Codeforces Round #632 (Div. 2) editorial

#	User	Rating
1	tourist	3985
2	jiangly	3814
3	jqdai0815	3682
4	Benq	3529
5	orzdevinwang	3526
6	ksun48	3517
7	Radewoosh	3410
8	hos.lyric	3399
9	ecnerwala	3392
9	Um_nik	3392

#	User	Contrib.
1	cry	169
2	maomao90	162
2	Um_nik	162
4	atcoder_official	161
5	djm03178	158
6	-is-this-fft-	157
7	adamant	155
8	awoo	154
8	Dominater069	154
10	luogu_official	150

I hope you liked problems!

Sorry for incorrect placement of problems. I had to do swap(E, F).

Tutorial is loading...

Solution C++

#include <bits/stdc++.h>

using namespace std;


void solve() {
    int n, m; cin >> n >> m;
    string black_row(m, 'B');
    vector<string> result(n, black_row);
    result[0][0] = 'W';
    for (int i = 0; i < n; ++i) {
        cout << result[i] << '\n';
    }
}


int main() {
    int t; cin >> t;
    while(t--) solve();
}

Tutorial is loading...

Solution C++

#include <bits/stdc++.h>

using namespace std;

void solve() {
    int n; cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    for (int i = 0; i < n; ++i) {
        cin >> b[i];
    }
    vector<int> good(2, 0);
    for (int i = 0; i < n; ++i) {
        if (a[i] > b[i] && !good[0]) {
            cout << "NO\n";
            return;
        } else if (a[i] < b[i] && !good[1]) {
            cout << "NO\n";
            return;
        }
        if (a[i] == -1) good[0] = 1;
        if (a[i] == 1) good[1] = 1;
    }
    cout << "YES\n";
}

int main() {
    int t; cin >> t;
    while(t--) {
        solve();
    }
}

Tutorial is loading...

Solution C++

#include <bits/stdc++.h>

using namespace std;


int main() {
    int n; cin >> n;
    vector<long long> prefix(n + 1, 0);
    for (int i = 0; i < n; ++i) {
        int x; cin >> x;
        prefix[i + 1] = prefix[i] + x;
    }
    int begin = 0, end = 0;
    long long ans = 0;
    set<long long> s = {0};
    while(begin < n) {
        while(end < n && !s.count(prefix[end + 1])) {
            ++end;
            s.insert(prefix[end]);
        }
        ans += end - begin;
        s.erase(prefix[begin]);
        ++begin;
    }
    cout << ans << endl;
}

Tutorial is loading...

Solution C++

#include <bits/stdc++.h>

using namespace std;

int n, k;

vector<int> find_steps(const vector<int>& a) {
    vector<int> steps;
    for (int i = 0; i < n - 1; ++i) {
        if (a[i] == 1 && a[i + 1] == 0) steps.push_back(i);
    }
    return steps;
}


int main() {
    cin >> n >> k;
    string s; cin >> s;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) a[i] = (s[i] == 'L') ? 0 : 1;
    int maxi = 0, mini = 0;
    int cnt = 0;
    int last = -1;
    for (int i = n - 1; i >= 0; --i) {
        if (a[i] == 0) {
            ++cnt;
        } else {
            if (cnt == 0) continue;
            maxi += cnt;
            mini = max(cnt, last + 1);
            last = mini;
        }
    }
    if (k < mini || k > maxi) {
        cout << -1;
        return 0;
    }
    bool is_min = false;
    vector<int> have_step;
    for (int i = 0; i < k; ++i) {
        if (!is_min) {
            auto steps = find_steps(a);
            cout << min(int(steps.size()), maxi - k + i + 1) << ' ';
            int cur = 0;
            while (k - i - 1 < maxi && cur < steps.size()) {
                cout << steps[cur] + 1 << ' ';
                a[steps[cur]] = 0;
                a[steps[cur] + 1] = 1;
                ++cur;
                --maxi;
            }
            if (maxi == k - i - 1) {
                is_min = true;
                have_step = find_steps(a);
            }
        } else {
            int v = have_step.back();
            have_step.pop_back();
            cout << "1 " << v + 1;
            a[v] = 0;
            a[v + 1] = 1;
            if (v > 0 && a[v - 1] == 1) {
                have_step.push_back(v - 1);
            }
            if (v + 2 < n && a[v + 2] == 0) {
                have_step.push_back(v + 1);
            }
        }
        cout << '\n';
    }
}

Tutorial is loading...

Solution C++

#include <bits/stdc++.h>

using namespace std;


int main() {
    int n; cin >> n;
    if (n < 3) {
        cout << -1;
        return 0;
    }
    vector<int> solution = {
        1, 3, 4, 8, 2, 7, 9, 5, 6
    };
    vector<vector<int>> table(n, vector<int>(n, 0));
    int cur = 1;
    for (int i = 0; i < n - 3; ++i) {
        if (i & 1) {
            for (int j = n - 1; j >= 0; --j) {
                table[i][j] = cur;
                ++cur;
            }
        } else {
            for (int j = 0; j < n; ++j) {
                table[i][j] = cur;
                ++cur;
            }
        }
    }
    if ((n - 3) & 1) {
        for (int j = n - 1; j >= 0; --j) {
            if (j & 1) {
                for (int i = n - 3; i < n; ++i) {
                    if (j > 2) {
                        table[i][j] = cur;
                        ++cur;
                    } else {
                        table[i][j] = solution[(2 - j) * 3 + i - n + 3] + n * n - 9;
                    }
                }
            } else {
                for (int i = n - 1; i >= n - 3; --i) {
                    if (j > 2) {
                        table[i][j] = cur;
                        ++cur;
                    } else {
                        table[i][j] = solution[(2 - j) * 3 + n - 1 - i] + n * n - 9;
                    }
                }
            }
        }
    } else {
        for (int j = 0; j < n; ++j) {
            if (j & 1) {
                for (int i = n - 1; i >= n - 3; --i) {
                    if (j < n - 3) {
                        table[i][j] = cur;
                        ++cur;
                    } else {
                        table[i][j] = solution[(j - n + 3) * 3 + n - 1 - i] + n * n - 9;
                    }
                }
            } else {
                for (int i = n - 3; i < n; ++i) {
                    if (j < n - 3) {
                        table[i][j] = cur;
                        ++cur;
                    } else {
                        table[i][j] = solution[(j - n + 3) * 3 + i - n + 3] + n * n - 9;
                    }
                }
            }
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            cout << table[i][j] << ' ';
        }
        cout << '\n';
    }
}

Tutorial is loading...

Solution C++

#include <bits/stdc++.h>

using namespace std;


vector<int> max_div;

void eratosthenes(int limit) {
    max_div.assign(limit + 1, 0);
    max_div[0] = limit + 10;
    max_div[1] = 1;
    for (int i = 2; i <= limit; ++i) {
        if (max_div[i]) continue;
        for (int j = i; j <= limit; j += i) {
            if (max_div[j]) continue;
            max_div[j] = j / i;
        }
    }
}


int main() {
    int n; cin >> n;
    eratosthenes(n);
    sort(max_div.begin(), max_div.end());
    for (int i = 1; i < n; ++i) {
        cout << max_div[i] << ' ';
    }
}

deque< vector<int> > indices; // indices[x] is the the list of indices that you can swap at time x int min_moves = 0; int max_moves = 0; while (true) { // should be at most O(N^2) min_moves += 1; // add a new set of moves to perform indices.push_back(vector<int>()); for (int i = 1; i < s.size(); i++) { // find all swaps we can make if (s[i] == 'L' && s[i - 1] == 'R') { indices.back().push_back(i); max_moves += 1; } } if (indices.back().size() == 0) { // we cant do any more swaps indices.pop_back(); min_moves--; break; } // perform the swaps for (int i : indices.back()) { swap(s[i], s[i - 1]); } }

int adjust = k - min_moves; while (adjust--) { // take one swap, then perform it separately in one move cout << "1 " << indices.front().back() << '\n'; indices.front().pop_back(); if (indices.front().size() == 0) { // we cleared a time step, therefore we should increase "adjust" indices.pop_front(); adjust++; } } // print the remaining steps for (auto v : indices) { cout << v.size() << ' '; for (int i : v) { cout << i << ' '; } cout << '\n'; }

include <bits/stdc++.h>

using namespace std;

long long n; map<long long,long long> ls;

int main() { long long i,k,sm=0,mx=0,z=0;

scanf("%lld",&n); ls[0]=1; for(i=1;i<=n;i++) { scanf("%lld",&k); sm+=k; mx=max(mx,ls[sm]); z+=i-mx; ls[sm]=i+1; } printf("%lld\n",z);

}

This works for problem C but what I don't understand is why do we have

mx=max(mx,ls[sm]); Is this not equivalent to if(ls.find(sm)!=ls.end())mx = ls[sm];

Comments (130)

Show archived | Write comment?

PuRpLe_FoReVeR

5 years ago, # |

+10

Auto comment: topic has been updated by PuRpLe_FoReVeR (previous revision, new revision, compare).

→ Reply

towhid1zaman

+14

Finally !! thanks for the editorial.

coolshubham

Can anyone help why i am getting wrong answer on testcase 7 -76004123

shubhammitt

5 years ago, # ^ |

← Rev. 4 →

Read the checker's comment. It is said in the question to print those indices where heads will turn from R to L.

Check this Testcase:
6 6
RLRLRL

← Rev. 2 →

I did few changes(but dont know whats the difference between this and previous code) but still not able to pass testcase 7-76027745

I got my error. Thanks a lot !!

fried-chicken

+76

I guess many people print something like chess-board on the problem A, so do I lol

Anyway, the editorial of the problem A is really amazing!

Love_Triangle

+15

i did the same like chess...,but figured out the correct answer after the contest(same as editorial)..baaah!!.

+20

Your answer is correct too :)

iam_anurag_9

if the chess board is correct, then why it was showing wrong answer.

like chess board, but a little difference.

EliasHridoy

But his Note for 1st test case is also confusing. where Black = 4 and white = 2. But he said B=3,W=2.

cshuenss

Just because there are 4 black squares in the example doesn't mean that B=4. If you read the question carefully you'll find that B is the number of black squares that border at least one white square, not the total number of black squares, and thus you can see that B=3 as the bottom right black square doesn't border any white squares.

lov.ish

-36

Another way to explain F is

GCD will be 1 when all the numbers are prime. So let x be the number of primes till n. Therefore for each i in [2, x] answer will be 1. After that one can observe we should include the smallest composite number every time.

Proof: If we include y before x such that y > x. Let pf(k) denotes the maximum prime factor of number k. So since we know that all prime numbers are included in the set, this condition will simply hold pf(y)>=pf(x). So it is optimal to select x before y to minimize imperfection.

mightymercado

+41

Consider n = 9

Of course, we first take all the prime number from 2 to 9 and 1 as well.

1 2 3 5 7

Hence, for k = 2 to 5, the minimal imperfection is 1 because the gcd is always 1.

For k = 6, we pick 4 so minimal imperfection is now 2.

For k = 7, we pick 6 so minimal imperfection is now 3.

For k = 8, we pick 9 so minimum imperfection is still 3.

For k = 9, we pick 8 so minimum imperfection is now 4.

In this example, it doesn't hold that we take the smallest composite number every time. Correct me if im wrong.

_pastor_

+18

The problems were very good but the tutorial is too damn late bro still you really did a good job thanks!

falconlover

Can someone please explain why my solution for problem A is failing. 76024216
Maybe, there's an issue with my declaration of the 2d array, as the output to some of the subcases has 'W' in the top left as well as right corner which is not what I intend to do.

Ebiarat

just declarate array before main and u will get AC.and if u will declarate it before main u not need to initialize array by zero

And how say Yuki726 "You are using i for iterating tests and rows both."

That would solve it. But my question is why is my code failing? Isn't declaring the array as int arr[n][m]={0} not the correct way of declaration?

Yuki726

You are using i for iterating tests and rows both.

Changing variable names didn't help 76026831.

.-._._-_.-_.-._-..

That wouldn't cause the issue. Even if it does, it should be a compile error. The access to arr[i][j] will use the closest declared 'i' in the scope (similar to how you can shadow a global variable of the same name inside a function).

ok, my above answer is not correct. initializing array with ={0} is not correct, try initializing with ={} and it passes, 76027021

I believe = {0} syntax is only valid for statically allocated arrays. Though I cannot find a reference for this. It doesn't even compile on my GCC.

+16

Do anyone know how to prove minimum bound for k in problem.D?

← Rev. 3 →

+11

Using dynamic programming, you can calculate for each 'R' what the maximum number of moves it will take to get its position in the final array. Formula:

dp[cur 'R'] = max(dp[next 'R'] + 1,'L' cnt on [i + 1, ..., n]).

It only remains to note that for the strategy in the solution, this estimate is achieved.

I just spoilered how to calculate this value :(

Thanks for the reply. Great contest btw.

ticmis

A bit confused about the formula.

What about: "LLLRRR" It is obviously that the minimum move is 0.

But according to the formula,"dp[next 'R'] + 1" force the value to add at least one for each 'R'. The first 'R' to calculate in dp is the first 'R' which is not on its right position? This makes me confused.

Im sorry. Its works only for 'R' after which there are 'L's in the string. We can drop out rightmost 'R's.

OK,got it. Thanks a lot! (*^▽^*)

woshinande

Does it mean that for consecutive 'R's, the formula should be: dp[cur 'R'] = max(dp[next 'R'], 'L' cnt on [i+1,...,n])?

abhinav12

can anyone help me? It says i failed on test2 token 175: 76028180

1
2
-1 -1
-1 1

try this case.

Hamim99

Is it possile to solve C using segment tree.If yes then please give your code or explain it.

spookywooky

Somebody want to explain F? I do not get it at all. What is the question, what is the idea to solve it?

There is obviously something with divisors because the imperfection is defined using gcd. And then?

dCoding

The problem: Given the set of integers from $$$1$$$ to $$$n$$$, $$$\forall$$$ $$$i$$$ $$$\epsilon$$$ $$$[2..n]$$$, find a set, $$$M$$$ of size $$$i$$$ such that max value of $$$\gcd (p,q)$$$ $$$\forall$$$ $$$p,q$$$ $$$\epsilon$$$ $$$M$$$ is minimized.

The solution: Firstly, note that $$$1$$$ must always be included in the optimal set. Next, note that if there are $$$P$$$ primes from $$$[2..n]$$$, then max value of $$$\gcd (p,q)$$$ $$$\forall$$$ $$$p,q$$$ $$$\epsilon$$$ $$$M$$$ is always $$$1$$$. Hence, for the first $$$P$$$ numbers, answer will always be $$$1$$$. Now from the $$$p+1$$$ th number onwards, if we include an integer $$$x$$$, the optimal set always contains at least $$$1$$$ divisor $$$d$$$ $$$(1\le d<x)$$$ of $$$x$$$. Why? Since we have already included all the primes from $$$[2..n]$$$, all the numbers we are left with are composite. Now if we add $$$x$$$ to the set $$$M$$$, max value of $$$\gcd (p,q)$$$ $$$\forall$$$ $$$p,q$$$ $$$\epsilon$$$ $$$M$$$ is the largest divisor of $$$x$$$, since $$$\gcd (\texttt {largest divisor of } x, x) = \texttt {largest divisor of } x$$$. Hence, we have to add elements to set greedily such that the largest divisor of $$$x$$$ $$$\forall$$$ $$$x$$$ $$$\epsilon$$$ $$$M$$$ is minimized. For this, we can find out the largest divisor, $$$d$$$ $$$\forall$$$ $$$i$$$ $$$\epsilon$$$ $$$[2..n]$$$, and store it in a list $$$v$$$. The answer is the sorted list $$$v$$$.

Code

#include <bits/stdc++.h>

#define ll long long int
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

#define MS(x,a) memset((x), (a), sizeof(x))
#define F0R(i,n) for(auto (i) = 0; (i) < (n); (i)++)
#define FOR(i,a,b) for(auto (i) = (a); (i) <= (b); (i)++)
#define ROF(i,a,b) for(auto (i) = (a); (i) >= (b); (i)--)

using namespace std;

const int INF = 1e9+2;
const ll LINF = 1e18+2;
const int MX = 2e5+5;
const ll MOD = 1e9+7;

int lf(int n) {
    int ans = 1;
    for(int i = 2; i*i <= n; i++) {
        if(n%i == 0) {
            ans = max(ans,i);
            ans = max(ans,n/i);
        }
    }
    return ans;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    int n; cin >> n;
    vector<int> oof(n); 
    oof[0] = 1;
    FOR(i,1,n-1) {
        int n2 = lf(i+1); 
        oof[i] = n2;
    }
    sort(all(oof));
    FOR(i,1,n-1) cout << oof[i] << " ";
}

Neel_Patel

In Second problem In this test case may be editorial's answer is wrong in

0 2 4 6

0 2 8 10

I have mad b from a. Like, 3rd element of b is made from 2 time add 2nd element of a, etc. I have also checked others code some of them gave answer as Yes, Other gave No .

If I am wrong, please reply me.

Array a consists only of {-1, 0, 1}

Thanks

AadityA_Singh

array a can contain elements -1,0,1 only so your test case is incorrect

Gilfoyl

Can someone explain why in problem A, for testcase 2 BWB WBW BWB is not an accepted solution

https://mirror.codeforces.com/contest/1333/submission/75973141

because in your output: BWB BBB there three B where a W is of them(2 in first row,1 in middle of second row).So B=3 here.Again a W has a adjacent B.So w=1.It doesn't maintain B=W+1.

thanks for pointing out my error

Welcome.

← Rev. 6 →

Here is how I up-solved D. Spoiler, it is long to be newbie friendly.

Note: if you are getting TLE in this problem. Use '\n' instead of endl to enabled output buffering because endl will flush the output. You don't want to flush the output everytime you need to output a newline because that means doing IO. The TL is very tight.

Spoiler

First, we consider the simpler case when the steps are not done simultaneously... I realized that we are just sliding any L to the left.

For example,

RL becomes LR

RRL becomes LRR

RRRL becomes LRRR

and so on.

The reason for this is because, when L is preceded by an R, then they effectively swap.

When there are two or more L's, then we just slide them one by one.

RRLRRRRL becomes LRRRRRRL then becomes LLRRRRRR

Note that this is just one strategy. Of course, we can also start with the rightmost L, then slide it to the previous L.

RRLRRRRL becomes RRLLRRRR by sliding the last L to the previous L.

Hence, there are multiple strategies of doing it in any order. But the number of operations stay the same becomes the n-th L will move to the n-th position by moving a fixed number of times.

Now, the number of moves that we incur when we do this strategy is the highest number of moves . So if the required number of moves k is larger, then the it's impossible to do it. However, if the required number of k is lower, then it's possible to reduce the number of moves by doing some of the swaps simultaneously.

You can approach this by simulating doing as many simultaneous swaps in each step. Here's how I implemented it:

After performing the simulation, we now that min_moves is the the minimum number of moves required and we know how to do it as well. So if k is less than this. Then, it's impossible (i.e. the answer is -1).

By now, min_moves <= k <= max_moves.

If k = min_moves or k = max_moves, then the answer is trivial.

So let's consider min_moves < k < max_moves.

By intuition, we CAN increase the number of moves from the minimum solution by selecting some swaps and not doing them simultaneously. We basically select k - min_moves of these swaps. Here's how I implemented it:

Some caveats: If you take a time step with L steps done simultaneously, and then you separate each of these steps in separate times. You might thinking you are ading L seconds, but you actually remove one second from the minimal solution. So, in the end, you are actually only adding L — 1 seconds to the minimum solution.

sh_maestro

Excellent explanation!

aman2gupta95

Dude... This is giving TLE..

See my submission. I didnt get TLE. And I checked your submission, it has many differences such that you are performing the swaps right away

teamcaproxx

does it matter if we perform the swaps right away. even i am getting TLE

I'm not saying performing swaps right away will (necessarily) cause TLE. I'm just saying there's a lot of differences in my solution to aman2gupta95's. And YES, it would matter.

Consider RRLL

How many swaps can you simultaneously in the first step?

Answer: 1

But if you perform swaps right away:

RRLL becomes RLRL then becomes RLLR in one loop. So, first, find all L's preceded by an R. Then only AFTER, swap them.

yes i have taken care of what you are saying but still i get tle

You are right, it seems weird that you are getting TLE. After checking, using \n instead of endl will get you AC. I think the TL is very tight and the output is huge, so input buffering must be enabled.

Here is your solution modified to get AC: https://mirror.codeforces.com/contest/1333/submission/76085125

thanks

Please check it.. I had incremented i value if any swaps is there..

You are right. You can perform the swaps right away if you increment twice during any given swap :)

OK...After changing endl to \n its not giving TLE.. maybe because of the large amount of output

SarthakBansal

4 years ago, # ^ |

I was getting TLE in Problem D, and I read "Change endl to \n " and I thought , "Well, sure OK! Why Not!" and it Passed (499ms).

Gotta say I'm Really Surprised how tight the time contraints are!

Anyways, Thanks !

The problem A says that B should be equal to W+1 but in the tutorial number of W is always 1 then how it is a good coloring what am i missing??Please help..

There are 2 black cells with a border to the white one, and the one white one.

So B=2, W=1

B is not equal to total number of B. B is equal to total number off such B whice has a adjacent W. I hope you understand.

kingmanas

What is the time complexity of this solution for problem C, by tmwilliamlin168?

This solution is similar to the solution at editorial. So nlogn

I am so sorry to bother you again, but could you please tell me how the log(n) part came into the time complexity. Is this some well known algorithm? I have seen such two pointer questions earlier too, but could never understand how it works internally.

from std::map. std::map::operatop[] cost log(n)

pk_2809

Can anyone help me to understand error in my code for (133C) my code

sare

Your code give 3 for this case, but it should give 2.

3
1 0 -1

time_14_07

Explain C in simple language and example.

FinalBoss_

In C we were asked to count number of subarrays that were good (i.e their subarrays must not have zero sum)

Now few observations before starting : 1. If a subarry is good that means all of 'its' subarrays are also good. 2. If a subarray is not good then all subarrays which will contain this subarray is also not good. example: 1 -1 3 4 now sum(1 -1) = 0 so it is a bad subarray so if you include (1 -1 3) this is also bad for same reason and so is (1 -1 3 4)

now main problem is to find subarrays with 0 sum in efficient manner for which we use idea of prefix sum. what we try to do is find sum of all elements coming before it. for example 1 2 3 4 is the array then it's prefix sum is 1 3 6 10. now we can say that there exist a subarray with 0 sum if : 1. prefix itself is 0. 2. prefix in some index is seen before.

let's say 1 2 -3 4 -1 it's prefix sum is 1 3 0 4 3 so there are two cases where subarray is 0 one is (1 2 -3) as prefix index was 0 here and second is (-3 4 -1) as prefix 3 in last index was seen before in 3rd index. we can use maps to find subarray with 0 sum to efficiently. Hope that helps.

Tirutu

I'm a bit confused right now. So you advise us to use set/map instead of unordered_map/unordered_set so we don't get TLE, because some "adorable community colleagues" added test cases to make it non-viable.

Does this recommendation stand only for this particular problem or we will have to use it from now on? Won't we get TLE on other problems for using an ordered data structure when it isn't generally needed?

filippodigra

Unordered set and unordered map use hashing, so if someone creates anti-hashing testcases ordered data structures are faster. In the worst case data structures that use hashing take O(N) complexity to access a element where N is the number of element in the data structure. Ordered data structures like set and map always have a O(logN) complexity. Use ordered data structures to decrease the risk of getting a TLE.

Sorry for my bad english.

What actually concerns me is: could it happen to get TLE using ordered map/set but pass with unordered(in case someone doesn't make anti-hash test cases)??

No. Just do not use unordered_set/map without modification of hashing.

yash_daga

I solved C using recursion. 75900316

alphonso1416

can u explain your code

Basically we can't include 2 indices the difference of whose prefix sums is 0. Let's say during the first call of the function we find difference of prefix sum of indices 4 and 11 is 0. So we return f(first_index, 10) + f(5, last_index).So that these two indices are not considered together. Here first_index and last_index are the values for that particular call of the function intially the function is called as f(0, n-1) where n is the length of the string.

Note that we also have to substract f(5, 10) as it has been counted twice. The value the function returns is n(n+1)/2 where n is the number of included elements.

The code is a bit untidy as I submitted it in a hurry during the contest . Hope this helps :)

In this tutorial,for problem C,Isn't the loop is amortize?Why not complexity is O(n+n)? Please anyone explain.

lazy_learner

Got it.

Code in Spoiler, please?

Cortex_guest

In problem C what should be the output of 1 2 0? for me it should be output-> 5 (1),(2),(1,2)(1,2,0),(2,0) but editorial solution gives 3

0's should be excluded, so answer is 3

nishant_raja

-17

Plz, write editorial in a more descriptive way. Write by assuming we don't know, not just write by assuming you are revising something, although this is for only last question's editorial.

MubtasimShahriar

The editorials are already well descriptive.
You will cope with understanding them gradually.
Keep reading until you understand and seek for help if you need.
And be patient.

N0N4M3

I tried to implement C before check author's code, but my implementation is wrong, and I don't see difference between my and author's code, can u help me find where is my solution going wrong


int n;
    cin >> n;
    for ( int i = 1; i <= n; i++ ) cin >> a[i];
    for ( int i = 1; i <= n; i++ ) pref[i] = pref[i-1] + a[i];

    int ans = 0, j = 0;
    for ( int i = 0; i <= n; i++ )
    {
        set<int>dup;
        for ( j; j <= n; j++ )
        {
            if ( dup.count(pref[j]) )
            {
                ans += j-1-i;
                break;
            }
            else dup.insert(pref[j]);
        }
    }

    cout << ans << endl;

set dup is empty at the start of every iteration by i. Also you need to use long long instead of int.

I do some fixes. Check it https://mirror.codeforces.com/contest/1333/submission/76047566

HelloWorld

Hello, I got WA on pretest 9 for problem Div2C. Can anyone please tell me what is pretest 9? My submission. Any help will be greatly appreciated.

LegendaryNewbie_

Can anyone please explain the editorial of C, specially the first solution with O(n^2 logn) solution.Here it is said that if a subarray with [ai.....aj] is good then [ai....aj-1] is also good, i'm not clear how it works.if i take an array of {1, 2, -3, 1, 0} then a subarray {1, 2, -3, 1} is good but {1, 2, -3(aj-1)} is not good, or i misunderstood ? Plz help anyone

here is an explanation, hope it will help you.

gyan_p_t

can anyone explain problem A?

AishwaryaR

Take an NxM board, it is allowed to color a cell as black or white. If a black cell has atleast one adjacent white cell, lets increment variable B. Same holds for W.

The task asks to color such that, B = W + 1. Now, the immediate way that strikes is to color like a chess board. This has one corner case to handle: when value of NxM is even, then we have B == W. So, we have to color one more white spot as black, preferably the position: 0,0 [Since I've assumed coloring all odd sum cells as black. My soln. for reference: link

The editorial provides a much simpler implementation, as in, color only the top-left cell as white and the rest as black. Why is this correct? Well, we have W = 1, since there's only one white cell, also, it has 2 neighboring black cells. We also have B = 2 always, this is because only two cells: (0,1) and (1,0) have neighboring white cell (other blacks only have black neighbors). Thus, the constraint: W = B + 1, always holds. I would recommend visualising/drawing out and seeing.

Bonus: As the editorial mentions, the problem gets a little tricky if 1 <= N,M! I realised why only after typing out this explanation.

Hritwik_gok

Can someone explain eugene one. I am unable to get it. I was relating this to count arrays with sum zero and using that approach. and then subtracting n(n+1)/2-count.

But I am not able to figure out how to include in count those subarrays of the given array whose subarrays are also not good.

mukesh_g7

You can check my solution here

aditya_sheth

+17

The spiral path idea in E is so nice!!

dmkz

For given constraits we do not need a sieve in problem F. We can find smallest prime divisor with naive brute force and get AC, 187 ms

Do you know the asymptotics of your solution (without sorting)? Its look like n*e operation so O(n), but I'm not sure.

It is $$$O(n \sqrt{n})$$$

UPD. Looks like time complexity is $$$O\left(\dfrac{n \sqrt{n}}{\log{n}}\right)$$$

Why?

We can run this function for some $$$n$$$ and calculate $$$\dfrac{n \sqrt{n}}{\text{#operations}}$$$. This is experiment.

Looks like time complexity is $$$O\left(\dfrac{n \sqrt{n}}{\log{n}}\right)$$$

Yes, it's looks like O(n sqrt(n) / log(n)), but it's interesting to find accurate complexity. New challenge!

thongnon1996

Just want to suggest that an O(n) solution of problem C is possible.

Python solution: http://mirror.codeforces.com/contest/1333/submission/76077853

Orion

Yes, I also solved for O(n). Here is my C++ solution: https://mirror.codeforces.com/contest/1333/submission/75888071

Mani

It is not O(n) , you have used map, so it's O(nlogn)

Lelouch_vi_Britannia

PuRpLe_FoReVeR In your solutions for 1333D - Challenges in school №41 there is a small mistake in computing mini(minimum possible k).

It's giving wrong answer for the case-

6 2

RLLRRR

Thank you! Fixed.

cs18b012

Hello, Can someone please explain C question duplicate prefix sum part of the tutorial? I didn't understand why that should be true?

Thanks in advance!

SJD1206

just a simple thing you have to maintain prefix array and then if any value repeats that show that the sum of elements including this indices is zero than you can easily calculate how much subarrays contains this subarrays after counting all subarrays which contain subarray with sum=0 subtract it from total number of subarrays that can be found by n*(n+1)/2

Thanks for the explanation, but can you explain why duplicates in a range (as mentioned in the tutorial) will create problem?

2powerofn

can anyone help me please, actually I think my code doesn't have a mistake but it fails on the test case 3 how can I improve this code https://mirror.codeforces.com/contest/1333/submission/76197328

this code can be done in O(n). you don't need to actually change the value in first array you can just check is it possible to make it equal to value at same index in second array

xyz_

can anyone identify what is wrong with my logic for div2 problem B- https://mirror.codeforces.com/contest/1333/submission/76225752

https://mirror.codeforces.com/contest/1333/submission/76229807 I have made some modifications to your code. 1) The first change is this if(a[i]==b[i] || (b[i]>0 && m[1]>0) || (b[i]<0 && m[-1]>0)) Because you want to convert a[i] to b[i], So comparing them is more intuitive. The second change is in for loop for(int i=n-1;i>=0;i--). Hope it helps ;)

HelloWorld In your second point, your for-loop is running from n-1 to 0 and so does mine. I think it's not a problem.

My logix is as follow-

if b[i]>0 and if we have at least one 1 in [0, i-1], we can converter a[i](-1, 0, 1) to b[i].
if b[i]<0 and if we have at least one -1 in [0, i-1], we can converter a[i](-1, 0, 1) to b[i].
if b[i]=0 we have following options- (i) if a[i]=0, handled by b[i]==a[i]. (ii) if a[i]=1, we need at least one -1 in [0, i-1]. (iii) if a[i]=1, we need at least one -1 in [0, i-1].

My code is handling 1,2, and 3i only. My updated code for handling 3ii and 3iii- https://mirror.codeforces.com/contest/1333/submission/76245364

Thanks for reply however.

prashiksahare110044

To make the code shorter for 1333E, I filled the spiral differently. https://mirror.codeforces.com/contest/1333/submission/77185747

mdallrosa

I believe this is an O(N^2) solution: submission

But i'm getting TLE on 1333D - Challenges in school №41 on test case number 10. Can anyone help me?

TheScrasse

You have to use "\n" instead of endl.

Thank you. Now i have got WA on test case number 76. I will try to move forward.

bitshar

My solution 77420751 for problem C uses the idea of two pointers. Broadly we can keep two pointers in the array denoting the start and endpoint of a valid subarray and iterate them to get the number of good subarrays. First, we store prefix sums in an array then we iterate the endpoint from 1 to an index such that the sum hasn't been seen before. The number of segments is end-pt- start-pt+1. Once we encounter an already found sum, we know that sum from startpt+1 to end-pt is 0, and any subarray starting before or equal to startpt+1 and ending after end-pt will contain this subarray. We thus increment our start-pt by 2 and repeat this process till we cover all array. We need to keep care of the fact that start-pt never decreases because this would mean over counting.

Chy_Chy

hey ... in your submission .... can you pls explain these lines: else { if(ind1<ma[sum[ind2]]+2) ind1=ma[sum[ind2]]+2; if(ind1>ind2) { ma[sum[ind2]]=ind2; ind2++; continue; } ans+=ind2-ind1+1; ma[sum[ind2]]=ind2; } i hope you will answer.... badly in need of that

Suppose you find a sum at some index j which you have already seen at some index i, this means that sum from i+1 to j (inclusive) is 0. Thus, you need to increase your pointer1 (ind1) (denotes the starting point of subarray in consideration) to i+2, as pointer2 (ind2) is at j and sum from i+1 to j is 0, so we need to go to i+2. The map stores the latest index at which this sum has been obtained. So, if ind1>ind2 by increase, we just assign map[sum[ind2]]=ind2 and continue with our loop. There will be no effect on ans in this case.

daayan

Can someone please tell me why this solution exceeds time limit? 77867414

G.O.K.U_Instinct

Can somebody help me out for ques C why this is getting wrong answer on test 8? https://mirror.codeforces.com/contest/1333/submission/78603617 is link to my submission. Thanks in advance.

I m Getting wrong answer at test 7 of problem D. Link to my submission https://mirror.codeforces.com/contest/1333/submission/78657016 Somebody plz tell me my mistake. Thanks in advance.

Ravan_2070

In Editorial of Problem D

Can anyone just explain this statement ,that we came to the point that we should now flip (U-k+1) pairs

Otherwise, we roll back to the previous iteration and use U−k+1 pairs in this move

vedang122

4 years ago, # |

7 0 0 1 2 0 -2 -1 In problem C I think monotonicity part is wrong. Consider above test case ( 0 based indexing in array) for which editorial's R = { 0 5 5 4 6 6 6} which is monotonic. Please correct me if I am wrong. Thank You.

Ashwanth.K

18 months ago, # |

I did problem F using linear sieve and some observations in $$$O(NlogN)$$$.
code: 210753600

PuRpLe_FoReVeR's blog

include <bits/stdc++.h>