%d", &n);
forn(i, n) {
int l, r, t;
scanf("%d%d%d", &l, &r, &t);
a[t - 1].pb(mp(r, l));
}
forn(i, 2) sort(all(a[i]), [](pt a, pt b) {
if (a.x == b.x) return a.y > b.y;
return a.x < b.x;
});
segtree t1(sz(a[0]) + 1), t2(sz(a[1]) + 1);
t1.upd(0, 0);
t2.upd(0, 0);
int ans = 0;
for (int i = 0, j = 0; i + j < n; ) {
if (i < sz(a[0]) && (j == sz(a[1]) || a[0][i].x <= a[1][j].x)) {
int pos = upper_bound(all(a[1]), mp(a[0][i].y, -INF)) - a[1].begin() + 1;
int cur = t2.get(0, pos) + 1;
ans = max(ans, cur);
t1.upd(i + 1, cur);
t2.add(0, pos, 1);
++i;
} else {
int pos = upper_bound(all(a[0]), mp(a[1][j].y, -INF)) - a[0].begin() + 1;
int cur = t1.get(0, pos) + 1;
ans = max(ans, cur);
t2.upd(j + 1, cur);
t1.add(0, pos, 1);
++j;
}
}
printf("%d\n", ans);
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
typedef long long li;
const int N = 300043;
bool is_bridge[N];
int w[N];
int c[N];
int v[N];
vector<pair<int, int> > g[N];
vector<pair<int, int> > g2[N];
int comp[N];
li sum[N];
li dp[N];
int cnt[N];
int fup[N];
int tin[N];
int T = 0;
li ans[N];
int v1[N], v2[N];
int n, m, k;
int dfs1(int x, int e)
{
tin[x] = T++;
fup[x] = tin[x];
for(auto p : g[x])
{
int y = p.first;
int i = p.second;
if(i == e)
continue;
if(tin[y] != -1)
fup[x] = min(fup[x], tin[y]);
else
{
fup[x] = min(fup[x], dfs1(y, i));
if(fup[y] > tin[x])
is_bridge[i] = true;
}
}
return fup[x];
}
void dfs2(int x, int cc)
{
if(comp[x] != -1)
return;
comp[x] = cc;
cnt[cc] += v[x];
sum[cc] += c[x];
for(auto y : g[x])
if(!is_bridge[y.second])
dfs2(y.first, cc);
}
void process_edge(int x, int y, int m, int weight)
{
li add_dp = dp[y];
if(cnt[y] > 0 && cnt[y] < k)
add_dp = max(0ll, add_dp - weight);
cnt[x] += m * cnt[y];
dp[x] += m * add_dp;
}
void link(int x, int y, int weight)
{
process_edge(x, y, 1, weight);
}
void cut(int x, int y, int weight)
{
process_edge(x, y, -1, weight);
}
void dfs3(int x, int p)
{
dp[x] = sum[x];
for(auto e : g2[x])
{
int i = e.second;
int y = e.first;
if(y == p)
continue;
dfs3(y, x);
link(x, y, w[i]);
}
}
void dfs4(int x, int p)
{
ans[x] = dp[x];
for(auto e : g2[x])
{
int i = e.second;
int y = e.first;
if(y == p)
continue;
cut(x, y, w[i]);
link(y, x, w[i]);
dfs4(y, x);
cut(y, x, w[i]);
link(x, y, w[i]);
}
}
int main()
{
scanf("%d %d %d", &n, &m, &k);
for(int i = 0; i < k; i++)
{
int x;
scanf("%d", &x);
--x;
v[x] = 1;
}
for(int i = 0; i < n; i++)
scanf("%d", &c[i]);
for(int i = 0; i < m; i++)
scanf("%d", &w[i]);
for(int i = 0; i < m; i++)
{
scanf("%d %d", &v1[i], &v2[i]);
--v1[i];
--v2[i];
g[v1[i]].push_back(make_pair(v2[i], i));
g[v2[i]].push_back(make_pair(v1[i], i));
}
for(int i = 0; i < n; i++)
{
tin[i] = -1;
comp[i] = -1;
}
dfs1(0, -1);
int cc = 0;
for(int i = 0; i < n; i++)
if(comp[i] == -1)
dfs2(i, cc++);
for(int i = 0; i < m; i++)
if(is_bridge[i])
{
g2[comp[v1[i]]].push_back(make_pair(comp[v2[i]], i));
g2[comp[v2[i]]].push_back(make_pair(comp[v1[i]], i));
}
dfs3(0, 0);
dfs4(0, 0);
for(int i = 0; i < n; i++)
printf("%lld ", ans[comp[i]]);
puts("");
}
```

Super Fast!!

Day by day I feel like nobody is getting the humor anymore they are just downvoting every comment.

In the Tutorial of Problem E: I think you mean yd + x instead of xd + y as the index of the x-day of the y-month in the year.

xd+y means index y-th day of x-th month which is corresponding pair of x-th day of y-th month whose index is yd+x.

Can someone tell me why my solution for D fails? I simply decide whether to gain intersection by joining two new segments or expanding others which already cover themselves totally. Counter-examples are especially appreciated.

Here's the submission: https://codeforces.com/contest/1389/submission/88410212.

The test, your solution fails when

n = 10, k = 297l1 = 741, r1 = 741l2 = 20, r2 = 770Try to fix it.

Got my error, thank you!

Is there any way to solve $$$D$$$ in $$$O(1)$$$.. maybe little case work ($$$chukles$$$), but I guess that could be done!

Waiting for Div 3 :) After so many difficult rounds to increase rating :(

What is

`Event Processing`

I see it here and there but never find some concise tutorial on it.https://www.geeksforgeeks.org/maximum-number-of-overlapping-intervals/

In that algo the beginning and ending of intervals is treates as "events" where the count of them is incremented or decremented.

Thanks man!

Can anyone explain this logic of problem B (solution 2): if i % 2 == k % 2: tmp = (k — i) // 2

Can we do the problem B with kadane algorithm in which I will find the max present in the first k position of the array, and then after reaching towards the max, I will choose whether to go beyond or go back ?

No way bro!!

I think we can just go through 45 combinations for C instead of the 100 given in the tutorial.

TLDR:0101010 has both a 101010 and a 010101 type strings in it.

DETAILED:Let us say the string we get after removing all other numbers but two be some 011100001001010.....

Then for the above we want it to reduce to the form : 01010101... Note, that if the reduced string ends with 1 then the answer is: size(original string)-size(reduced string). Else, the answer is: size(original string)-{size(reduced string)-1}.

Can someone explain the dynamic programming approach to problem E?

Problem E doesn't use dynamic programming. It's just math and number theory.

Cool contest!Though I've lost some points :(

if(max(l1,l2)>=min(r1,r2))

When their is no intersection... then this code works ... but i confuse in one thing ll dist=max(l1,l2)-min(r1,r2); ll x=max(r1,r2)-min(l1,l2);

i am not understanding properly these two lines.. why we did this. this x can be negative.. like take an example

10 225

100 19

80 15

now how this code is giving correct answer for this... please someone explain me in detail.. what i am not able to see what is happening?

Firstly.What does these numbers mean? Test containing 4 numbers, but u give 6. It's not 1 test and not 2.

SecondlyX can't be negative.

In both pairs {l, r} l <= r, so

max(r1, r2) >= max(l1, l2)=>max(r1, r2) >= min(l1, l2)I hope you understand

10 225

100 19

80 15

take this example-> k is 225 and N is 10 so if we do max(19,15)-min(100,80) this is x as in code, so this is -ve now... now how this code gives correct answer for this.

Sample are incorrect and occasionally it's work (but it shouldn't)

OMG sorry brother. i dont saw that constraint.. thanks. I wasted my 5 hours shit

Why this code is giving TLE? https://ideone.com/wXdX3A

Lets calculate O(10 * 10 * n + 10 * n) = O(100*n) n = 2 * 10^5. We get O(2 * 10^7). Its about 10 secs on Python and about 0.2-0.3 secs on C++ (I have solve with the same asymptotics) Friendly advice. Use C++ in real competitions.

Same code gives AC in c++ but tle in python :| btw thank you.

С++ ~100 times faster than python

always use fast I/O in Python. normal I/O in python takes a lot of time. https://www.geeksforgeeks.org/python-input-methods-competitive-programming/

Hey @adedalic I have one doubt in D, You have said,

EDIT:`Now I get it, u have used a for loop (i to n), that checks by joining first i pairs and compares it with minimum each time . Thanks.`

Just iterate over n and find mininum value.

got it.

I wrote a recursive solution using memoization where I checked for each I it's i-1 and i+1 and updated my answer with maximum sum Does my solution not violate the condition given in the question :. " you can't perform two or more moves to the left in a row" Can anyone please explain this that how my function goes for 2 or more left moves in a row if z is>=2 and does not violate this and gets accepted

__88418038

Could anyone please explain the dp with segment tree approach for problem F?

Here's how I understand it (I did not solve in contest), feel free to correct me if I'm wrong or there is an easier way to think about this.

First, let us sort all segments in increasing order of their right endpoints in the list $$$segs$$$. Split the segments based on type and sort them in the same fashion in the lists $$$s[t]$$$. Let $$$dp[i][j][t]$$$ be the answer assuming that the problem is restricted to the first $$$j$$$ segments in $$$segs$$$, along with the condition that we can only select the first $$$i$$$ segments in $$$s[t]$$$, and that we have used segment $$$i$$$ in $$$s[t]$$$ in our solution.

Now, to transition to $$$segs[j+1]$$$ , assuming $$$segs[j+1]$$$ is of type 1 and is the $$$k+1$$$'th segment in $$$s[1]$$$, we have the following:

Now, notice that for any new segment, the segments that do not touch it must be a prefix of the sorted segment list. Thus, these transitions are exactly range add modifications and maximum queries on two segment trees. We can easily find what range to add by using a binary search. Finally, we can eliminate the 2nd dimension with $$$j$$$ just by directly modifying the segment trees.

See my submission for implementation details.

So, for implementation purpose, we can ignore the $$$2nd$$$ dimension of the dp state you mentioned, right? So our dp state will be:

$$$dp(i,j): $$$ maximum segments if we have only considered till the ith segment of type $$$j$$$ and this one is included. And we traverse the segments on the sorted order of their right endpoints.

Yes, exactly. When I was trying to figure the solution out based on other people's code, I found the 3 dimensions to be easier (as a solution that I could actually come up with), but you have it exactly right.

Okay, sounds good. Thanks a lot.

https://codeforces.com/contest/1389/submission/88444945 Why my code is wrong? Can anyone help? for length = odd, I check for the most frequent character. for length = even, I iterate i,j and the pattern is i,j,i,j, .... ,i, j ( end with j )

Because you don't need to start with p1=0 & p2=0 Try this example char1: 3 4 9 13 15 char2: 5 7 11 18 Your algorithm will answer 2 ( 3, 5 ) But the correct answer is 6 ( 4,7,9,11,13,18 ) You can see that it doesn't need to always start with p1 = 0. Thanks to myself ! I do this beacuse I can't delete my comment.

I am trying to hack 88307545 solution during the

hacking phase.Byinput datawasThe code will

output(on local PC and on online compiler):I write the second test case to make codeforces think "the first line is for the first test case and the others are for the second", but I got

unsuccessful attempt.I didn't understand why this happens?Can someone explain to me?Thanks in Advance!O(1) solution for D 88447760

+1 88435747 in Kotlin

Could somebody please explain the dp approach to B?

Dp approach of B will be like that we have two choices at every index either we can move left or right, we will explore the moves and the maximum between these two options will be considered. base cases are:

1 . when we don't have any moves left (k==0), in this case, we will return the current index value.

2 . when the last move taken were left or we don't have any left moves remained or we are at the starting index of the array, then we have only one option to move right at this index.

3.else we have two option to move right and left, and take the maximum among those to the answer. you can check my solution for more clarity, 88714266.

anurag122898 could you please have a look at this? With the same approach, I'm getting two different results just because I'm swapping the two options' sequence. I never knew something like this could make a difference. I'm putting the SS of the comparison between 92397103 and 92402059 for convenience.

Can anyone help me with my code for Problem C ? Getting a Wrong Answer.

Your code is checking if i is at even and j is at odd positions, which is wrong. You need to find the longest SUBSEQUENCE of alternating i and j of even length. INPUT: 1 2025 YOUR OUTPUT: 1 CORRECT OUTPUT : 2

Can you please explain it a bit more clearly?

Same approach as Editorial still can't be accepted in python3 due to TLE on test 2 , what is this?

1389C-Good String 88530443

First: Submit in PyPy instead of CPython

Second: Convert the input to a list of ints once instead of converting it in each iteration.

O(1) solution for problem D Segment Intersections.

If anyone is interested in the explanation then comment.

interested

There can only be three initial configurations for the two intervals:

1> One completely overlaps the other.

2> There is some partial intersection b/w them.

3> They are disjoint.

Now, let us tackle them one by one.

1>> We will have an initial I which can be calculated easily, thereafter if the current I is smaller than k we can increase the intersection length by one for all n pair of segments in just 1 step and we can do this untill all the pairs have become equal It is obvious that we cannot increase I by more than 1 in a single operation, So this is optimal. Now if still I is less than k we only have one way to increase I by increasing the segment lengths of a pair this will give us a +1 for each 2 step. Also, we can do this indefinitely.

2>> Its the same as the first one except that the initial I has to be calculated differently.

3>> Initial I=0 , Now how can we increase I? we first have to make a pair of segment meet each other while doing this our I remains the same, lets call this cost as gap. Now as soon as the segments meet we can start having a bunch of +1s . Let us calculate how many +1s we can get in a row, Suppose the worst case when initially both segments are points, then we will have no of +1s equal to the no of +0s (we used to fill the gap) . And in every other case the no of +1s will be greater than +0s , Remember this. Now let us see what choice do we have after using up these two operations , At this point we reach a situation analogous to one explained earlier in 1>> where we will be getting +1 for every 2 steps. We have all the tools ready . Let us expand this for n. I claim that in an optimal final solution we will have a situation where x of the n segments are completely overlapping and at most 1 pair of segments which has been extended by a 2 step operation or has a partial overlap , Also x will be maximum. Suppose a case where x is not the maximum then we have a case where x-1 segments have complete overlap and 1 segment with a partial overlap or an extended 2 step operation segment. We can simply rule out the first case because we can never compensate for the loss of one completely overlapping segment with a partial one. As for the other case the no of 2 step operation will be greater than the complete overlap of the segments as compensation. This is where we use the fact we proved earlier, that we can get a complete overlapping with total no of steps less than or equal to twice the size of the segments, which will always be at least as efficient as the +2 step operation and hence we can carve out a completely overlapping segment from this segment without loosing optimality (for the lack of better word).

Thanks, man for the effort! stating 3rd case more mathematically -

suppose we have to make 'rem' out of 'n' segs, let 'ex' be the amount to 'activate' the seg (= max(l1,l2)-min(r1,r2)), and 'tot' (= max(r2,r1)-min(l1,l2)) we can get by doing 1 inc per 1 move operation (+1/1 ops) after 'activation'.

Its oBvious that we have to 'activate' the 1st seg and do +1/1 ops If still we have something remaining (let it be 'x')

now if 'x' >= 'tot' then its better to activate another seg and get tot why? because 'tot'+'ex' <= 'tot'*2 as 'ex' <= 'tot'

if 'x' < 'tot' then we decide on the basis of which is bigger, 'x'+'ex' or 'x'*2

so how is this o(1) -> after the 1st activation, we can activate segs until reqd amount < 'tot' and then we decide between 'activation'+'+1/1 ops' and '+1/2 ops' for the last one.

Yeah at one point I thought the way I described it isn't going to help anyone, happy to see that you got it... P.S. yours seems much formal and clear.

Hello Can you check my submission of D problem..I tried the O(1) solution.I cannot find any mistake in it.I tried with over 100 cases. https://codeforces.com/contest/1389/submission/88646237

Found my mistake I didnt consider the case separately in disjoint where k is less than r2 -l1

__

This is in no way a criticism, but I am indeed slightly surprised to find 3 different languages in the implementation section.

Can anyone please explain the second solution code of problem B?

88381812

Try to write the code in the code snippet. Yours is very unreadable and takes a lot of space in the comments

EDIT: For your problem, a string is good if it has all same characters in case or even length, or same parity indexed characters are same in case of odd length.

A rather random question though, for the explanation for question B, it says any position between 1 and k-2(t) + 1 will be visited. But if k == 4 and t == 2, supposedly i can visit position 1 and 2, but the formulas will suggest I only visit between 1 and {4 — 2(2) + 1}, which means visit between 1 and 1?

I get what it is trying to say now, thanks!

in the div2 b problem whats the intution for this"it is optimal to choose exactly the same i for all the pairs (right, left). And that i should be such that ai+1+ai is maximum possible."

Can someone please explain the dp approach for B?

Can we do the problem B with kadane algorithm in which I will find the max present in the first k position of the array, and then after reaching towards the max, I will choose whether to go beyond or go back ?

Really nice editorial for Problem B really liked it

Can anyone provide the Test case 2 — 19th input for task B. Most people had this error "expected 218 found 216". If not exact same test case, then maybe something similar?

I don't know the exact test case but you should also handle for the case k=0 explicitly because my solution got accepted after I make a case for k==0.

In that case the answer would be just the 0th element which I've handled. Thanks for the help!

Man, I am getting confused in problem A editorial. It says if 2l>r then its not possible else its l and 2*l. Then how can in the example l=1 and r=1337 produce an output of 6 and 7. Can anyone please explain?

in problem B, the tutorial assumes that the left steps will always be accompanied by right steps as they have been taken in pairs but it may so happen that the last step will be a left step, so in that case the tutorial fails and so does the solution in the following test case

18 11 4 11 19 18 19 19 5 14 15 17 4 10 9 8 17 9 2 15 10

The tutorial assumes left steps are

always precededby a right step; Hence the order (right,left).oh thanks... i had misunderstood it

hey can you help me in this i'm trying it since last 7 hours

The idea of G is quite similar to this problem Museum Tour

Very interesting:

Undirected Graph -> find Bridges -> Tree -> Can be solved using DP

Directed Graph -> find SCC -> DAG -> Can be solved using DP

The *2100 Prob. D seems much more difficult than the *2200 Prob. E.

I'm trying the array walk since 7 hours still not able to get it can someone please help me with this code where i'm making mistake seriously i just wanted to solve it at any cost please help!!!!

can anyone explain the dp approach to problem B?

there are things on youtube which explains the dp approach of B, you just type codeforces educational round 92 then you will get what you want :)

For the Good String question: I read the tutorial and worked on it myself for practice. Below is my solution in C++. It passes the test 1. But for test 2. my output is wrong for some cases. Could you please let me know where is wrong? Thanks a lot!

using namespace std;

int main(int argc, char** argv) {

}

In problem B editorial

"You have t pairs of moves (right, left) to insert somewhere inside the sequence of k-2t moves to the right."

Why k — 2*t ?

Not expecting DP in second problem.

88352400 for probblem C , can someone please tell me what's wrong in my code?

Spoiler## include

## include

using namespace std; int main(){ ios::sync_with_stdio(0); cin.tie(0); int T; cin >> T; while(T--){ list adjlist[10]; string s; cin >> s; for (int i = 0 ; i < s.size() ; i++) adjlist[s[i] — '0'].push_back(i); int ans = 0; for (int i = 0 ; i < 10 ; i++) ans = max(ans , (int)adjlist[i].size()); for (int i = 0 ; i < 10 ; i++){ for (int j = 0 ; j < 10 ; j++){ if (i == j) continue; int cnt = 0; for(auto it1 = begin(adjlist[i]),it2 = begin(adjlist[j]); it2 != end(adjlist[j]) && it1 != end(adjlist[i]); ++it2){ if (*it2 > *it1) ++cnt , ++it1; } ans = max(ans , 2 * cnt); } } cout << s.size() — ans << '\n'; } }

Hey, Can somebody please help in problem B, I am getting TLE. I have used DP.

Submission Link

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

https://www.youtube.com/watch?v=S_-BaaH7P80&list=PLrT256hfFv5X1GNpiqEh5njN_WJmQu8Q6

Great Tutorial!!

Why in Problem G we have to find biconnected components? Any counterexamples to SCC?

In Problem B, Solution 1. Why are they adding arr[i+1] to the maximum pair sum. I mean we can go till pos, so we can add the pairs till pos but because of arr[i+1] pair , the last iteration will also add arr[pos+1]. Also, if we need to add that, why they not added (arr[i+1]) into the 's' variable.

Can anyone explain me what the line "Also, you can't perform two or more moves to the left in a row" in the problem 1389B-Array Walk, is meant to be? Can anyone elaborate this line, please?

I used a dynamic programming approach to problem $$$B$$$, but I get an output of $$$18$$$ instead of $$$19$$$ for the inputs

Here is my code

I don't know why this is happening, since for every possible number of total moves and left moves, we could have previously had one less left move or the same number of left moves. This is the main logic in my idea. Any hints or help would be appreciated.