UNIQUE VISION Programming Contest 2024 Summer (AtCoder Beginner Contest 359) Announcement

6 месяцев назад, # |

← Rev. 2 →

Idk how I passed C after 4 WAs, just did some +/- math. Idea was to move diagonally as much as possible and align the x coordinate or y coordinate with the target node.

Code

#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define all(v) v.begin(),v.end()
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define pii pair<int,pi>
#define rep(i,n) for(int i=0;i<n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;--i)
#define cusrep(i,a,b,j) for(int i=a;i<b;i+=j)
#define rng(i,a,b) for(int i=a;i<b;i++)
#define rrng(i,a,b) for(int i=a;i>b;--i)
#define ar array

void run_case() {
  ll sx, sy, tx, ty;
  cin >> sx >> sy >> tx >> ty;
  ll alpha = min(abs(sx-tx), abs(sy-ty));
  ll nx = (sx>=tx) ? sx - alpha : sx + alpha;
  ll ny = (sy>=ty) ? sy - alpha : sy + alpha;

  if(ty%2==0){
      if(abs(tx)%2!=0){
        --tx;
      }
    } else {
      if(abs(tx)%2==0){
        --tx;
      }
  }
  if(sy!=ty){
    if(ty%2==0){
      if(abs(nx)%2!=0){
        --nx;
      }
    } else {
      if(abs(nx)%2==0){
        --nx;
      }
    }
  }
  ll ans = alpha + (1LL*abs(nx - (tx))/2LL) + 1LL*abs(ny - ty);
  cout << ans << "\n";
}
 
int main() {
  fast;
  int t = 1;
  while(t--){
    run_case();
  }
}

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_trie_again

6 месяцев назад, # |

+25

what the fck was C

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6 месяцев назад, # ^ |

Idk man, I just figured some +/- math after moving diagonally as much as possible and aligning resultant x or y with target node. After that it was tricky to move in the same direction where I did some guessing.

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6 месяцев назад, # ^ |

I felt like leaving the contest .

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ExplodingFreeze

6 месяцев назад, # |

Is there a simpler solution to G than Small to Large merging? Or did people just manage to implement that approach in like 5 mins @_@

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Pirate_King

6 месяцев назад, # ^ |

-8

small to large merging is simple ??

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andrewtam

6 месяцев назад, # ^ |

+18

It's a classic euler tour + monostack problem.

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qiuzx

6 месяцев назад, # ^ |

There are indeed simpler solutions, but small to large merging is also simple enough. It's only about 30 lines and 5 mins or so is enough to implement it (my solution in contest was small to large merging).

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kelzzin2_fan

6 месяцев назад, # ^ |

Could you roughly explain your approach using small-to-large merging?

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ok12

6 месяцев назад, # |

can someone please explain the idea behind the 3rd question import math a,b=map(int,input().split()) c,d=map(int,input().split()) t=math.ceil(1.5*abs(b-d))

if t<=abs(a-c) and t!=0: print(int(abs(a-c) -1.5*abs(b-d))//2 + abs(b-d)+1) else: a=abs(b-d) print(a) please explain me why my code is incorrect please

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TrendBattles

6 месяцев назад, # ^ |

you can try to move the initial and the end points to adjacent ones (where (x + y) mod 2 = 0) such that each one is still in its tile and then draft some examples

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ok12

6 месяцев назад, # ^ |

thanks

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udta_kouwa

6 месяцев назад, # |

← Rev. 3 →

My solution for problem-D is to store last K characters as state. AC

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Ishaan_23

6 месяцев назад, # ^ |

Same here, I used a bitset and converted it using to_ulong

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skill_Issued

6 месяцев назад, # ^ |

nice mate. I used bits instead of using strings. Submission

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Dkboss3103

6 месяцев назад, # ^ |

this solution is really good.

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then00bprogrammer

6 месяцев назад, # |

Great contest. D was a really good problem, wish I would've solved E a bit quicker though. Also, How to do C?

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Solved ABDEFG. So strange.

6 месяцев назад, # ^ |

Idea is to move diagonally as much as possible and then align target with the resultant x or y coordinate. Tricky part was to move in the same direction in x. In y its just that difference of y coordinates as height of tiles are 1.

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someanth

6 месяцев назад, # |

how can we get atcoder contest testcase at which our solution got failed...

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Parzival18s

6 месяцев назад, # |

Anybody who found C,D tougher than E,F or just me?

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FFTotoro

6 месяцев назад, # |

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FFTotoro

6 месяцев назад, # ^ |

And G is the easy version of CF1725E.

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shove_it

6 месяцев назад, # ^ |

Can you explain the complexity of your E solution?

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6 месяцев назад, # |

Anyone can give hints on how to solve D?.

My approach to count all the possibilities such there is atleast 1 subarray of K length that is a palindrome and then the subtract that from 2^q and get the answer, but I couldn't calculate it properly(due to overcounting).

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6 месяцев назад, # ^ |

bitmask dp

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6 месяцев назад, # ^ |

Damn bro!! Pease explain a bit further, what will you store using bitmasking? and time complexity of your solution

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https://atcoder.jp/contests/abc359/submissions/54847829

6 месяцев назад, # ^ |

← Rev. 2 →

in mask I will store values of last $$$K$$$ question marks

$$$O(2^K * N * K)$$$

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gurnam771

6 месяцев назад, # ^ |

i think for each index we only need last k-1 elements and check if they are forming palindrome or not

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6 месяцев назад, # ^ |

same thing

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6 месяцев назад, # ^ |

thanks

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ExplodingFreeze

6 месяцев назад, # ^ |

Let $$$dp[i][m]$$$ (for $$$k - 1 \leq i \lt n$$$) represent the number of ways of constructing a good prefix of the first $$$i$$$ characters ending with mask $$$m$$$.

For $$$i = k - 1$$$, this is fairly direct, check all $$$2^k$$$ possible masks and see if they work (are possible and don't contain any palindromes). If they do, set $$$dp[k - 1][m] = 1$$$, otherwise set it to $$$0$$$.

Now notice that the masks for position $$$i$$$ and $$$i - 1$$$ share $$$k - 1$$$ characters. So for any $$$i \geq k$$$, we can check each of the $$$2^k$$$ masks again. If a mask $$$m$$$ is valid, we simply calculate $$$dp[i][m] = dp[i - 1][m'] + dp[i - 1][m' + 1]$$$ where $$$m'$$$ is $$$m$$$ with its highest ($$$(k - 1)$$$-th) bit disabled.

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6 месяцев назад, # ^ |

thanks a lot

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comingsoon.cpp

6 месяцев назад, # |

I think problem B's statement could have been made a lot easier to understand.

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shviv1

6 месяцев назад, # |

My code failed for last three testcases in C.Can someone point out where the mistake lies?

Code

ll sx,sy;
cin>>sx>>sy;

ll tx,ty;
cin>>tx>>ty;



ll ans=(abs(ty-sy));

ans+=(max(0LL,((abs(sx-tx)-(abs(ty-sy))))/2));


cout<<ans<<endl;

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6 месяцев назад, # ^ |

answer for

4 0
-1 0

should be 3 and not 2.

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shviv1

6 месяцев назад, # ^ |

But -1 is invalid input

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6 месяцев назад, # ^ |

2 1
200 98

Expected is 148 but your code gives 147

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Kannayya

6 месяцев назад, # |

← Rev. 2 →

"Can anyone explain C, please? I should have at least tried d, as I've spent so much time with C, actually, not so much time but significant time. I'm feeling bad about it now."

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6 месяцев назад, # |

D is good.

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polosatic

6 месяцев назад, # |

← Rev. 4 →

I used small to large in G. But I can't understand what's wrong in my solution. mp[v][c] contains two numbers: the number of vertices of color c in v's subtree and the total length of paths to them.

Code

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#pragma GCC optimize("unroll-loops,O3")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define cok cout << (ok ? "YES\n" : "NO\n");
#define dbg(x) cout << (#x) << ": " << x << endl;
#define dbga(x,l,r) cout << (#x) << ": "; for (int ii=l;ii<r;ii++) cout << x[ii] << " "; cout << endl;
#define int long long
#define pi pair<int, int>
const int N = 1e6+9;
int n, ans, a[N], sz[N];
vector<int> vec[N];
map<int, pi> *mp[N];
int add[N];
void dfs(int v, int p = -1) {
    sz[v] = 1;
    pi mx = {-1, -1};
    for (int u: vec[v]) {
        if (u == p) continue;
        dfs(u, v);
        sz[v] += sz[u];
        mx = max(mx, pi{sz[u], u});
    }
    if (mx.se == -1) {
        mp[v] = new map<int, pi>;
        (*mp[v])[a[v]] = {1, 0};
        return;
    }
    mp[v] = mp[mx.se];
    add[v] = add[mx.se] + 1;
    ans += (*mp[v])[a[v]].se + (*mp[v])[a[v]].fi * add[v];
    (*mp[v])[a[v]].fi++;
    (*mp[v])[a[v]].se -= add[v];
    for (int u: vec[v]) {
        if (u == mx.se || u == p) continue;
        for (auto [color, para]: *mp[u]) {
            ans += ((*mp[v])[color].se + (*mp[v])[color].fi * add[v]) * para.fi + (para.se + para.fi * (add[u] + 1)) * (*mp[v])[a[v]].fi;
        }
        for (auto [color, para]: *mp[u]) {
            (*mp[v])[color].fi += para.fi;
            (*mp[v])[color].se += para.se + para.fi * add[u] - para.fi * add[v];
        }
    }
}
signed main() {
    cin.tie(0); ios_base::sync_with_stdio(0);
    cin >> n;
    for (int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        a--;b--;
        vec[a].pb(b);
        vec[b].pb(a);
    }
    for (int i = 0; i < n; i++) cin >> a[i];
    dfs(0);
    cout << ans;
}

UPD: submission

UPD2: I found issue, AC

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Chimpanzee

6 месяцев назад, # |

+63

If C has a million haters I'm one of them. If C has a hundred haters I'm one of them. If C has one hater it's me. If C has zero haters I have died. If the world is against C I am with the world, if the world is for C I am against the world.

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6 месяцев назад, # ^ |

D E easier than C

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Noobish_Monk

6 месяцев назад, # ^ |

The whole contest was easier

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nichke

6 месяцев назад, # ^ |

+25

Tbh C was cute problem. I didn't like it either, but it's really simple and solution is pretty concise. I feel like CP turned into mindlessly rewriting same algos over and over, especially on AtCoder ABC. Problems like this humiliate you in a good way.

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https://www.luogu.com.cn/problem/P7840

6 месяцев назад, # ^ |

I was helllbent on solving C, finally did and missed on other problems D,E,...

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Chimpanzee

6 месяцев назад, # ^ |

I don't particularly enjoy writing math casework

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pkhaustov

6 месяцев назад, # ^ |

+33

I also don't like problems like this one. But since ABC is supposed to be educational, I try to make use of such problems. For example, I've learned the following lessons (in this particular order):

Skip nasty non-standard grid problems whenever it's possible. Especially if there are problems with more points to get.
Try to find out "simple observation" (as they will call it in editorial for sure) before considering shit tons of edge cases.
Test your solution like crazy. After the first WA, write stress test immediately, don't wait for luck.
Solving more and more similar problems make your more resistant and prepared to them.
Competitive programming is 100% pain.

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Chimpanzee

6 месяцев назад, # ^ |

Last point is real

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harvey_wzy

6 месяцев назад, # ^ |

If C has n haters ($$$ 3 \le n $$$) I'm one of them. If C has two haters that's you and me. If C has one hater I have AFOed. If the world is for C, I'm with you.

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lazy-guy-1618

6 месяцев назад, # |

E was stack and $$$\text{O(N)}$$$ ? (Asking cuz constraints had 2e5)

Code

void solve() {
    int n;
    cin >> n;
    vector<int> h(n), dif(n);
    inc(i, n) cin >> h[i];
    stack<pair<int, int> > st;
    st.push({1e9 + 1, 0});
    inc(i, n){
        auto val = st.top().first;
        if(val < h[i]){
            while(val < h[i]){
                st.pop();
                val = st.top().first;
            }
        }
        dif[i] = st.top().second;
        st.push({h[i], i+1});
    }

    vector<ll> final(n + 1);
    for(int i = 1; i<=n; ++i){
        final[i] = final[dif[i-1]] + (1LL*h[i-1]*(i - dif[i-1]));
    }

    for(int i = 1; i<=n; ++i) cout << final[i] + 1 << " ";
    cout << "\n";
    return;
}

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0x0002

6 месяцев назад, # |

So typical problem G. Maybe it will be better to swap some orders of the problem.

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Misaka16172

6 месяцев назад, # |

Could you guys just STOP making "greedy in grid" problems such as C plz? :(

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someanth

6 месяцев назад, # |

how can we get atcoder contest testcase at which our solution got failed...

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gza

6 месяцев назад, # |

+113

this is the original problem of F?

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6 месяцев назад, # ^ |

Why ABC usually has an original problem?

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incra

6 месяцев назад, # ^ |

My code which get AC in AT also get AC in LG!

But I suppose it wasn't on purpose, as there weren't many easy ideas:(

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CF_xyy

6 месяцев назад, # ^ |

But I think ABC is a good way to let us find the same problem so that get AC.

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gza

6 месяцев назад, # ^ |

so that get AC is wrong in grammar,you should modify it to be "so that you can get AC on the original problem."

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CF_xyy

6 месяцев назад, # ^ |

Yes,you are right.

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6 месяцев назад, # |

This problem is same as F!

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DukhiAatma420

6 месяцев назад, # |

← Rev. 3 →

+21

C was exactly same as https://mirror.codeforces.com/contest/1421/problem/D. Just consider the right point of tiles (moving between left and right didn't encounter any cost), and those right points forms the hexagon.

It can be solved via linear algebra, no need to go into greedy

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Coder2020HCY

6 месяцев назад, # |

+25

Problem F has appeared here previously : https://www.luogu.com.cn/problem/P7840

However, Luogu is a website that currently mainly used in China. So I think it is a coincidence. I only point out for notice.

Here is the translation by Copilot:

There are $$$n$$$ servers.
Each server has a busyness level denoted by $$$v_i$$$.
The servers are connected using $$$n−1$$$ network links.
Each server has a connection count $$$d_i$$$.

The total running time of the server network is given by the expression:

$$$ \sum_{i = 1}^n d_i^2 v_i $$$

The goal is to minimize this value.

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6 месяцев назад, # ^ |

I think so.

I wonder who finds this original problem?

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ziyistudy

6 месяцев назад, # |

-29

Didn't Atcoder check for if there have been the same problem？？？

This problem from Luogu same as today's G.

And this problem is from the year 2021.

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Coder2020HCY

6 месяцев назад, # ^ |

+26

It is a Chinese website and it only support Chinese... It is extremely difficult to check EVERY CP website and search whether there is a same problem. So It can be only coincidence. BTW it didn't cause any significant effect, because it is just an ABC contest.

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Coder2020HCY

6 месяцев назад, # ^ |

+29

They do nothing wrong. Try not to be too aggressive.

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gsczl71

6 месяцев назад, # ^ |

Is it hard to check weight? Inexplicably! Only a fool thinks so. The questions are exactly the same, and luogu is a great site for brushing up on questions, and the bottom line as a question writer is to not come up with exactly the same questions. Even though the questions are in Chinese, can't you learn Chinese? You can't read Chinese, can you? And ziyistudy is stating facts and you're defending a shit group of question writers. I think you are very wrong.

Translated with www.DeepL.com/Translator (free version)

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