Codeforces Round #925 (Div. 3) Разбор

№	Пользователь	Рейтинг
1	tourist	3690
2	jiangly	3647
3	Benq	3581
4	orzdevinwang	3570
5	Geothermal	3569
5	cnnfls_csy	3569
7	Radewoosh	3509
8	ecnerwala	3486
9	jqdai0815	3474
10	gyh20	3447

№	Пользователь	Вклад
1	maomao90	174
2	awoo	164
3	adamant	162
4	TheScrasse	160
5	nor	158
6	maroonrk	156
7	-is-this-fft-	152
8	orz	146
9	pajenegod	145
9	SecondThread	145

1931A - Восстановление маленькой строки

Идея: myav Разработка: myav

Разбор

Решение

#include<bits/stdc++.h>
using namespace std;
 
 
void solve(){
    int n, sz = 26;
    cin >> n;
    string mins = "zzz", cur;
    for(int i = 0; i < sz; i++){
        for(int j = 0; j < sz; j++){
            for(int k = 0; k < sz; k++){
                if(i + j + k + 3 == n){
                    cur += char(i + 'a');
                    cur += char(j + 'a');
                    cur += char(k + 'a');
                    mins = min(cur, mins);
                }
            }
        }
    }
    cout << mins << "\n";
 
}
int main(){
    int t;
    cin >> t;
    while(t--) {
        solve();
    }
}

1931B - Сделай равными

Идея: MikeMirzayanov Разработка: Vladosiya

Разбор

Решение

def solve():
    n = int(input())
    a = [int(x) for x in input().split()]
    k = sum(a) // n
    for i in range(n - 1):
        if a[i] < k:
            print('NO')
            return
        a[i + 1] += a[i] - k
        a[i] = k
    print('YES')
    
    
for _ in range(int(input())):
    solve()

1931C - Снова сделай равными

Идея: senjougaharin Разработка: senjougaharin

Разбор

Решение

def solve():
    n = int(input())
    a = list(map(int, input().split()))
    i1 = 0
    i2 = 0
    while i1 < n and a[i1] == a[0]:
        i1 += 1
    while i2 < n and a[n - i2 - 1] == a[n - 1]:
        i2 += 1
    res = n
    if a[0] == a[n - 1]:
        res -= i1
        res -= i2
    else:
        res -= max(i1, i2)
    print(max(0, res))
 
 
t = int(input())
 
 
for i in range(t):
    solve()

1931D - Делимые пары

Идея: MikeMirzayanov Разработка: Vladosiya

Разбор

Решение

def solve():
    n, x, y = map(int, input().split())
    a = [int(x) for x in input().split()]
    cnt = dict()
    ans = 0
    for e in a:
        xx, yy = e % x, e % y
        ans += cnt.get(((x - xx) % x, yy), 0)
        cnt[(xx, yy)] = cnt.get((xx, yy), 0) + 1
    print(ans)
    
    
for _ in range(int(input())):
    solve()

1931E - Аня и подарок на День святого Валентина

Идея: Gornak40 Разработка: Gornak40

Разбор

Решение

#include <bits/stdc++.h>
 
#define all(arr) arr.begin(), arr.end()
 
using namespace std;
 
const int MAXN = 200200;
 
int n, m;
string arr[MAXN];
int len[MAXN], zrr[MAXN];
 
void build() {
    memset(zrr, 0, sizeof(*zrr) * n);
    for (int i = 0; i < n; ++i) {
        len[i] = arr[i].size();
        for (auto it = arr[i].rbegin(); it != arr[i].rend() && *it == '0'; ++it) {
            ++zrr[i];
        }
    }
}
 
string solve() {
    int ans = 0;
    for (int i = 0; i < n; ++i) {
        ans += len[i] - zrr[i];
    }
    sort(zrr, zrr + n);
    reverse(zrr, zrr + n);
    for (int i = 0; i < n; ++i) {
        if (i & 1) ans += zrr[i];
    }
    return (ans - 1 >= m ? "Sasha" : "Anna");
}
 
int main() {
    int t; cin >> t;
    while (t--) {
        cin >> n >> m;
        for (int i = 0; i < n; ++i)
            cin >> arr[i];
        build();
        cout << solve() << '\n';
    }
}

1931F - Скриншоты чата

Идея: senjougaharin Разработка: senjougaharin

Разбор

Решение

#include <iostream>
#include <vector>
#include <set>
#include <queue>
#include <algorithm>
 
using namespace std;
 
typedef long long ll;
 
int timer = 0;
 
void dfs(int v, vector<vector<int>> &g, vector<bool> &vis, vector<int> &tout) {
    vis[v] = true;
    for (int u: g[v]) {
        if (!vis[u]) {
            dfs(u, g, vis, tout);
        }
    }
    tout[v] = timer++;
}
 
void solve() {
    timer = 0;
    int n, k;
    cin >> n >> k;
    vector<vector<int>> a(k, vector<int>(n));
    vector<int> authors(k);
    for (int i = 0; i < k; ++i) {
        for (int j = 0; j < n; ++j) {
            cin >> a[i][j];
            a[i][j]--;
        }
        authors[i] = a[i][0];
    }
    vector<vector<int>> g(n);
    for (int i = 0; i < k; ++i) {
        for (int j = 1; j + 1 < n; ++j) {
            int i1 = a[i][j], i2 = a[i][j + 1];
            g[i1].push_back(i2);
        }
    }
    vector<int> tout(n, -1);
    vector<bool> vis(n);
    for (int i = 0; i < n; ++i) {
        if (tout[i] == -1) {
            dfs(i, g, vis, tout);
        }
    }
    for (int i = 0; i < k; ++i) {
        for (int j = 1; j + 1 < n; ++j) {
            int i1 = a[i][j], i2 = a[i][j + 1];
            if (tout[i1] < tout[i2]) {
                cout << "NO";
                return;
            }
        }
    }
    cout << "YES";
}
 
int main() {
    int t;
    cin >> t;
    for (int _ = 0; _ < t; ++_) {
        solve();
        cout << "\n";
    }
}

1931G - Одномерный пазл

Идея: senjougaharin Разработка: senjougaharin

Разбор

Решение


#include <iostream>
#include <vector>
#include <set>
#include <queue>
#include <algorithm>
 
using namespace std;
 
typedef long long ll;
 
const int mod = 998244353;
 
ll pow_mod(ll x, ll p) {
    if (p == 0) {
        return 1;
    }
    if (p % 2 == 0) {
        ll y = pow_mod(x, p / 2);
        return (y * y) % mod;
    }
    return (x * pow_mod(x, p - 1)) % mod;
}
 
ll inv(ll x) {
    return pow_mod(x, mod - 2);
}
 
vector<ll> fact = {1};
 
ll cnk(ll n, ll k) {
    ll res = fact[n];
    res = (res * inv(fact[k])) % mod;
    res = (res * inv(fact[n - k])) % mod;
    return res;
}
 
ll calc(int n1, int n2, int n3, int n4) {
    return (cnk(n1 + n3 - 1, n3) * cnk(n2 + n4 - 1, n4)) % mod;
}
 
void solve() {
    int n1, n2, n3, n4;
    cin >> n1 >> n2 >> n3 >> n4;
    if (n1 + n2 == 0) {
        cout << (n3 == 0 || n4 == 0 ? 1 : 0) << '\n';
        return;
    }
    if (abs(n1 - n2) > 1) {
        cout << "0\n";
        return;
    }
    ll res = 0;
    if (n1 <= n2) {
        res += calc(n1 + 1, n2, n3, n4);
    }
    if (n2 <= n1) {
        res += calc(n1, n2 + 1, n3, n4);
    }
    cout << res % mod << '\n';
}
 
int main() {
    for (ll i = 1; i <= 4e6; ++i) {
        fact.push_back((fact.back() * i) % mod);
    }
    int t;
    cin >> t;
    for (int _ = 0; _ < t; ++_) {
        solve();
    }
}

sort(trailnums.begin(), trailnums.end(), [&](pair<long long int, long long int> a, pair<long long int, long long int> b) { return to_string(a.first).length() < to_string(b.first).length(); });

long long sum=0; for(auto i:v) { sum+=i; } int req=sum/n; int flag=0; long long exc=0; for(int i=0;i<n;i++) { if(v[i]>=req) { exc+=(v[i]-req); } else{ int req_am=req-v[i]; if(exc>=req_am){ exc-=req_am; } else{ cout<<"no"<<endl; } } } cout<<"yes\n";

Комментарии (98)

Написать комментарий?

Wasif_Shahzad

3 месяца назад, # |

Can anyone explain why for Problem. F this submission gets TLE.

246345877

and this submission gets accepted?

246347175

I don't feel like there's any difference in logic, just the way of implementation is different.

→ Ответить

VitaliiV

3 месяца назад, # ^ |

← Rev. 8 →

In the second submission there is an optimization of the DFS function: do not run a search if the vertex has already been visited

My submissions:

246247300

246250491

You can see the difference more clearly here

are you talking about the external dfs optimization from main function or internal dfs optimization?

I mean the lines

le(i, 1, n){
			if(!vis[i]){
				if(dfs(i)){
					cout << "NO" << endl;
					flg = 1;
					break;
				}
			}
		}

	for(auto c: graph[v]){
		if(dfs(c)){
			return true;
		}
	}

le(i, 1, n){ if(!vis[i]){ if(dfs(i)){ cout << "NO" << endl; flg = 1; break; } } } I'm talking about this

How can this be the issue? can I get explanation? Won't the if(!vis[i]) take care of it?

← Rev. 4 →

DELETED

robostac

In the TLE submission your first loop (to clear graph / vis / dfsStack) will run 10000(t) * 200005(NxM) operations. The accepted one only clears the memory needed for each test case.

so I don't need to initialize it with size NxM? and resize it to n+1 in every test case while clearing it in the end of each test case?

ProveMeRight

The solution in which you are getting TLE. Coz:

You are unnecessary initializing array of maxN again and again, even when the Value of N be small.

gubidubi

In the second submission the graph is a array of sets, so when you insert a repeated edge it doesn't get duplicated. In the first submission you are duplicating a lot of edges (because you are pushing them in a vector), and going through all of them during the dfs.

I will try it a little while. Hope it works.

I'm wrong, this still would not cause TLE (the total amount of edges over all tests is less than 2*10⁵), robostac gave the right answer.

jkulanko

A different way to do F:

Each element (after the 0th element in the kth list) can only be it's current position or its current position+1
Store both possibilities in a vector
Remove possibilities for elements when they're no longer possible (i.e. the current element's position in the kth list conflicts with the elements position in a previous list)
The answer is no if an element has no possibilities, or if its only possibility conflicts with another elements only possibility
Otherwise the answer is yes

Code: https://codeforces.com/contest/1931/submission/246250979

Donbcheap

I think my solution is near from this but it got WA test 2

The Submission : 246363856

How was my solution hacked? https://codeforces.com/contest/1931/submission/246212609

I see it was probably due to collisions in the map but does that mean I need a better hashing function for pairs? If so can someone suggest one?

https://codeforces.com/blog/entry/62393

The other option is you can almost always use set / map in c++ as any attempt to use unordered with a custom hash will often have enough constant overhead to be about the same overall time.

martin0327

Using hash map in this problem is a bad idea. Still, you may want to read the following:

and_t

← Rev. 3 →

can somebody help me understand why this solution got TLE Problem D

Shahzod1207

probably, for multiset's count function, if I ain't mistaken, it works in O(logn) time

lucasap

I think it is because of s.count( {cx,cy} ), that is not O(1) but O(count_of({cx,cy}) + log(N)), so worst case in an array with all ones and x=2, you get O(N^2) time complexity. I think in C++ you should be going for a unordered_map<pair<int,int>, int> in which the values are the count so you can do the addition in O(1).

-8

Good contest but again I will most likely stay with newbie.

polsage

Hey, I have submitted a java code for problem C but it gave me error on test case 3 whereas when i submitted the exact same code in c++ it passed all test cases why

java link: https://codeforces.com/contest/1931/submission/246230264

c++ link: https://codeforces.com/contest/1931/submission/246276255

WORTH

Don't use $$$==$$$ for comparing Integer

Read this.)

ooooh my god thank you

your dp shows my pain

matislu

In G we don't need to calculate factorials to $$$4 \times 10^6$$$ since the biggest factorial we need in combinations is actually $$$2 * max(c_{i}) \leq 2 \times 10^6$$$

I_Love_Huong_Thao

i think something wrong in solution code of problem A ._.

Forestmy17

Thanks for good Round. Good F

zc2707

Why D's solution is correct? It uses a hash table (dict), so is it possible to hack this hash table?

← Rev. 2 →

Great question! Causing collisions for dicts since Python 3.3+ is very challenging. You can see in the note of https://docs.python.org/3/reference/datamodel.html#object.__hash__ that the seeding of the hash function (used by dict) are not predictable since they are "“salted” with an unpredictable random value", unless you set the PYTHONHASHSEED.

0x3F

Video explanation of Problem G (Chinese)

rhymehatch

A lot of dict/unordered hacks this round. Although I still see quite a lot of solutions not hacked that use the same approach for problem D.

Nice round. Had no idea toposort is available in standard Python library.

amirisokokok

could any explain why in F , having cycle in graph means there is no logical order of numbers ?

consider the following case

1 2 3 4 2 3 1 4 3 4 3 2

the graph would be

2 -> 3 -> 4 and 3 -> 1 -> 4 and 4 ->3 ->2 if you construct graph for these nodes then there's a cycle. now answer to why does it make sense: we have to follow the order no matter what. here 2 comes before 3 and 4, similarly 3 comes before 1 and 4 but the 3rd one, which creates the cycle, has that 4 comes before 3. it is possible to have 3 before 4 and 4 before 3 in the line simultaneously. so that's why having cycle means no ordering for the data.

second_saturday

consider these two statements,

1 comes before 2

2 comes before 1

they can't be true at the same time,

ramentaneja

Was I the only one who had an unusually hard time in this round? T_T

MAT1U5

same

Farsh_LE

Nice problems , had fun solving them

I really hope more rounds like this exist

Stuck_On_Gray

why the code of problem A uses the string cur outside loop so I think this isn't true; please anyone can review it?

m_nature

It's correct because the first triples that it finds is guaranteed to be the minimum, so everything after that affecting the cur variable doesn't matter

(That also means it doesn't need to continue the loop to find the minimum here and can return immediately when it finds the first triples)

But I agree that appending to the cur variable after that is definitely unnecessary and the logic is also wrong IF the minimum is not the first triples lol)

I still don't understand anything in G, can someone please explain me

han_hyo_joo

This was my thought process when upsolving G.

The main observation needed is to notice that 1 + 3 = 1 and 2 + 4 = 2, so we can condense all 3 and 4 pieces. However, it is possible to start the sequence with a 3 or 4 piece, so for this reason we can imagine having an "extra" 1 or 2, respectively. Now there are a few cases:

No correct puzzle sequence can have consecutive 1's or 2's, so if $$$ |c_1 - c_2| > 1 $$$, the answer is 0.
If $$$ c_1 = c_2 = 0 $$$, the answer is 1 if $$$ c_3 = 0 $$$ and/or $$$ c_4 = 0 $$$ and 0 otherwise. This is true because 3 cannot mesh with 4.
If $$$ c_1 = c_2 $$$ and $$$ c_1, c_2 > 0 $$$, either 1 or 2 can be used to start the puzzle sequence, so we can fix the first number and calculate the rest of the sequence. Counting the # of ways to condense $$$ c_3 $$$ 3 pieces into $$$ c_1 + 1 $$$ 1 pieces is the same as counting the # of ways to put $$$ c_3 $$$ indistinguishable balls into $$$ c_1 + 1 $$$ boxes, which is $$$ \binom{c_1 + c_3}{c_1} $$$ (this can be visualized with the stars and bars technique). Note that 1 is added to $$$ c_1 $$$ for that "extra" piece, if we fix 1 to be the first puzzle. As such, the answer is $$$ = \binom{c_1 + c_3 - 1}{c_1 - 1} \cdot \binom{c_2 + c_4}{c_2} + \binom{c_1 + c_3}{c_1} \cdot \binom{c_2 + c_4 - 1}{c_2 - 1} $$$
If $$$ c_1 = c_2 + 1 $$$ or $$$ c_1 + 1 = c_2 $$$, then we are required to fix either 1 or 2 to be the first piece, respectively. Then it's the same exact idea as the previous case

your explanation is better than editorial, thanks...

zijun0502

Could you clarify why we need to add one? I'm not entirely clear on what you mean by an "extra piece." Could you elaborate further on that point?

We assume that every 3 and 4 piece must be in a sequence of 1333... or 24444.... However it is also possible to place a sequence of 3's or 4's at the beginning of the sequence, without a corresponding 1 or 2 behind it. So adding an invisible 1/2 lets us strictly associate every 3/4 with a corresponding 1/2, which makes calculation easier.

Thank you! Your explanation was really helpful.

hoangthichviolympic

Anyone can check my wrong in this code for problem E? It's call wrong in test 3, but i think i have same ideal with propers.. 246215307

phukaoz

why my solution TLE on system test.
is unordered_map of unordered_map took significantly more time to find than unordered_map that pair as key?

Akshay184

Can some one explain the approach for problem G in detail?

the_hyp0cr1t3

+11

Problem G:

Let the four types be denoted by $$$\color{purple}{b_1}$$$, $$$\color{blue}{b_2}$$$, $$$\color{olive}{b_3}$$$ and $$$\color{teal}{b_4}$$$ respectively.

Lemma 1: Between any two $$$\color{purple}{b_1}$$$, there must be at least one $$$\color{blue}{b_2}$$$. It is easy to see that one cannot fit only $$$\color{olive}{b_3}$$$ and/or $$$\color{teal}{b_4}$$$ between two $$$\color{purple}{b_1}$$$.

Lemma 2: Between any two $$$\color{blue}{b_2}$$$, there must be at least one $$$\color{purple}{b_1}$$$. It is easy to see that one cannot fit only $$$\color{olive}{b_3}$$$ and/or $$$\color{teal}{b_4}$$$ between two $$$\color{blue}{b_2}$$$.

It follows that between any two $$$\color{purple}{b_1}$$$, there must be exactly one $$$\color{blue}{b_2}$$$. Likewise, between any two $$$\color{blue}{b_2}$$$, there must be exactly one $$$\color{purple}{b_1}$$$. In other words, they must follow an alternating pattern $$$\color{purple}{b_1}, \color{blue}{b_2}, \color{purple}{b_1}, \color{blue}{b_2}, \ldots$$$ or $$$\color{blue}{b_2}, \color{purple}{b_1}, \color{blue}{b_2}, \color{purple}{b_1}, \ldots$$$, with possibly some $$$\color{olive}{b_3}$$$ and $$$\color{teal}{b_4}$$$ in between.

Lemma 3: If a block exists to the left of a $$$\color{olive}{b_3}$$$, it must either be another $$$\color{olive}{b_3}$$$ or a $$$\color{purple}{b_1}$$$. If a block exists to the right of a $$$\color{olive}{b_3}$$$, it must either be another $$$\color{olive}{b_3}$$$ or a $$$\color{blue}{b_2}$$$.

Lemma 4: If a block exists to the left of a $$$\color{teal}{b_4}$$$, it must either be another $$$\color{teal}{b_4}$$$ or a $$$\color{blue}{b_2}$$$. If a block exists to the right of a $$$\color{teal}{b_4}$$$, it must either be another $$$\color{teal}{b_4}$$$ or a $$$\color{purple}{b_1}$$$.

Putting everything together, any valid arrangement will be of the form

$$$\{\color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \ldots\}$$$ or
$$$\{\color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \ldots\}$$$

For such an arrangement to exist, it must hold that $$$\mathrm{abs}(c_1 - c_2) \leq 1$$$. Then, the number of valid arrangements is the number of ways to arrange $$$\color{olive}{b_3}$$$ and $$$\color{teal}{b_4}$$$ into the 'slots' between alternating $$$\color{purple}{b_1}$$$ and $$$\color{blue}{b_2}$$$. This can be done using the stars and bars technique.

Submission: 246362383

Cloud_Z3

Another idea for Problem F (indeed, what first came into my mind is not the solution in tutorial):

The first screenshot show the main order of users except the one (say, i) to which the screenshot belong. We can use the remaining screenshots to narrow the range of the i's place: if there's only one possible place, good; if there are multiple possible places, just choose one; if there's contradiction, then answer is "NO" directly. Adding i's place to the main order gives the whole order, compare it to remaining screenshots gives final answer (if contradictions exist, then "NO", otherwise, "YES"). Time complexity is OK.

https://codeforces.com/contest/1931/submission/246229823

i3mr125

what are the prequesites for G . I didn't understand anything, and why am i still green and i hit the 1400 mark

Topic wise, you only need to know the "stars and bars" combinatorics concept. You can read about it here: https://cp-algorithms.com/combinatorics/stars_and_bars.html.

ooh yeah thank you so much , i am not very good at combinatorics so i guess that's why i am not familiar with it. i will definitely look in to it thanks

CodeFlexer

For the problem F, I do understand why existence of topological sort (and therefore absence of cycles) of such graph is necessary condition for existence of the order. However, why is it sufficient? I.e. What is exact proof that absence of cycles guarantee that such order of chat participants exists?

You have a set of statements where $$$a_i < a_{i+1}$$$ that you can get by ignoring $$$a_0$$$.

That's an ordering. You cannot have $$$a_1 < a_2$$$, $$$a_2 < a_3$$$ and then another $$$a_3 < a_1$$$ due to transitivity of the ordering on integers. If $$$a_1 < a_2$$$ and $$$a_2 < a_3$$$ then $$$a_1 < a_3$$$. That's what cycle check is for. It finds the case where transitivity is broken if that case exists.

If you're asking about missing statements to have an ordering where every $$$a_i$$$ is at its true position, then you can easily see that if transitivity holds, you can invent your own statements where it can still hold and create a full order.

UzuHa_NaruSuke

If there are no cycles in the directed graph that means that you can run the topological sort algorithm on it.

The correct sequence will be the sequence after we topologically sort it as it will be satisfying all the conditions ie all the edges in the graph are not forming a cycle.

Therefore it is enough to no if we can topologically sort the graph or no ie check for any directed cycles.

Ylandolsi

Video Solution for problemD : https://www.youtube.com/watch?v=FPRiSmmyfiE

sun08

Can anyone explain for me what wrong in my details explanation of problem D: D_practice_Modular
I don't understand why a[i] mod x = (x - a[j] mod x) mod x instead of a[i] mod x = x - a[j] mod x

vabbisht17

The value of a[i] mod x should always be in range [0, x - 1] (should be less than x). In the equation a[i] mod x = x - a[j] mod x this property will not always be true as if a[j] mod x becomes zero then a[i] mod x = x which is contradicting the property. For example

a[j] = 10 , x = 5

a[i] mod 5 = 5 - 10 mod 5

a[i] mod 5 = 5

thus to prevent this and keep the value of a[i] mod x in range we use the equation a[i] mod x = (x - a[j] mod x) mod x

I don't quite understand, one more than is, in range [0, 2x) we have two multiple of x are 0 and x. Why we only choose the case x in the expression instead of both cases? When a[i] mod x + a[j] mod x = 0 ==> a[i] mod x = a[j] mod x = 0

That case a[i] mod x = a[j] mod x = 0 actually gets considered on its own when we use a[i] mod x = (x - a[j] mod x) mod x.

Let's say a[i] mod x = a[j] mod x = 0 thus a[i] mod x = (x - a[j] mod x) mod x LHS becomes 0 and RHS becomes (x - 0) mod x which is also 0. Hence LHS = RHS. So both cases are being considered in this single equation.

minion30031

Can someone please figure out what the mistake with my submission is? Any help is appreciated. https://codeforces.com/contest/1931/submission/246419122

AkiLotus

2 месяца назад, # ^ |

I think this part is the issue. You are sorting the numbers with trailing zeros by their absolute lengths in increasing order (instead of their numbers of trailing zeros in decreasing order). Seems like test #2 was only of integers from 1 to 10, thus this solution will be automatically be correct for just that case (but not when the integers are higher).

For example: 4000000 should be prioritized more than 711100 since it has 6 trailing zeros (compared to 2 of 711100), but your sorting will make 711100 stand first.

Hi AkiLotus, I also tried that approach before using the number of trailing zeros in decreasing order:

Submission:https://codeforces.com/contest/1931/submission/246418992

But even that does not seem to work.

Appreciate your effort in looking into my code.

Thanks

I see now. The remtrail function isn't correct either, specifically this part.

    string ans = "";
    for(int i=0;i<n;i++) {
        if(s[i] == '0') break;
        else {
            ans += s[i];
        }
    }

Take a guess of what the output of remtrail(170011000) would be with this code. Correct answer is supposed to be 170011.

Spoiler (open only when you figured out the bug)

Ahhh. I see!!! understood the mistake. Thanks a lot for the help

nikich340

From tutorial D: Since (ai + aj) mod x = 0, it follows that (ai mod x + aj mod x) mod x = 0. Since (ai − aj) mod y = 0, it follows that ai mod y − aj mod y = 0.

Quite not clear, how it follows, and why it's not (ai mod y − aj mod y) mod y = 0 for 2nd case then.

RedDreams

$$$

0\ \le\ ai\ mod\ y\ \le\ y-1 \newline

0\ \le\ aj\ mod\ y\ \le\ y-1 \newline

-(y-1)\ \le\ (ai\ mod\ y − aj\ mod\ y)\ \le y-1 $$$

So if $$$\ (ai\ mod\ y − aj\ mod\ y)\ mod\ y = 0$$$

Then $$$\ (ai\ mod\ y − aj\ mod\ y) = 0$$$

Solution for D without hash:

We need to count number of pairs equal with respect to $$$(mod\ y)$$$ and sum of which equal to $$$0$$$ with respect to $$$(mod\ x)$$$.

For each $$$a[i]$$$ store pair $$$ \{ a[i] (mod\ y), a[i] (mod\ x) \} $$$. Sort array of pairs.

Now pairs can be divided into continuous blocks of numbers equal with respect to $$$(mod\ y)$$$. And inside each block elements are sorted by their $$$(mod\ x)$$$.

So inside each block we can count number of pairs which add up to $$$0$$$ with respect to $$$(mod\ x)$$$ using 2 pointers in $$$O(n)$$$.

Complexity of solution $$$O(n * log(n))$$$. (Because of sorting)

My submission 246421822

lowest_rating12

Could someone tell me why my code for Question D always returns compilation error? My submission: 246426188

nvm I got it fixed

VincentVanGogh

Could Someone explain why the following code fails in problem F. My logic for the problem is completely different compared to what the tutorial talks about, but i feel it should work.

I have first constructed a pattern using the first screenshot, using every element except the first, in the first screenshot.

Then I check, if all the other screenshots after this first screenshot follow that pattern or not.

https://codeforces.com/contest/1931/submission/246430531

variety-jones

Take a look at Ticket 17368 from CF Stress for a counter example.

Linuef

E题代码没看懂啊，有什么小白能看懂的吗

exposecheaters650

lnight121

In Problem D,when i use unordered_map,i got TLE on test 30.But if use map,this test only need 70ms.Is this intentional design by the author？

The unordered map uses a certain hashing algorithm and the author has put in testcases that cause collisions to it.

You can read this for more information.

So answering your question, yes it is intentional design by the author as the testcases are set in such a way.

Jagnath_Reddy

why tle? D link

saisumanchitturi

int arr [][]= new int [ x ][ y ];

Constraints: $$$1 \le x, y \le 10^9$$$

You can't just create such a big matrix.

parag776

in G, why do we need this condition? It is guaranteed that the sum of ci for all test cases does not exceed 4⋅1000000.

alexiuscp

https://codeforces.com/contest/1931/submission/246604571

Please help, I am getting TLE for problem D test 5. My code works by iterating k to find (a[i]+yk) % x == 0, a[i]+yk<=a[n-1]. Is it possible to make this solution faster?

Lorazhang

What does 'tout' mean in problem F's answer? Can't understand

not_a_failure

Getting RUNTIME_ERROR (Exit code is -1073741819) on large inputs for this solution https://codeforces.com/contest/1931/submission/246676323 for Problem E. Does anyone know what am I doing wrong here?

1xx55

I tested your code and I think error occured when you sort vector b.

This is an AC submission based on your code. 247704863

I don't know what wrong on your code too (maybe Segment fault?).You can make some further study.

thank you

joynik1243

Question — 1931D

My solution on external java compiler is giving the correct result, but the codeforces judge is saying wrong answer. I am frustated as i cannot find the problem. Plz someone review my code and tell show me the problem:

import java.util.*;

public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-->0){ int n=sc.nextInt(); int x=sc.nextInt(); int y=sc.nextInt(); int[] arr=new int[n]; for(int i=0;i<n;i++) arr[i]=sc.nextInt(); int count=0; HashMap<Integer, HashMap<Integer, Integer>> hm=new HashMap<>(); for(int j=0;j<n;j++){ int a=(x-arr[j]%x)%x; if(hm.containsKey(a)){ HashMap<Integer, Integer> hy=hm.get(a); if(hy.containsKey(arr[j]%y)){ count+=hy.get(arr[j]%y); } } if(!hm.containsKey(arr[j]%x)){ hm.put(arr[j]%x, new HashMap<>()); } HashMap<Integer, Integer> hy=hm.get(arr[j]%x); if(!hy.containsKey(arr[j]%y)){ hy.put(arr[j]%y, 1); } else hy.put(arr[j]%y, hy.get(arr[j]%y)+1); } System.out.println(count); } } }

My solution is giving the correct result on my external compiler but the codeforces judge is getting wrong answer. I am frustated looking for the problem. Can someone plz show me the problem with my code?

https://codeforces.com/contest/1931/submission/246796190

246796190

Mr.Hakimov

thanks for the problem E! really enjoyed solving it :)

huangyuheng

Can G use dp?

rampalyadavdeokali

#include<bits/stdc++.h> using namespace std; void solve(){ int n; cin>>n; vectorv(n); for(auto &i:v) { cin>>i; } if(n==1) { cout<<"yes\n"; return; }

}

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

not run all test case

problem name = B.Make Equal

help me for run it

Isaac_Newton__

In E can somebody explain why have we used ans -1 instead of ans.

ans : the amount of digits that we have in final number.

but how many digits does $$$10^m$$$ have ? It's $$$m+1$$$.

So we need to check if ans >= m+1 ,ans it's the same as ans-1 >= m.

jafirislam10

1931C can anyone tell me there is i>=1 that means first index of the array is not changeable in order to make the whole array equal . all the other elements of array needs to be equal as first index isn't it ?

select three integers i, j, x(1≤i≤j≤n)

I think array index range from 1 to n , not 0 to n-1 , and it's ok. first element's index is 1.

abhinavm1435

Can somebody pls help to identify the cause of runtime err in my soln : https://codeforces.com/contest/1931/submission/248125344

2 месяца назад, # |

In tutorial for D., why is the condition for y not the same as the condition for x? Since it a[i] - a[j] is either equal to y or 0 as well.

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