Codeforces Round #835 (Div. 4) Editorial

#	User	Rating
1	Benq	3783
2	jiangly	3666
3	tourist	3611
4	Um_nik	3536
5	inaFSTream	3477
6	fantasy	3468
7	maroonrk	3464
8	QAQAutoMaton	3428
9	ecnerwala	3427
10	Ormlis	3396

#	User	Contrib.
1	Um_nik	185
2	adamant	177
2	awoo	177
4	nor	169
5	maroonrk	165
6	-is-this-fft-	164
7	antontrygubO_o	155
8	ko_osaga	151
8	dario2994	151
10	SecondThread	149

We hope you enjoyed the contest! Sorry for the late editorial.

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int MAX = 200007;
const int MOD = 1000000007;

void solve() {
	int a[3];
	cin >> a[0] >> a[1] >> a[2];
	sort(a, a + 3);
	cout << a[1] << '\n';	
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
	// solve();
}

1760B - Atilla's Favorite Problem

Idea: SlavicG

Tutorial

Solution

#include "bits/stdc++.h"
using namespace std;
 
using 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; string s; cin >> n >> s;
    sort(all(s));
    cout << s.back() - 'a' + 1 << "\n";
}   
 
int32_t main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t = 1;
    cin >> t;
    while(t--) {
        solve();
    }
}

1760C - Advantage

Idea: Errichto

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

#define forn(i, n) for (int i = 0; i < int(n); i++)

int main() {
    int t;
    cin >> t;
    forn(tt, t) {
        int n;
        cin >> n;
        vector<int> a(n);
        forn(i, n)
            cin >> a[i];
        vector<int> b(a);
        sort(b.begin(), b.end());
        forn(i, n) {
            if (a[i] == b[n - 1])
                cout << a[i] - b[n - 2] << " ";
            else
                cout << a[i] - b[n - 1] << " ";
        }
        cout << endl;
   }
}

1760D - Challenging Valleys

Idea: mesanu

Tutorial

Solution

#include <bits/stdc++.h>
using namespace std;
void solve()
{
    int n;
    cin >> n;
    vector<int> a;
    for(int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        if(i == 0 || x != a.back())
        {
            a.push_back(x);
        }
    }
    int num_valley = 0;
    for(int i = 0; i < a.size(); i++)
    {
        if((i == 0 || a[i-1] > a[i]) && (i == a.size()-1 || a[i] < a[i+1]))
        {
            num_valley++;
        }
    }
    if(num_valley == 1)
    {
        cout << "YES" << endl;
    }
    else
    {
        cout << "NO" << endl;
    }
}

int32_t main(){
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}

1760E - Binary Inversions

Idea: SlavicG

Tutorial

Solution

#include "bits/stdc++.h"
using namespace std;
 
using 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()

ll calc(vector<int>& a) {
    ll zeroes = 0, ans = 0;
    for(int i = sz(a) - 1; i >= 0; --i) {
        if(a[i] == 0) ++zeroes;
        else ans += zeroes;
    }
    return ans;
}
void solve() {
    int n; cin >> n;
    vector<int> a(n);
    forn(i, n) cin >> a[i];
    ll ans = calc(a);
    forn(i, n) {
        if(a[i] == 0) {
            a[i] = 1;
            ans = max(ans, calc(a));
            a[i] = 0;
            break;
        }
    }
    for(int i = n - 1; i >= 0; --i) {
        if(a[i] == 1) {
            a[i] = 0;
            ans = max(ans, calc(a));
            a[i] = 1;
            break;
        }
    }
    cout << ans << "\n";
}   
 
int32_t main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t = 1;
    cin >> t;
    while(t--) {
        solve();
    }
}

1760F - Quests

Idea: flamestorm

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int MAX = 200007;
const int MOD = 1000000007;

void solve() {
	int n, d;
	long long c;
	cin >> n >> c >> d;
	long long a[n];
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	sort(a, a + n, greater<long long>());
	int l = 0, r = d + 2;
	while (l < r) {
		int m = l + (r - l + 1) / 2;
		long long tot = 0;
		int curr = 0;
		for (int i = 0; i < d; i++) {
			if (i % m < n) {tot += a[i % m];}
		}
		if (tot >= c) {
			l = m;
		}
		else {
			r = m - 1;
		}
	}
	if (l == d + 2) {cout << "Infinity\n"; return;}
	if (l == 0) {cout << "Impossible\n"; return;}
	cout << l - 1 << '\n';
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
	// solve();
}

1760G - SlavicG's Favorite Problem

Idea: SlavicG

Tutorial

Solution

#include "bits/stdc++.h"
using namespace std;
 
using 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()

const int N = 1e5 + 10;
vector<pair<int, int>> adj[N];
set<int> s;
bool ok = true;
int n, a, b;
void dfs1(int u, int par, int x) {
    if(u == b) return;
    s.insert(x);
    for(auto e: adj[u]) {
        int v = e.first, w = e.second;
        if(v == par) continue;
        dfs1(v, u, x ^ w);
    }
}

bool dfs2(int u, int par, int x) {
    if(u != b && s.count(x)) return true;
    for(auto e: adj[u]) {
        int v = e.first, w = e.second;
        if(v == par) continue;
        if(dfs2(v, u, w ^ x)) return true;
    } 
    return false;
}

void solve() {
    s.clear();
    cin >> n >> a >> b; --a, --b;
    forn(i, n) adj[i].clear();
    for(int i = 0; i < n - 1; ++i) {
        int u, v, w; cin >> u >> v >> w; --u, --v;
        adj[u].pb({v, w});
        adj[v].pb({u, w});
    }
    dfs1(a, -1, 0);
    if(dfs2(b, -1, 0)) 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();
    }
}

#include <bits/stdc++.h> using namespace std; int t; int n, a, b; void dfs(int u, int now, int father, vector<vector<pair<int, int>>>& g, unordered_set<int>& st) { if (father != -1) { st.insert(now); } for (auto [nxt, w]: g[u]) { if (nxt != father) { dfs(nxt, now ^ w, u, g, st); } } } bool dfs2(int u, int now, int father, vector<vector<pair<int, int>>>& g, unordered_set<int>& st) { if (st.count(now)) { return true; } for (auto [nxt, w]: g[u]) { if (nxt != father && nxt != b and dfs2(nxt, now ^ w, u, g, st)) { return true; } } return false; } int main() { ios::sync_with_stdio(false); cin >> t; while (t--) { cin >> n >> a >> b; vector<vector<pair<int, int>>> g(n + 1); unordered_set<int> st; for (int i = 0; i < n - 1; i++) { int u, v, e; cin >> u >> v >> e; g[u].push_back({v, e}); g[v].push_back({u, e}); } dfs(b, 0, -1, g, st); if (dfs2(a, 0, -1, g, st)) { cout << "YES" << endl; } else { cout << "NO" << endl; } } return 0; }

#include <bits/stdc++.h> using namespace std; typedef long long ll; void solve(){ int n, mx = 0; cin>>n; string s; cin>>s; for (int i = 0; i < n; i++){ mx = max(mx, (int)s[i]); } cout << mx - 96 << endl; } int main(){ int tc; cin >> tc; while(tc--){ solve(); } }

Comments (75)

Write comment?

mercury.16

6 months ago, # |

Such a elegant implementation of problem G

→ Reply

samdavydov

← Rev. 2 →

I believe C can be done in O(n).

Read everything into array a
Find two maximum values — max and pre_max
Do a run:

for (i in a) 
if (i == max) print(i - pre_max) else print(i - max)

Lolka555

6 months ago, # ^ |

Yes, this one is work. You can check my solution that use idea of remember max and pre_max: 181961801

ciopphit

4 months ago, # ^ |

can you post it cause it says I'm not allowed to view it

mnkp

I had the exact same idea for G as the tutorial, But I screwed up the implementation in so many ways

yash1402_

I find 'screwing up in so many ways' very relatable and funny, maybe this should be my spiritual quote!!

white_harmony

Why is this code for problem C, getting TLE? Code

ssvb

"Arrays.sort" has a bad worst case time complexity in Java and can be hacked. See https://codeforces.com/blog/entry/102175 and maybe some other blogs on this topic.

strokeme

Damn the problem ratings are so inflated

AlphaMale06

I think only G and F are inflated, G should be 1500/1600, but then again, for some 1500s and 1600s I think trees are a new topic, so it is normally going to be inflated, while F is just a binary search with a few more steps, so I think it should be 1300/1400.

tibinyte

cap, SlavicG problems are proven to be $$$3500$$$

It's actually a really nice problem, I just don't think it's that hard

TrendBattles

In problem F, I also sort the array in descending order and then I made a prefix sum of n elements. (Let's call it p) We have 3 situations:

1.If (c + p[1] - 1) / p[1] > d then there's no way that we can get c coins in d days.

2.If there is p[i] (i <= d) such that p[i] >= c then any value k satisfies.

3. Brute-force k from d to 0 as k >= d + 1 in this situation is not possible anymore.

Time complexity is O(n log n + d)

sarafsanskar468

Hi, @TrendBattles, If the sum of the entire array is greater than the c coins needed, will that not result in infinity? In case 2: you wrote only for a particular element, why so.?? What if a particular element is not greater than c but the entire array sum is greater?

Well, I made a prefix sum of n elements and p[i] is the sum of the first i elements.

lab_practise

https://ideone.com/3eEgf3 whats the problem in this code. wa on test 2... problem D

l1l1l1l1

For problem G，unordered_map will give you TLE

zbl

why does my code get TLE for problem G.

haiender288

I changed from unordered_set to set and it got accepted

182220698

why unordered_set slower than set......

While it's true that unordered_set is faster than set on average, the worst case time complexity of unordered_set can get to O(n^2) instead of O(log n) in a normal set (specifically on big prime numbers). It's likely that the test you got TLE on was deliberately constructed for solutions like yours

ok thx～

Bryce3D

Must be beethoven97 again

vantaablackk

How does G solution take care of 2 things?

1.) We may never need to teleport and just reach end node with value 0. In that case inside dfs2, we need to have a condition for that.

2.) I don't see where is the condition to not go beyond end node in dfs2 when we don't get true before reaching end node.

gaspar.kristijan

1.) We may teleport from a to a. Consider the following graph Let's say we start at node 1 and need to end at node 3.

(n) node

-w- connection with weight w

(1) -2- (2) -2- (3)

It isn't optimal to teleport to any node however we can compare path from 1 to 2 having XOR value = 2 and path from 3 to 2 having the XOR value equal to 2. In that case we don't teleport but if we had to, we would teleport from node 2 to node 2.

Alternatively what the functions actually checks is all the possible XOR values starting from a and b. If there is a match answer is yes, else no. In our case we have 2 matches, 0 and 2.

We get 0 at the start because that's our starting value and 2 by traveling from node 1 to 2. When starting from b we get 0 by traveling from node 1 to 3 and value 2 by traveling from 3 to 2.

2.) dfs2 starts at node b

    if(dfs2(b, -1, 0)) cout << "YES\n";
    else cout << "NO\n";

It checks if it can hit any optimal values calculated in dfs1. In dfs1 all possible values are calculated where we have the limitation in place.

heyTamjid

In problem C, I guess you do not need to sort the entire array. What you need is just to find out the largest and 2nd-largest element from the array.

(2nd-largest element might be identical to the largest element if the largest value appears 2 or more times in the array)

altforminuscontribution

Awesome round.

Rahil

any idea why I'm getting wa2 in e?

submission

The_Ahmed

It getting WA because of two things:-

You need to handle the case in which you will not do any operations.
Function cal() needs to return long long, because in the worst case ans = (10^10), this will cause overflow of course.

I didn't notice that we can not use the operation thank you <3

k0dev

← Rev. 6 →

"The minimum value of k is 0, and the** maximum value is n** (for larger k, we won't be able to do the same quest multiple times anyways, so it's useless to consider them). "

Why the maximum value for k is n? Isn't it d? Then for larger k than d we won't be able to do the same quest multiple times anyways because there are only d days right? So the complexity should be O(n log(n) + d log(d)) ? Am i missing something or?

HNOONa

maximum value of k depends on d and the entries supplied, it may also exceed d

ex: A=[3,2,1], d=1,c=3 k can increase infinitely, look, i have a single day to make profit and if i complete the first quest i get the sum>=c, i dont need to use that quest again or any other quest so you can increase k as much as you like

← Rev. 3 →

ofc if K can increase infinitly then you return Infinite, you dont infinite loop xd. What i mean is computationally, to find k, you work with order of d not n (or at least in my solution i do that)

tushar_kumar

can anyone tell me my mistake in problem D 181949662

ok_booomer

yes. you are right. The code shown is same as what you are saying

variety-jones

Take a look at Ticket 16544 from CF Stress for a counter example.

can someone kindly point out error in my solution here for problem G? https://codeforces.com/contest/1760/submission/182250273

test

Output

answer

PANIC_

← Rev. 4 →

manu.chittora

someone please explain "Also, we cannot pass b on the path from a→c." in tutorial of G.

It refers to the fact we cannot pass through the node b (end node) when starting dfs from a. Because it's impossible to enter b without our value being 0.

See the problem statement

you are allowed to enter node b if and only if after traveling to it, the value of x will become 0.

In other words, you can travel to node b only by using an edge i such that x XOR wi=0.

okay:) Thank You

demonstrator

Can somebody please explain why i am getting WA 182023139 in G . I am storing xor from starting to ending node in a variable then from every node we can go to ending node . and xor will be xor of node from starting to xor of starting to end and then checking if it exists in our set or not.

Made some changes in your code and got AC.

Biggest change I did was the starting node in dfs2 and some changes in the functions themselves.

Submission: 182291486

Take a look at Ticket 16545 from CF Stress for a counter example.

Gisu

In 1760D - Challenging Valleys you could also insert INF at the start and at the end of the processed array (without successive equal elements) to simplify the job with edges, then check every triplet inside for a[i-1]>a[i]<a[i+1], 182365763

dwaraganathan15

Can somebody help me with my problem G code. I am getting runtime error in testcase 20.

https://codeforces.com/contest/1760/submission/182632177

InfinitePath

I have submitted this problem like 20 times, but I keep getting wrong answer on token 162 on test 1. I've literally tried debugging every nuance of the code and compared it with the solution, but I don't know what I'm doing wrong. Here's my code: https://codeforces.com/contest/1760/submission/182718256

You never know, 21st time's the charm.

Take a look at Ticket 16543 from CF Stress for a counter example.

ultra_viz

In the tutorial of problem D it says: i=0 or bi−1<bi and i=n−1 or bi>bi+1

But shouldn't it be like: i=0 or bi−1>bi and i=n−1 or bi<bi+1 or am I missing something?

harous

did anyone solve G with bfs ? i got stuck in test 40 . any idea why i'm getting wrong answer ? 183283417

Take a look at Ticket 16546 from CF Stress for a counter example.

KFPanda

Could someone point out where my code fails? ato[i] is xor value of path from a to i. 184338941

Take a look at Ticket 16551 from CF Stress for a counter example.

Sreedhar3210

why does this code fails 185000763 for G.

Take a look at Ticket 16564 from CF Stress for a counter example.

SJMormo

5 months ago, # |

Solved F in O(n) using prefix sum — Code

chrisdiamond10

2 months ago, # ^ |

you sorted at the beginning tho --> O(nlogn)

thevaibhavdixit

Could anybody please tell me what might be the problem in this code for problem E. Its failing on second test case for 2653rd iteration. Link to submission

5 months ago, # ^ |

Take a look at Ticket 16572 from CF Stress for a counter example.

Thanks man finally figured it out. That site is a lifesaver.

Pro_Noob_25

why my solution is failing on test 2 iteration number 53 code

Take a look at Ticket 16617 from CF Stress for a counter example.

gmontalvo404

Can someone help me? The example case of problem F in my local works but in Codeforces it does not. 188250142

spkumar9952

Problem G unordered_set gives TLE but set solution Accepted. I also faced similar issues in other problems as well. Can anyone explain me why this is happening ?

tgp07

I have tried stress testing my solution to problem G, and I haven't yet been able to find a test case that breaks my code. Can anyone please take a look at this? 188826168