Indian ICPC 2022-23 Regionals — Qualifier Rounds

20 months ago, # |

+17

Best of Luck everyone!

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xox0

20 months ago, # |

+141

Kanpur Mathura prelims has to be the worst ICPC contest, probably ever. Imagine setting 4 problems for a team of 3, and so easy that all of them could easily be solved in $$$\lt$$$ 20 minutes total. And then to add icing on the cake the penalty for a single wrong submission is 20 minutes as well

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This is just sad. Wait, no, not just sad. Sad and enraging. Great job!

20 months ago, # ^ |

← Rev. 2 →

+33

Agree.... feeling sad after making wrong submissions blindly :(

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jaglike_makkar

20 months ago, # ^ |

+29

And that surprise "Case 1:" twist was the ultimate test :)

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satyam343

20 months ago, # |

+32

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tanmay.raj.29

20 months ago, # |

+53

Was this ICPC qualifier or leetcode contest? 0_0

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yash_daga

20 months ago, # ^ |

+75

No, leetcode contests are harder.

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blitztage

20 months ago, # ^ |

+54

Enough with the direspect bro, this can’t be accepted,

leetcode contests are much better than this

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islingr

20 months ago, # |

+33

speedforces

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magga

20 months ago, # ^ |

+30

full shit sherlock

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tamajitbuba

20 months ago, # ^ |

← Rev. 2 →

+31

| ||

|| |_

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vrintle

20 months ago, # ^ |

+22

Feeling sorry for your 1 2 2 50.

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18o3

20 months ago, # |

Can anybody tell why this got TLE?

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satyam343

20 months ago, # ^ |

They added extra tests and rejudged your submission. After rejuding, execution time has not been updated.

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islingr

20 months ago, # ^ |

just write it iteratively, and/or use booleans rather than max

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blitztage

20 months ago, # |

+111

If you take 10 lakh rupees as cumulative fees for just the prelims, it is just downright shameful coming up with such a problemset. Probably should reassess whether your priority is advancement of CP or just a pitiful money grab

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l1ability

20 months ago, # ^ |

+27

mathura laxmi chit fund

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o.O.o

20 months ago, # |

+93

Refund.

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20 months ago, # |

Can someone please tell me why this got TLE ?

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vrintle

20 months ago, # ^ |

You can't rely on 2d-vectors where your complexity is $$$\mathcal{O}(N^2 \cdot T)$$$ which is probably $$$8 \cdot 10^8$$$ iterations, with time limit of 3s.

Use 2d-arrays instead.

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20 months ago, # ^ |

Thanks a lot!

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SilverWing05

20 months ago, # ^ |

I'm about to cry

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from Imgflip Meme Generator

20 months ago, # ^ |

Already did :) Four TLE's because of this lol.

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SilverWing05

20 months ago, # ^ |

Guess what

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angry_pikachu

20 months ago, # ^ |

idk why, my solution got accepted , i have used 2-D vectors too.

Any idea how many slots are there in kanpur-mathura? I hope there should be 120 seats :/

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SilverWing05

20 months ago, # ^ |

Lucky you :')

As always, no official info. from them. But if we see last year, Kanpur had 70 seats, so Kanpur + Mathura should be around 120 to 140. Just a guess though.

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yash_daga

20 months ago, # |

+58

How to make a 1600 rated problem 1800.

Give input in floating points.

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xox0

20 months ago, # ^ |

More like 1200 into 1400 but yeah, that was so annoying (:

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shiv_codegen

20 months ago, # |

+14

Speedforces af.

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rakesh272611

20 months ago, # |

worst contest I have seen

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tanmay.raj.29

20 months ago, # |

← Rev. 2 →

+23

Preparing for ICPC prelims:

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VS-Codes

20 months ago, # |

-11

Can anyone give a testcase that fails this code for B? Cannot understand what is wrong with this approach.

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_Aryan_

20 months ago, # ^ |

What when s[i] is equal to value at both pointers!

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_CHEEMS_

20 months ago, # ^ |

I used the same approch but sill not able to understand why equal value would effect

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dedsec_29

20 months ago, # ^ |

Try this test case
1
5
abdcd
dcabd

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Runtime-Terr0r

20 months ago, # |

← Rev. 2 →

What happened to that tradition of ICPC where one problem is solved by all teams and one problem is not solved by any team Why having prelims as div 69420 contest

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s_prateek26

20 months ago, # |

+14

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Bosscoder

20 months ago, # |

+11

How many teams will be selected from each institute ?

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WAtoAC2001

20 months ago, # |

Kanpur regionals qualification rules have been updated here

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_aka5h

20 months ago, # ^ |

do you have the same for Amritapuri regionals?

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-214748364

20 months ago, # |

Does anyone know how many slots are there for Mathura Regionals?

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20 months ago, # |

Why is nlogn solution giving TLE for XOR sort?

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DCSekhara

20 months ago, # ^ |

For me it didn't

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20 months ago, # ^ |

This gave TLE

Spoiler

#include <bits/stdc++.h>
using namespace std;
typedef int ll;

vector <ll> compress (vector <ll> v){
        vector <pair<ll,ll>> w((ll)v.size());
        for(ll i=0;i<(ll)v.size();i++){
                w[i].first = v[i];
                w[i].second = i;
        }
        sort(w.begin(),w.end());
        vector <ll> res((ll)v.size());
        for(ll i=0;i<(ll)w.size();i++){
                res[w[i].second] = i;
        }
        return res;
}

void solve(){
        ll n;
        cin >> n;
        vector <ll> v(n), order(n), pos(n);
        for(ll i=0;i<n;i++){
                cin >> v[i];
        }
        v = compress(v);
        for(ll i=0;i<n;i++){
                pos[v[i]] = i;
                order[i] = v[i];
        }
        vector <pair<ll,ll>> ans;
        sort(order.begin(),order.end());
        for(ll i=0;i<n;i++){
                if(pos[order[i]] == i){
                        continue;
                }
                ll x = i;
                ll y = pos[order[i]];
                ans.push_back(make_pair(x,y));
                ans.push_back(make_pair(y,x));
                ans.push_back(make_pair(x,y));
                pos[v[x]] = y;
                pos[v[y]] = x;
                swap(v[x],v[y]);
        }
        cout << (ll)ans.size() << endl;
        for(auto x : ans){
                cout << x.first+1 << " " << x.second+1 << endl;
        }
        cout.flush();
}

int main(){
#ifndef ONLINE_JUDGE
        //freopen("input.txt","r",stdin);
        //freopen("output.txt","w",stdout);
#endif
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        int tc;
        tc = 1;
        cin >> tc;
        for(int i=1;i<=tc;i++){
                //cout << "Case #" << i << ": ";
                solve();
        }
        return 0;
}

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aryanc403

20 months ago, # ^ |

+28

#define endl "\n"
195 ms

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20 months ago, # ^ |

+53

I’m jumping off the terrace

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meet.05

20 months ago, # ^ |

after 1st tle even optimized it to o(n), still didn't make it. Our team is jumping off too.

code

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
#include <complex>
#include <queue>
#include <set>
#include <unordered_set>
#include <list>
#include <chrono>
#include <random>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <stack>
#include <iomanip>
#include <fstream>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> p32;
typedef pair<ll, ll> p64;
typedef pair<double, double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int>> vv32;
typedef vector<vector<ll>> vv64;
typedef vector<vector<p64>> vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
ll MOD1 = 998244353ll;
ll MOD2 = 998244353ll;
double eps = 1e-12;
#define forl(i, e) for (ll i = 0; i < e; i++)
#define forls(i, s, e) for (ll i = s; i < e; i++)
#define rforl(i, s) for (ll i = s; i >= 0; i--)
#define rforls(i, s, e) for (ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout << #x << " = " << x << endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin()                    \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
#define endl "\n"

void solve()
{
    ll n;
    cin >> n;

    ll v[n];
    ll s[n];
    unordered_map<ll,vector<ll>> m;
    ll curr = n - 1;

    forl(i, n)
    {
        cin >> v[i];
        s[i] = v[i];
    }

    sort(s, s + n);

    forl(i, n)
    {
        if (v[i] != s[i])
        {
            m[s[i]].pb(i);
        }
    }

    vector<pair<ll,ll>> ans;

    while (curr >= 0)
    {
        if (v[curr] == s[curr])
        {
            m[v[curr]].pop_back();
            curr--;
        }
        else
        {
            ll x = curr + 1;
            ll y = m[v[curr]].back() + 1;

            ans.pb({x, y});
            ans.pb({y, x});
            ans.pb({x, y});

            // ll el = v[y - 1];
            m[v[curr]].pop_back();
            swap(v[curr], v[y - 1]);
        }
    }

    cout << ans.size() << endl;

    for (auto x : ans)
    {
        cout << x.fi << " " << x.se << endl;
    }
}

int main()
{
    fast_cin();

    ll t = 1;
    cin >> t;

    while (t--)
    {
        solve();
    }

    return 0;
}

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VaibhaveS

20 months ago, # |

How to solve central subset?

link

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20 months ago, # ^ |

-9

Centroid decomposition ig

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ExplodingFreeze

20 months ago, # ^ |

← Rev. 2 →

+23

Let sz(x) mean the size of the subtree rooted at x. Formally $$$sz(x) = 1 + \sum sz(v)$$$ for all children $$$v$$$ of $$$x$$$.

Choose an arbitrary root and walk through the tree normally with a dfs calculating values of $$$sz$$$. For any node where $$$sz(x) \geq \sqrt{n}$$$, add $$$x$$$ to the subset and consider $$$sz(x) = 0$$$ for further calculations.

Why does this work?

For any child $$$v$$$ of $$$x$$$, $$$sz(v) \lt \sqrt{n}$$$, so the distance between any node in this subtree from $$$v$$$ must also be $$$\lt \sqrt{n}$$$. Since $$$v$$$ is a child of $$$x$$$, the distance of any node in the subtree from $$$x$$$ is $$$\lt \sqrt{n} + 1$$$ or in other words $$$\leq \sqrt{n}$$$.
Since $$$sz(x) \geq \sqrt{n}$$$ whenever we perform this, we add at most $$$\frac{n}{\sqrt{n}} = \sqrt{n}$$$ nodes to the subset.

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gupta_nitin

20 months ago, # ^ |

← Rev. 2 →

1) Pick any spanning tree and root at node $$$1$$$.

2) Pick the node having the max depth, go up $$$\sqrt{n}$$$ for that node following parents, push this one in the answer, and remove that subtree.

3) Do step 2 until the tree is not empty.

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ExplodingFreeze

20 months ago, # |

← Rev. 2 →

+43

Kudos to Amritapuri online round setters, it felt like a fairly nice and balanced round (except for maybe a bit of an aggressive difficulty jump after the first problem). I really liked how cute the solution to problem A is.

Not at all biased by rank 11 in Amritapuri qualifier vs rank 227 in Mathura qualifier :P

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jeal0uspengu1n

20 months ago, # ^ |

Yeah Amritapuri was relatively better, getting a lower rank didn't hurt in Amritapuri but it did in Mathura ;(

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WeaponizedAutist

20 months ago, # |

What are the qualification rules for amritapuri?

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crimsonred

20 months ago, # |

Did anyone manage to get non-TLE with $$$n \sqrt{n}$$$ on A?

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shrapnelzz

20 months ago, # ^ |

Ya I guess my worst case would be n^(3/2), you can refer here

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pvtr

20 months ago, # |

Will there be additional system testing on the Amritapuri prelims?

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_aka5h

20 months ago, # ^ |

is that common?

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xennygrimmato

20 months ago, # ^ |

Nope, full tests are run during contest time.

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shreyas_v

20 months ago, # |

← Rev. 4 →

For the mathematicians and physicists problem of amritapuri region we did dp ( very similar to this problem), but got mle even after replacing 2 d input array by a map of pairs.Any suggestions?

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20 months ago, # |

+13

How to solve The number of good subarrays

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sobhagyaSD

20 months ago, # ^ |

For each L you can find the smallest R by binary search and if L,R is good then L,(R+1) is also good.

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20 months ago, # ^ |

Could you please elaborate your checker function of binary search

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sobhagyaSD

20 months ago, # ^ |

← Rev. 2 →

So for checking if a subarray is good after removing [l,r] you need to count the f(rem) array, for that you can take the contribution of each bit for example if ith bit is set in k subarrays then the contribution will be 2^i*k for finding the number of subarray that has ith bit set you can do is total_subarray-number_of_subarrays_where_ith_bit_is_off. For achieving this you can keep some prefix suffix arrays.

Refer this code for solution: here

PS: Sorry for my alien language :(

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20 months ago, # ^ |

Thank you

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20 months ago, # |

Kanpur-Mathura Online Qualifier Round unofficial list here. (made by me :))

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Newtech66

20 months ago, # ^ |

What are the numbers next to the college name? The number of teams qualifying?

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20 months ago, # ^ |

Nope, they're just the count, stating the no of times that college is present in under 100.

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swapbeast

20 months ago, # ^ |

← Rev. 2 →

Our team has rank 45 from Kanpur-Mathura and we are the 2nd team from our college and the 1st team has rank 10 so do we stand a chance for regionals

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20 months ago, # ^ |

Yes, ofc!

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wannaBe

20 months ago, # ^ |

If one team is selected from a college, 2nd team wouldn't be allowed, right ?

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adityagamer_simp_edition

20 months ago, # ^ |

That is the rule for Amritapuri regionals, not Mathura regionals. You can check last year's criteria here.

I can't believe that such a great singer is on Codeforces. I am your big fan.

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Jeffin

20 months ago, # |

+15

Are qualification rules for Amritapuri same as last year? Can more than 3 teams qualify from an institute?

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AAK

20 months ago, # |

+29

Any ideas as to when the results would be announced? As the regional dates are close approaching, and most of us would need to book flights to get to Bangalore/Amritapuri.

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20 months ago, # ^ |

Results for Amritapuri have been announced: link

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Panther339

20 months ago, # |

gwalior pune region info anyone

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CodeKar

20 months ago, # |

When will mathura region results gonna be declare?

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angry_pikachu

20 months ago, # ^ |

← Rev. 3 →

today or tomorrow, there are 150 seats

Source — codedrills telegram channel

(someone claimed to got reply from icpc mathura-kanpur via email )

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