1846A - Rudolph and Cut the Rope

Author: Sasha0738

**Hint 1**

What if the nail with the rope is the only one?

**Hint 2**

How long should be rope to reach the ground?

**Tutorial**

**Solution**

```
#include <iostream>
using namespace std;
int main() {
int test_cases;
cin >> test_cases;
for (int test_case = 0; test_case < test_cases; test_case++) {
int n;
cin >> n;
int ans = 0;
for (int i = 0; i < n; i++) {
int a, b;
cin >> a >> b;
if (a > b)
ans++;
}
cout << ans << endl;
}
return 0;
}
```

1846B - Rudolph and Tic-Tac-Toe

Author: Sasha0738

**Hint 1**

What is the win condition to be checked?

**Tutorial**

**Solution**

```
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main() {
int test_cases;
cin >> test_cases;
for (int test_case = 0; test_case < test_cases; test_case++) {
vector<string> v(3);
for (int i = 0; i < 3; i++)
cin >> v[i];
string ans = "DRAW";
for(int i = 0; i < 3; i++) {
if (v[i][0] == v[i][1] && v[i][1] == v[i][2] && v[i][0] != '.')
ans=v[i][0];
}
for (int i = 0; i < 3; i++) {
if (v[0][i] == v[1][i] && v[1][i] == v[2][i] && v[0][i] != '.')
ans=v[0][i];
}
if (v[0][0] == v[1][1] && v[1][1] == v[2][2] && v[0][0] != '.')
ans=v[0][0];
if (v[0][2] == v[1][1] && v[1][1] == v[2][0] && v[0][2] != '.')
ans=v[0][2];
cout << ans << endl;
}
return 0;
}
```

1846C - Rudolf and the Another Competition

Author: vladmart

**Hint 1**

What is the optimal order of task solving?

**Hint 2**

t_{1} ≤ t_{2} ≤ t_{3} ≤ ... ≤ t_{m}

**Tutorial**

**Solution**

```
#include <bits/stdc++.h>
#define int long long
using namespace std;
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int ttt;
cin >> ttt;
while(ttt--){
int n, m, h;
cin >> n >> m >> h;
pair<int, long long> rud;
int ans = 1;
for(int i = 0; i < n; i++){
vector<int> cur(m);
for(int j = 0; j < m; j++){
cin >> cur[j];
}
std::sort(cur.begin(), cur.end());
int task_cnt = 0;
long long penalty = 0, sum = 0;
for(int j = 0; j < m; j++){
if (sum + cur[j] > h) break;
sum += cur[j];
penalty += sum;
task_cnt++;
}
if (i){
if (make_pair(-task_cnt, penalty) < rud) ans++;
} else rud = {-task_cnt, penalty};
}
cout << ans << '\n';
}
return 0;
}
```

1846D - Rudolph and Christmas Tree

Author: vladmart

**Hint 1**

Consider all the triangles in order from bottom to top.

**Hint 2**

What if the current triangle intersects with the next one?

**Hint 3**

Trapezoid

**Tutorial**

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
cout.precision(10); cout.setf(ios::fixed);
int ttt;
cin >> ttt;
while (ttt--) {
int n, d, h;
cin >> n >> d >> h;
vector<int> y(n);
for(int i = 0; i < n; i++){
cin >> y[i];
}
long double ans = (long double)d * h / 2.0;
for (int i = 0; i + 1 < n; ++i) {
if (y[i + 1] >= y[i] + h) ans += (long double)d * h / 2.0;
else{
long double d2 = (long double)d * (y[i] + h - y[i + 1]) / h;
long double nh = y[i + 1] - y[i];
ans += (d + d2) / 2.0 * nh;
}
}
cout << ans << '\n';
}
return 0;
}
```

1846E1 - Rudolf and Snowflakes (simple version)

Author: natalina

**Hint 1**

1 + k + k^{2} + k^{3} + ... + k^{p}

**Hint 2**

What maximum value of k can be reached for the current constrains on n?

**Hint 3**

Can we precalculate for each n if there is appropriate k?

**Tutorial**

**Solution**

```
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
set<long long> nums;
int main() {
for (long long k = 2; k <= 1000; ++k) {
long long val = 1 + k;
long long p = k*k;
for (int cnt = 2; cnt <= 20; ++cnt) {
val += p;
if (val > 1e6) break;
nums.insert(val);
p *= k;
}
}
int _ = 0, __ = 1;
cin >> __;
for (int _ = 0; _ < __; ++_) {
long long n;
cin >> n;
if (nums.count(n)) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}
```

1846E2 - Rudolf and Snowflakes (hard version)

Author: natalina

**Hint 1**

1 + k + k^{2} + k^{3} + ... + k^{p}

**Hint 2**

What maximum value of k can be reached for the current constrains on n?

**Hint 3**

Can we precalculate for some n and snowflakes with 4 or more levels if there is appropriate k?

**Hint 4**

How can we check snowflakes with 3 levels separately?

**Tutorial**

**Solution**

```
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
set<long long> nums;
int main() {
for (long long k = 2; k <= 1000000; ++k) {
long long val = 1 + k;
long long p = k*k;
for (int cnt = 3; cnt <= 63; ++cnt) {
val += p;
if (val > 1e18) break;
nums.insert(val);
if (p > (long long)(1e18) / k) break;
p *= k;
}
}
int _ = 0, __ = 1;
cin >> __;
for (int _ = 0; _ < __; ++_) {
long long n;
cin >> n;
if (n < 3)
{
cout << "NO" << endl;
continue;
}
long long d = 4*n - 3;
long long sq = sqrt(d);
long long sqd = -1;
for (long long i = max(0ll, sq - 5); i <= sq + 5; ++i) {
if (i*i == d)
{
sqd = i;
break;
}
}
if (sqd != -1 && (sqd - 1) % 2 == 0 && (sqd - 1) / 2 > 1)
{
cout << "YES" << endl;
continue;
}
if (nums.count(n)) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}
```

Author: Sasha0738

**Hint 1**

How number of objects of each type will change after the mimic transformation?

**Hint 2**

What if all the current objects have the same value?

**Tutorial**

**Solution**

```
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int main() {
int test_cases;
cin >> test_cases;
for (int test_case = 0; test_case < test_cases; test_case++) {
int n;
cin >> n;
vector<int> v(n);
map<int, int> m;
for (int i = 0; i < n; i++) {
cin >> v[i];
m[v[i]]++;
}
vector<int> elements_to_erase;
int ans;
for (int i = 0; i < 5; i++) {
if (v.size() - elements_to_erase.size() == 1) {
cout << "! " << ans << endl;
break;
}
cout << "- " << elements_to_erase.size() << " ";
for (int j = 0; j < elements_to_erase.size(); j++) {
cout << elements_to_erase[j] << " ";
}
cout << endl;
vector<int> new_v;
map<int, int> new_m;
for (int j = 0; j < v.size() - elements_to_erase.size(); j++) {
int x;
cin >> x;
new_v.push_back(x);
new_m[x]++;
}
elements_to_erase.clear();
int tm = -1;
for (auto& k : new_m) {
if (k.second > m[k.first]) {
tm = k.first;
}
}
if (tm != -1) {
for (int j = 0; j < new_v.size(); j++) {
if (new_v[j] != tm)
elements_to_erase.push_back(j + 1);
else
ans = j + 1;
}
m.clear();
m[tm] = new_m[tm];
}
v = new_v;
}
}
return 0;
}
```

Author: vladmart

**Hint 1**

Bitmasks

**Hint 2**

Weighted graph

**Hint 3**

Dijkstra's algorithm

**Tutorial**

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int ttt;
cin >> ttt;
while (ttt--) {
int n, m;
cin >> n >> m;
bitset<10> tmp;
cin >> tmp;
int s = (int) tmp.to_ulong();
vector<pair<pair<int, int>, int>> edges(m);
for (int i = 0; i < m; i++) {
cin >> edges[i].second;
cin >> tmp;
edges[i].first.first = ((1 << n) - 1) ^ (int) tmp.to_ulong();
cin >> tmp;
edges[i].first.second = (int) tmp.to_ulong();
}
vector<int> dist(1 << n, INT_MAX);
dist[s] = 0;
set<pair<int, int>> q = {{0, s}};
while (!q.empty()) {
auto [d, v] = *q.begin();
q.erase(q.begin());
for (int i = 0; i < m; i++) {
int to = v & edges[i].first.first;
to |= edges[i].first.second;
if (dist[to] > d + edges[i].second) {
q.erase({dist[to], to});
dist[to] = d + edges[i].second;
q.insert({dist[to], to});
}
}
}
if (dist[0] == INT_MAX) dist[0] = -1;
cout << dist[0] << '\n';
}
return 0;
}
```

HacksForces

Hahaha

Hahaha Hahaha

For E2 there is a simpler solution to check if there exists a $$$k$$$ such that $$$1 + k + k ^2 = n$$$.

You can rearrange the equation to $$$k(k + 1) = n - 1$$$. It follows that the only $$$k$$$'s you need to check are $$$\lfloor \sqrt{n - 1}\rfloor$$$ and $$$\lfloor \sqrt{n - 1}\rfloor - 1$$$. This is a modified version of the official solution : code

EDIT: as davi-v pointed out,there is no need to check for $$$\lfloor \sqrt{n - 1}\rfloor - 1$$$.Great!

can you explain why are checking the equation for k=2 and why we only floor(sqrt(n — 1)) and floor(sqrt(n — 1)) — 1 ?

Sure! I assume you meant $$$exponent = 2$$$,not $$$k = 2$$$. If you deal with all $$$k$$$ $$$<=$$$ $$$10 ^ {6}$$$ (for example by precalculating a set of valid numbers),for all other $$$k$$$ $$$>$$$ $$$10 ^ {6}$$$ the exponent must be

exactly$$$2$$$. That is because the minimmum exponent is $$$2$$$ and if the exponent was $$$3$$$ the sum would be larger than $$$10 ^{18}$$$.Now we have to check that there exists a $$$k$$$ such that $$$1 + k + k ^2 = n$$$. This is the same as checking if there is a $$$k$$$ such that $$$k(k + 1) = n - 1$$$. Now, let $$$r = \sqrt{n - 1}$$$. It is easy to see that $$$k < r$$$. However, $$$k$$$ must be as large as possible so $$$k$$$ is either $$$\lfloor r \rfloor$$$ or $$$\lfloor r \rfloor - 1$$$ when $$$n - 1$$$ is a perfect square.I want to say like for p=2 the equation would be 1+k+k^2. So why we have taken for p=2 if the no. Is not found in precalculated array.

Because for n <= 10

^{18}and p = 2 the value of k could be approximately sqrt(n) = 10^{9}. But we can precalc values only for k <= 10^{6}1+k+k^2=n how do we get to this ?

This is formula for the number of vertexes in minimal snowflake. 1 vertex at the initial value, k vertexes at the second layer and k^2 vertexes at the third one.

انا

Very inefficient approach. It can be done with binary search sir !!!

I did think about BS in the cts, but I couldn't go with that, can you tell me that solution ?

can u explain "if (p > (long long)(1e18) / k) break;" Why this statement is necessary?

The if statement is necessary because otherwise the sum could overflow. If you have to check if $$$a \cdot b > val$$$ , a neat way of doing this and avoiding overflow is checking if $$$a > val / b$$$. Hope this helps!

Thank u for your reply, i got it:)

You don't need to check $$$\lfloor{\sqrt{n-1}}\rfloor - 1$$$. Let $$$m = n - 1$$$. For $$$k(k+1)=k^2+k$$$ to be $$$m$$$, $$$k < \sqrt{m}$$$. Let $$$x = \lfloor\sqrt{m}\rfloor$$$. The question is: which numbers of the form $$$k=x-c$$$, $$$c \ge 0$$$ do we need to check?

Substituting in $$$k(k+1)$$$, we get $$$(x-c)(x-c+1)=x^2+x(-2c+1)+c^2-c$$$. Note that $$$x(-2c+1)+c^2-c$$$ must be $$$\ge 0$$$, as $$$x^2 \le m$$$. We need to show that if $$$c > 0$$$, $$$x(-2c+1)+c^2-c < 0$$$, so $$$x^2+x(-2c+1)+c^2-c$$$ can't equal $$$m$$$.

Note $$$c \le x$$$, otherwise $$$k$$$ would be negative. Let $$$x = c + d, d \ge 0$$$. Substituting, we get

If $$$c > 0$$$, $$$-c^2-2cd+d=-c^2+d(-2c+1) < 0$$$, as $$$-c^2 < 0$$$ and $$$-2c+1 \le 0$$$.

You are correct! Thank you for pointing out my mistake, I will make sure to correct it shortly. Also, I found a simpler proof to prove we don't need to check $$$ \lfloor \sqrt{n - 1}\rfloor$$$ :

I claim that for any number $$$a = k ^ {2},k\in\mathbb{N},\nexists b \in \mathbb{N}$$$ such that $$$b \cdot (b + 1) = a$$$. When $$$b = k,b \cdot(b + 1) > a$$$ and when $$$b <= k - 1,b \cdot(b + 1) < a$$$. Because $$$b \cdot(b + 1)$$$ is either $$$>$$$ or $$$<$$$ than $$$a$$$,we can conclude that $$$b \cdot(b + 1) \neq a$$$.

E2 is a superb binary search question, liked it

I did think about BS in the cts, but I couldn't go with that, can you tell me that solution ?

If you still haven't got the idea a more simpler bs solution is consider 1+k+k^2+..k^p, observe that this is less than k^(p+1) and as well observe that the maximum value of p can be 63 since any power greater than p and the sum>1e18, so from here we can bruteforce for all powers and for this power we can upper bound the value of k by n^(1/p), and binary search which value of k satisfies 1+k+...k^p==n, if any confusion feel free to ask.

Can you please explain the concept of n^(1/p). I am not getting this part.

So if it's something like if 1+k+k^2+..+k^p=n then k^(p+1)>n, you can easily deduce that I think(correlate it with binary recall that bit set at position i is bigger than all combined from i-1 to 1), so from k^(p+1)>n, we get got any k greater than n^(1/(p+1)) we definitely can't get the answer so that is the upper bound, hope it helps.

got it,thanks.

Hi, can anyone help me debug my submission for E2? Don't know why I am getting WA.Thanks in advance.212818653

I don't know why I can't pass test9 :(((

I tried the similar approach in python (no overflow) and got WA on tc 16 after increasing the max value upto 1e25... (was AC initially and finally got hacked :(( )

Further increment gave TLE... maybe the constraints weren't designed for the binary search solution... (unless I'm wrong, in which case I would appreciate if someone pls provided me with an AC solution)

P.S: My initial approach in C++ was returning LLONG_MAX in the binpow function whenever overflow occured (used a check_overflow() function...) which gave WA on tc 5... Idk why...

why is the D in editorial AC without setprecision? what differs from my submission . ik i messed up the setprecision in my code but still why

Precision is set before the solution code at the beginning of the main function

wow didn't notice. nice info ty

At the time of contest 1846C is accepted so I have done 4th but next day it is showing wrong on test case 10. it's your fault not mine.

it's your fault

It's plateform fault because it is showing correct how can I know it fails on some test case

After contests end more tests are added

From the announcement: "The round will be hosted by rules of educational rounds (extended ICPC). Thus,

solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of open hacks."This is part of typical D3 rounds: you will not know whether your solution passes all tests during the contests, and this challenges you to think carefully about special and corner cases and thus helps to improve your skills.

For E2, there is a much simpler code arising from the result of an inequality.

We can just iterate through the values of p. For each p, we need to check for only a single value of k which is floor(n^(1/j)). This is the only possible value of k that may satisfy the condition for a given value of p. This is because:

k^p < 1+k+(k^2)+....+(k^p) < (1+k)^p implies k < (1+k+(k^2)+....+(k^p))^(1/p) < 1+k

If n is equal to 1+k+(k^2)+....+(k^p), then k should be less than n^(1/p) and as per the above inequality k should be floor(n^(1/p)).

Here's the code.

A very effective and simple approach, many thanks !!

e2 is yituodabian

e2 hao nan

So we can take one medicine multiple times in G?

It doesn't matter if it's allowed or not since it would never actually be useful to take the same medicine multiple times anyway.

I claim that if there is a sequecne of medicine that removes all symptoms and medicine $$$x$$$ is used multiple times, we can keep the last occurrence of $$$x$$$ and remove all earlier ones, and the sequence stays valid.

Proof: Consider the symptoms medicine $$$x$$$ cures. It doesn't matter if these symptoms are cured or not before the last occurrence of $$$x$$$ since the last time of taking medicine $$$x$$$ cures these symptoms anyway. Thus, taking medicine $$$x$$$ also earlier has no effect.

Now consider the symptoms medicine $$$x$$$ gives. It doesn't matter if these symptoms are cured or not before the last occurrence of $$$x$$$ since taking this medicine will give these symptoms anyways, and they need to be cured later. Thus, taking medicine $$$x$$$ also earlier has no effect.

This means that taking medicine $$$x$$$ also earlier than the last occurrence has no effect on any symptoms. $$$\square$$$

Thanks a lot for the proof, its crystal clear!

this round has really weak testcases. It's not only for B and C, but also for G. This is my submission for G, in no way should it pass but still it passes.

Your solution is fine (why I don't know), check the announcement page someone explained why this works

ummm... the solution takes at most 20 medicine.

Yes, but your solution is still correct. It can be shown that we never need more than 10 medicine to get fully cured, here is a short explanation (without proof). I can provide a proof if you can't see why that is true.

I failed five times on E2. But G is a piece of cake,because UVA658 is quite similar to it.

My Solution Pretty sure , I have the same solution as that of the editorial . Dunno why I got the TLE though

use c++

Use fast input. Passed in 202 ms. Code: 215151784

Any idea why this submission is giving a WA for D? I have just subtracted the overlapping area instead of using the trapezium formula.`

https://mirror.codeforces.com/contest/1846/submission/212715188

setprecision() goes before output we set precicion for

In problem G, what if Rudolf cannot transition from the initial state to a specific state using a particular medicine, but can still alleviate symptoms one by one?

1846-C - Rudolf and the Another Competition

Any idea why my submission gets a TLE here. Time complexity is O(n*mlogm). 212937677

Spoilerimport sys

from os import path

input = sys.stdin.readline

if(path.exists('input.txt')): sys.stdin = open("input.txt","r") sys.stdout = open("output.txt","w")

def listInp(): return list(map(int,input().split()))

def mapInp(): return map(int,input().split())

for _ in range(int(input())):

can python sqeeze through C and D without TLE?

here's my code and they seem to not work even tho the complexity is exactly identical to the editorial

https://mirror.codeforces.com/contest/1846/submission/212613024 (C)

https://mirror.codeforces.com/contest/1846/submission/212645926 (D)

an explanation or help would be wonderful.

For D you can see my submission here 212968791 (I just change your code a little and get accepted)

thanks man. Apparently storing it in an array doesn't work as fast as I thought

Del.

My solution in B does not require a review of 4 cases.

well , i am still a beginner, but in problem 3 ... i don't get how did we store the solution times for each student by this :

for(int i = 0; i < n; i++){ vector cur(m); for(int j = 0; j < m; j++){ cin >> cur[j]; }

while this should only store the last student only since it overwrites the existing values in each new outer loop ?

You're correct; this only stores the solution times for the last student. But notice that right after reading this, we already handle this student by calculating the optimal number of problems and minimum time penalty for this student. Thus, we don't need to store the problem-specific times for any longer and they can be overwtitten by new data.

how does this work in C solution given in the editorial?

make_pair(-task_cnt, penalty) is used to create a pair with two values: -task_cnt and penalty. The negative sign before task_cnt is used to ensure that when comparing pairs, the pair with a smaller task_cnt value takes higher priority. If the -task_cnt is equal for two pairs, the comparison will be based on the penalty value. This pair is used for comparison in the following line:

212951522 For problem F, why does this code give Idleness Limit Exceeded

You output mimic position before reading full response from OJ

In E2, can somebody tell me the value of k that satisfy 29th test case i.e. n = 64000160000400001, I think this test case is wrong.

400,000

Dial's algorithm can be used in G to achieve $$$O(2^n \cdot (m + MAXd))$$$ complexity. 212963777

Cuz the range of weight of a edge is 10^3(d) and there is 1024 vertices at max, so Dial's algorithm work fine but not more effective than Dijkstra's algorithm with STD in C++.

Amazing problemset. Anyone easily can hacks.(hahaha)

I did think about BS at E2 in the cts, but I couldn't go with that, someone please tell me the editorial of that solution, I really need it. Thank you.

Assume that we have an array {$$$a_k$$$} satisfied $$$a_k = k^2+k+1$$$ , then we can see $$${a_k}$$$ is increasing when k increases. So BS works on this array. We can use BS to check whether the given $$$n$$$ is in {$$$a_k$$$}.Pick $$$l$$$=2 and $$$r$$$=minimum $$$k$$$ satisfied $$$a_k > 10^{18}$$$ and BS.Then we can check $$$b_k = k^3+k^2+k+1 $$$ , $$$c_k = k^4+k^3+k^2+k+1 $$$ ... till $$$A_k = k^{63}+k^{62}+...+k+1$$$ .If the given $$$n$$$ doesn't stay in the arrays mentioned above , the answer is "NO".

You don't really need to construct these arrays and store the value of them (MLE) , just calculate $$$a_{mid}$$$ and compare $$$a_{mid}$$$ with the given $$$n$$$ .

Thank you so much, I wish you best in the next contest <3

In problem G, you should have mentioned that you can use a medicine several times because without it the problem seems quite complicated. By the way, does anyone know if it's possible to solve the problem if we can use each medicine only once?

this is my submission for this problem submission. When we use one medicine at a time then we just have to make the following change in the code

I just swapped the two loops of dp just like we do in coin change dp problem.

check this

For problem E? How do we obtain the maximum possible value for k

Try to solve E2 with binary search, Don't know what I've missed

https://mirror.codeforces.com/contest/1846/submission/212862518

Line 11: i<62

It should be i<64 i think

this solution is accepted in e1 but not in e2,provide me some reason, let n=1+x+x^2......+x^i ------ eq(1) now, we know, x^i<n<(x+1)^i (by binomal theorm) that, means x<n^(1/i)<x+1, now x can be gif(n^(1/i)), we will satisfy eq(1) with x, if it is true ans exist. we will do it for i between 2 to 61.

Hey your solution is too slow, the bound for the largest power is incorrect as well, check for i<=63, since 2^63~1e18, it will still tle with the current solution though, check this for an alternate approach my approach

it would be 60, and now it's tle, but complexity for each case woould be logn*logn*60

By your approach right, mine is a simple binary search that's just logn*63

my solution is giving tle now :(

your apporach is logn*logn*60 too

How is this 212715312 logn*logn*60?

leave it i have used inbuilt pow function in python

Oooh, F was sooo easy, why I didn't read it during the contest:(

I'm not sure about this, but for E1 (and maybe E2 for some $$$n$$$ not too large), since $$$n = 1 + k + k^2 + k^3 + \dots$$$, we would have $$$k$$$ be a divisor $$$(n - 1)$$$, and we can iterate all such k, noting that for some integer $$$p \geq 2$$$ (if it exists):

We only need to check if $log_k{(n(k - 1) + 1)} - 1$ is an integer and if it is larger to equal to $$$2$$$ using repeated division, since the float log function can be expensive:

how does this work in C solution given in the editorial?

有人吗

为什么使用等比数列预处理会WA？

In E2 , for n greater 1e12 we can directly use Shree Dharacharya formula to know whether a k exist or not. And for k from 1 to 1e6 we can make a set of sum of G.P . I got this intuition in the contest but was getting TLE while calculating the sum of G.P . :( Got accepted in 1 try post contest.

need help of E2 , don't know why can't pass test 9 :(( please help , thx a lot! 213588968

if (p > (long long)(1e18) / k) break; Why this statement is necessary in E2

Why there is Rudolph and Rudolf?

Can somebody help me to spot why is giving Idleness Limit Exceeded?

https://mirror.codeforces.com/contest/1846/submission/213728138

I've tried several things, but still can't find the issue.

Same bro, did you find the issue?

No, I haven't found it yet. I've stopped debugging it for some days as well

my very easy to understand solution for E2,#213938261

I think this thing is simpler than editorial in G problem. Call the current status is S, the good effect is A, the bad effect is B. So when you take a medicine, the status transform

`S & (~A) | B`

I am getting WA on test case 2 for n = 7 in my submission for E1. I have tried my solution on my local also on the online available gnu c++ compiler it is giving correct answer that is "YES" while when I run it on here it give WA. I am not able to get why so? If anyone here can help me, I will be very thankful to him.

Sorry for posting this late, can someone please tell why this gives wrong answer in problem E1 test 2, my approach instead of searching for the exponent I used log to find it, any help would be appreciated, Thanks in advance.

Hi, could anyone help debug my submission? This is for 1846B. My submission is https://mirror.codeforces.com/contest/1846/submission/214305302

Thanks in advance!

Note that after finding a winner when you check rows and columns, you just break for loop, meaning you are checking diagonals even if you found the winner.

Yup, got it, thanks so much!

What a deceitful problem C!

what is k for 1000015000057 in 11th test in E2?

K=1000007 level=1

thanks

Why do I get a Idleness Limit Exceeded on my code for F? [Code] Please help

Your code didn't link

Hello guys, in 1846F - Rudolph and Mimic, continously getting time limit exceeded, even though I have verified my solution with editorial, and have tried to match it to the author's solution. Can anyone please let me know what could be the issue here :( -> 216464518

Can anyone tell me why my Code is giving an output of 3 whereas the required output is supposed to be 2 for the following test case(Test Case number 2 — Problem 13).

217202106

Test Case:

1

5 5 14

7 2 9 10 3

5 3 2 9 7

6 1 10 5 7

2 6 9 10 4

10 5 7 8 9

why this 221034056 isn't working for c am i missing something

I did solve G without using Dijkstra is it hackable?

my idea to use dp is either i take the element either not i know sometimes you'll be forced to take medicine two times or 3 ...

but what i did that i did push the element n*m times so that max number we take that element is n times no more that that is that true? https://mirror.codeforces.com/contest/1846/submission/223492698

alternate dp solution for G

code1846G is really a nice question,lovely