Thanks for participating!

Idea: flamestorm

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
void solve() {
int a, b, c;
cin >> a >> b >> c;
cout << ((a + b == c || c + a == b || b + c == a) ? "YES\n" : "NO\n");
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
```

Idea: mesanu

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
void solve()
{
int n;
cin >> n;
int x[n];
set<int> a;
for(int i = 0; i < n; i++)
{
cin >> x[i];
}
for(int i = 0; i < n; i++)
{
if(a.find(x[i]) != a.end())
{
cout << "NO" << endl;
return;
}
a.insert(x[i]);
}
cout << "YES" << endl;
}
int main()
{
int t;
cin >> t;
while(t--)
{
solve();
}
}
```

Idea: flamestorm

**Tutorial**

Tutorial is loading...

**Bonus**

How do you write a validator for this problem? (Given a grid, check if it is a valid input.)

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
void solve() {
char g[8][8];
vector<int> r;
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
cin >> g[i][j];
if (g[i][j] == 'R') {r.push_back(i);}
}
}
for (int i : r) {
bool ok = true;
for (int j = 0; j < 8; j++) {
if (g[i][j] != 'R') {ok = false; break;}
}
if (ok) {cout << "R\n"; return;}
}
cout << "B\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
```

**Tutorial**

Tutorial is loading...

**Solution**

```
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define forn(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
vector<int> pairs[1001];
void solve() {
int n; cin >> n;
vector<int> id[1001];
for(int i = 1; i <= n; ++i) {
int x; cin >> x;
id[x].push_back(i);
}
int ans = -1;
for(int i = 1; i <= 1000; ++i) {
for(int j: pairs[i]) {
if(!id[i].empty() && !id[j].empty()) {
ans = max(ans, id[i].back() + id[j].back());
}
}
}
cout << ans << "\n";
}
int32_t main() {
for(int i = 1; i <= 1000; ++i) {
for(int j = 1; j <= 1000; ++j) {
if(__gcd(i, j) == 1) {
pairs[i].push_back(j);
}
}
}
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
```

Idea: mesanu

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
void solve()
{
int n, q;
cin >> n >> q;
vector<long long> pref;
pref.push_back(0);
vector<int> prefmax;
for(int i = 0; i < n; i++)
{
int x;
cin >> x;
pref.push_back(pref.back()+x);
if(i == 0)
{
prefmax.push_back(x);
}
else
{
prefmax.push_back(max(prefmax.back(), x));
}
}
for(int i = 0; i < q; i++)
{
int k;
cin >> k;
int ind = upper_bound(prefmax.begin(), prefmax.end(), k)-prefmax.begin();
cout << pref[ind] << " ";
}
cout << endl;
}
int main()
{
int t;
cin >> t;
while(t--)
{
solve();
}
}
```

Idea: SlavicG

**Tutorial**

Tutorial is loading...

**Solution**

```
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define forn(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
void solve() {
int q; cin >> q;
bool otherA = false, otherB = false;
ll cntA = 0, cntB = 0;
while(q--) {
ll d, k; string x; cin >> d >> k >> x;
for(auto c: x) {
if(d == 1) {
if(c != 'a') otherA = 1;
else cntA += k;
} else {
if(c != 'a') otherB = 1;
else cntB += k;
}
}
if(otherB) {
cout << "YES\n";
} else if(!otherA && cntA < cntB) {
cout << "YES\n";
} else {
cout << "NO\n";
}
}
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
```

Idea: SlavicG

**Tutorial**

Tutorial is loading...

**Solution**

```
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define forn(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
void solve() {
int n; cin >> n;
vector<int> a(n);
forn(i, n) cin >> a[i];
//we care at most about first log2(a) values
int cur_or = 0;
vector<bool> vis(n, false);
for(int i = 0; i < min(31, n); ++i) {
int mx = 0, idx = -1;
for(int j = 0; j < n; ++j) {
if(vis[j]) continue;
if((cur_or | a[j]) > mx) {
mx = (cur_or | a[j]);
idx = j;
}
}
vis[idx] = true;
cout << a[idx] << " ";
cur_or |= a[idx];
}
forn(i, n) if(!vis[i]) cout << a[i] << " ";
cout << '\n';
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
```

B was just a few lines with the help of STL :D (Submission: 176051852)

Nice!

Or

which I find even simpler.

But the complexity of your algorithm being O(n*log(n))?? It can easily be solved using O(n) using maps. The logic is simple if any number is present more than once then it is impossible to rearrange in increasing order Here is mine: Submission 177430404

Problem C,When I judge the horizontal and vertical directions of a color,my submission-->175928516, I can't pass the test. But when I only judge the corresponding direction of a color, it passed,my submission-->176020169 Can someone tell me why

Look at the title,some horizontal rows have been painted red, and some vertical columns have been painted blue.

I also didn't read it properly costed me more than an hour to cause it took more than 15 mins to run on pretests :'(

me too!!

now i understand,thx,i ignore the difference between "some" and "all"

Consider this case

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBRRRRRRRRAnswer : 'R' Output : 'B'

some horizontal rows have been painted red, and some vertical columns have been painted blue

some horizontal rows have been painted red, and some vertical columns have been painted blue,it means not all points of a color are painted horizontal or vertical,so the case is legal

thank u :) now i understand!

I was also coding like that.. like R, B in both direction but it was failing.175991481. and at last I was that logic in given test cases.... and cracked it... this was not given anywhere in question 176049904

During the contest, I thought that the rows and columns it said refer to the rows and columns in the sample...

I have the same logic for problem C as mentioned in the tutorial but it fails on tc 2, My submission is https://codeforces.com/contest/1742/submission/176081297 . Could anyone please take a look at why this is failing.

your logic is correct but when you are checking rows ch should be only

Rand while checking column ch should only beB. ****this**** is the corner case.for problem C, many of us missed this case :(

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBRRRRRRRRAnswer : 'R' Output : 'B'

Oh damn!! thanks a lot!!

Thank you

I think the answer should be R, as it has the occurrence 8 consecutive times in a row. Please tell, why are there so many messages related to this?????

Because many of us just checked whether the entire row is same color rather than checking entire row is Red Or Not. This also follow for column also.

Can you explain this testcase?? This would be really helpful. Even i have a doubt in the question

why is my 176088793 getting WA on 11 (problem D)

when the input array is 2 6 10 15 your output is 0 because when you reach 6 the gcd will be 1 but actually there is no coprime number between 6 , 10 , 15 but it is true that their gcd is 1 The correct output should be 5 by taking 2 and 15

I got TLE in problem E. My submission is 176039360 . How can I improve this?

As per your solution I suggest you to have a read on

Time Complexity AnalysisIt would be helpful for you to understand why your code is getting TLEHello Bro, can you tell we why we have subtracted prefmax.begin() in the upper bound function?

Upper bound function returns an iterator. But I need the index of the array. That is why I need to subtract the initial address of the array so that I can find the index. To know more and learn more about this you can follow this.

thank you very much bro :)

use binary search for finding the element. Time complexity will be better in that case. Since the time complexity of the binary search is O(log N) and that of linear search is O(N). Also, you can apply binary search since your lower array is non decreasing one.

try to learn about binary search and Upper_bound function.... That would help you to solve these types of query related problems in a feazeble time...

inspite of brute_force you should use

binary_searchI got TLE on Problem D. Not sure why. How can I improve this solution

Instead of keeping track of every index , you can consider the values of the array elements , that can go up to only 10^3 . Now , if you run two for loops , then time taken O(10^3*10^3) and for T=10 test cases , O(10*10^3*10^3) = O(10^7) [neglecting time complexity for __gcd(a,b) stl ].

Handle duplicates and number 1. Since if 1 is present then gcd with any element will be (1,N)=1 and also do not find gcd of the smaller index. Only find gcd if the index sum of elements is greater than that of previous sum(store sum of index). My solution: Problem D

Thanks it helped me

Video Editorial for Chinese：

Bilibili

Is that Chinese youtube?

YES，and better than YouTube

I think there is an easier way to implement the algorithm in G's tutorial. 176054281

Weak pretest in problem F.

Yeah, got an overflow error. I think that should have been accounted for in pretests.

Thank you for such a great contest and for a great editorial

I got wrong on Problem E. Not sure why. How can I improve this 176044919

try to think about it by easiest way like

if u want way how i solve itjust store in array call freq what is biggest number until now in i and after it prefix the main array and BS about it answer in freq array

u will BS about query in freq array by ur way and handle it , after that output the matchable index from arr

thx,i will try.

I am getting the Wrong answer for F on test case 2 .. my submission- here

plz someone help me out ::) I got my mistake

can u explain a bit more please! :)

sadly, i misread the "OR" as "XOR" and failed to ak

oh bad luck

it's more like oh bad concentration

flamestorm is the solution of bonus of problem C like this :

we go through each row except the ones with full red then check whether if a column is painted blue whether it is painted red in any of the previous rows and if a column is painted red whether it is painted blue in any of the previous rows and if a column is white whether it is painted blue in any of the previous rows; If any of these three happens then input grid is invalid otherwise valid

Is it correct ?

Hey flamestorm,

In problem C, the valid input must contain

exactly oneof theses situations:1- Have one or more rows full of R's

2- Have one or more columns full of B's

otherwise, the input is invalid

Is it correct ?

Well, those are a part of the conditions, but there seems to be more invalid cases to consider. The following is one example.

I don't see any reason of downvoting him , he made a correct point.Maybe he has bad record in past in terms of commenting & posting but that doesn't have any relation with this,And why am I saying this? I am saying this because we all make mistakes in life ,doesn't mean we can't move forward

I got TLE in F question for Test Case #2, any ideas/hints how can I improve the code? submission

You need to save the count of every character, as k can be 10^5 every time and there are 10^5 queries the size of the string can exceed 10^10. You can check out my submission : https://codeforces.com/contest/1742/submission/175994173

can anybody help with the error in my code (Submission :176138148)

Is there any solution to solve D. Coprime better than O(E*E) where E is the maximum possible element. Can it be solved in O(N)?

The solution of problem E is implemented very nicely. Thanks for the editorial.

Hello Bro, can you tell we why we have subtracted prefmax.begin() in the upper bound function?

problems D has some bugs. see https://codeforces.com/blog/entry/107979

I cannot believe it

Problem D. Who knows, why this solution is wrong? (fails on 11th test, see submission).176022793

Basic idea: gcd(a1, a2, a3) = gcd(a1, gcd(a2, a3)); so I find first number from the right, for which gcd([number, …, an]) = 1, and then find first ai from right, for which gcd(number, ai)=1.

Take a look at Ticket 16350 from

CF Stressfor a counter example.I can't understand the concept of the problem "F" solution. Can anyone explain me??

The empty line between each test case of problem C cost me 50 minutes. Since the input box has different colored stripes for any two consecutive inputs, there was no need for that empty line. I thought reading the input section is a waste of few seconds. I should have read the input section carefully.

176168579 Please help I am getting WA on test 10 in problem F but I think that my output is correct

me wa in tc10 too, read statement carefully: Alperen has two strings, s and t which are both initially equal to "a". Just add 2 lines mp1['a']++, mp2['a']++ your AC code: https://codeforces.com/contest/1742/submission/176174906

Got this now. Thanks

In problem D I have gotten wrong answer on test 24 on an invalid test int the statement they said that n must be greater than 2 but I have had wrong answer on a test with n=1 so please rejudge the problem SlavicG , mesanu , flamestorm please do something 175973439 .

Help him pleez

Help him, he deserves achieving pupil :(

Agree

## Help_Rida

that is not fair, please help him .

it is unfair not to rejudge his submission

totaly agreeee

hiz frendz need mental aid

In $$$C$$$, why didn't they put $$$n$$$ and $$$m$$$ $$$(1$$$ $$$\le$$$ $$$n,m$$$ $$$\le$$$ $$$1000)$$$ or something like that?

I think when $$$n=m=8$$$ problem is more like Div.4 $$$B$$$ problem, this could make problem $$$C$$$ little harder, just as hard as a Div.4 $$$C$$$ should be.

I don't think constraints matter at all in C ,n,m<=1000 doesn't make any difference in solution

In some solutions it actually does matter.

Example: 175929785 from ltunjic

how does it matter? you can just create RRRRR....RRRR string just by appending R m times

Right, but I still think problem would be little harder for beginners.

Using n,m<=1e3 . Even a O(n^2) solution can pass ,still the approach will remain same

Hey [user:flamestorm]I submitted the solution of D during contest time. During the contest it got AC but after system test it is still in queue. Why so? It's not also counted in my solution. It just got skipped. What is the reason for that? [submission:176039715]

In Problem B, the checker log is telling me that I gave a wrong answer in test case #2 while my answer is the same as the output. As you can see in the submission, the 25th token is correct. Am I missing something? 176210112

Your code is dereferencing the value of $$$a[n]$$$. This is undefined behaviour, and may have caused a WA any moment.

bro why already decide the size of your array , and after initialization you made mistake , that's why you are getting WA , before solving these problems just recall//revisit on the basics of c++ , and then solve problems ,

good luck

Can dp be used in problem E?

it's basic question we can't use dp because the range of question is 10^9 where array limit is only about 10^6 or you can use map to save previous element but searching in map complexity is log n where we can search for the same in input in same complexity with binary search. so dp can only increase your space complexity and code complexity so yeah dp can be use in problem e but its useless to use it. This is my thinking maybe i am wrong.

In problem-G for test second test case :

input : 5 1 2 3 4 5 5

my_outupt: 5 5 2 5 4 3 1

prefix or of input: [5, 5, 7, 7, 7, 7, 7]

prefix or of myout:[5, 5, 7, 7, 7, 7, 7]

isn't it a correct output. jurys output: 5 2 1 3 4 5 5

with prefix_or: [5, 7, 7, 7, 7, 7, 7]

if my output is not correct then for input : 5,2 the accepted output is 5,2 with prefix_or [5,7] and [5,7] . so what is the problem actually wants . could any one please help? Thank you.

anyone please explain 4 solution

can anyone help me i don't know what i'm doing wrong here 176423898 in question c stripes

I solved C by getting the 8x8 array as a string input line by line and querying each line of input. For example, if a line is equal to 'R' 8 times as in "RRRRRRRR", then the last stripe painted was red, i.e R. Because if there's even one row of full red paint, it becomes futile for the column to ever be the last stripe painted. Here is my solution reference (C#): 176457546

I faced some issue in the coprime problem. can anyone please identify the flaw in the code. Thanks in advance.

you are not moving j.while(i>=0){ if(isCoPrime(arr[j],arr[i])){ found=true; ans=i+j+2; break; } i--; }

In this loop you are not decreasing j.this approach is wrong according to me if you try with two loop then it will show TLE. see Editorial.

check for: 1 2 3 6 according to your code ans should be 4 but real ans is 6 because 2 and 3 are coPrime so, 2*3 = 6.

Have a problem with G

176665365

but the test is really huge and I can catch the problem/

Maybe someone has the idea?

Hey can anyone explain me this one-liner python code for

`1742B`

?Ok I think I got it!

`[*open(0)]`

: here`0`

is file descriptor for`stdin`

. So every line of stdin is put into list.`[2::2]`

: Probably just taking the array of n into consideration.`'YNEOS'[0::2]`

= YES`'YNEOS'[1::2]`

= NO, so if the condition`len(a:=s.split())>len({*a})`

is true then`NO`

else`YES`

.The condition itself is comparing the length of input array and set of the array elements (by

`{*a}`

). If all the array elements are unique then the condition is false, hence`YES`

. But if there is repeating elements in the array, then the length of the set becomes less than the length of array, hence`NO`

.D. Coprime：

## include

using namespace std; int gcd(int a,int b){ if(b == 0) return a; return gcd(b,a%b); } int main(){ int t; cin >> t; while(t--){ int n; int a[200005]; cin >>n; for(int i =1; i <= n; i++){ cin >> a[i]; } int c =a[1]; for(int i =2; i <= n; i++){ c = gcd(c, a[i]); } if(c!=1){ cout << -1<<endl; continue; } c = a[n]; int ans =0; for(int i =n-1; i >= 1; i--){ c = gcd(c, a[i]); if (c==1){ for(int j = n; j >= i+1;j--){ if(gcd(a[i],a[j])==1){ ans = i+j; break; } } } if(ans!=0)break; } cout << ans<< endl; }

} Who can help me?Why my test 1 is wrong?

D, why am I getting TLE in 4th Test case? 176940826can anyone please help me in debugging this code for problem C, i am not getting the correct output at one of the testcases.

my submission : 177701574 https://codeforces.com/contest/1742/submission/177701574

Take a look at Ticket 16377 from

CF Stressfor a counter example.Thanks i modified a little but still i m getting error at one of the newtestcase

new submission :https://codeforces.com/contest/1742/submission/178271786

Well, you can always purchase a subscription if you need more help. Your new submission fails on Ticket 16379

Okay, thanks i didnt know about it

cant Problem E be done in this way ?

We first calculate the upper bound of each K in the array a

Then ans = prefix_sum[location of upper_bound of K — 1]

Please help me if I am getting something wrong

great！！！

Can anyone tell me what i'm ddoing wrong for problem G. Here is my submission i'd

In

problem D (Coprime), would anyone help me to understand that how its time comlexity becomesO(a2i (logai+n))I guess I didn't understand F correctly...

The is the 10th test:

Official answer is:

YES

YES

YES

YES

YES.

Why it isn't:

YES

NO

YES

NO

YES?

UPD: solved it

trash round

192675897 My code shows wrong in the 10th word. But I can't see the 10th word. It finished after the 6th. How can I find all test examples?

Take a look at Ticket 16727 from

CF Stressfor a counter example.