### chokudai's blog

By chokudai, history, 2 years ago,

We will hold AtCoder Beginner Contest 212.

The point values will be 100-200-300-400-500-500-600-600.

We are looking forward to your participation!

• +171

| Write comment?
 » 2 years ago, # |   +36 Why downvote this blog? It will be a nice contest!
•  » » 2 years ago, # ^ | ← Rev. 3 →   +41 I think the reason being is:5:30 — 7:10 PM IST: AtCoder Beginner Contest 2126:00 — 8:30 PM IST: ICPC Amritapuri 2020 Mock Round (For Indians)6:00 — 9:00 PM IST: July 2021 CodeChef Lunchtime (With Extra Prizes for Indians)Collision at the peak :)
•  » » » 2 years ago, # ^ |   +24 Simple choice for me.Because I love AtCoder more than any problem solving website
•  » » » » 2 years ago, # ^ |   -21 For Indians, the simple choice would be ICPC mock since ICPC >>>> any problem solving website :)
•  » » » » » 2 years ago, # ^ |   +27 yes i took ICPC option and now regret it ;-;
 » 2 years ago, # |   0 Clash with CodeChef Lunchtime!
•  » » 2 years ago, # ^ |   +9 No Codechef Lunchtime is clashing with ABC not the other way around
 » 2 years ago, # | ← Rev. 4 →   0 How to solve E?
•  » » 2 years ago, # ^ | ← Rev. 2 →   +3 just do the following K times: suppose dp[i] is the number of paths that end at i, initially only dp[1] = 1.For every day, all vertices can go to i except those that are broken(and i, of course). So you can just do $nextdp[i] = \sum dp[i] - \sum _{j \in adj[i]} dp[j] - dp[i]$ , then just swap nextdp and dp.
•  » » » 2 years ago, # ^ |   +8 I thought it would be TLE . Is not TLE only because size of all adj[i] won't be 5000 ?
•  » » » » 2 years ago, # ^ |   0 yes
 » 2 years ago, # |   +10 I used Fast Walsh for H. Is it needed?
 » 2 years ago, # |   +24 tourist is too fast.
•  » » 2 years ago, # ^ |   +39 Well this was the only choice he has! Since he has to top the Codechef Lunchtime also and He manage to get 2 min break in between.
 » 2 years ago, # |   +7 E was a great problem, thank you for teaching me dp on trees ( atleast a part of it ).
•  » » 2 years ago, # ^ |   +8 Can anyone tell me the reason for downvote? Was it not "dp on trees" or is something else the reason for downvote?
•  » » » 2 years ago, # ^ |   +9 I doubt it has a flavor of "DP on tree," but still I don't see any reason on downvoting it... (Some of them may be downvoting you just because you're gray, which attitude I don't like)
•  » » » » 2 years ago, # ^ |   +8 Yes I realised it later.....the graph might not even be a tree so dp on trees is wrong...thank you.
 » 2 years ago, # |   +24 I feel like implementing F was way harder than G.
 » 2 years ago, # | ← Rev. 2 →   0 Code#include using namespace std; int mod = 998244353; int main(){ int n,m,k;cin>>n>>m>>k; vector>arr(n,vector(n,false)); for (int i = 0;i>a>>b; a--;b--; arr[a][b]=true; arr[b][a]=true; } vector>adj(n); for (int i = 0;idp(n,0); for (auto x:adj[0]){ dp[x]=1; } for (int i = 0;idp2(n,0); for (int j = 0;j
•  » » 2 years ago, # ^ |   0 The issue with your code is that in worst case, it runs in approximately $O(n^2k)$ because there could be maximum ${n(n-1)/2} - m$ valid edges, so for each iteration of $i$, the inner loop runs in $O(n^2)$, hence the TLE.
 » 2 years ago, # |   0 Thank You for the wonderful problemset. The number of tasks was a little too much considering only 100 mins to solve them.
 » 2 years ago, # |   +20 As a writer of ABC(sadly this time I'm not), I'm curious about how do participants feel with the new format.
•  » » 2 years ago, # ^ |   +25 i wish there was infinity of such genuine,geniosity,pure beauty,attractive,elegant,gorgeous,inspiring,exciting... problems :) thank you .
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 So you are saying this was not a one time thing ? Anyways, if the 8 problem ABC is going to be adapted for further ABC too. I wish these 2 points to be implemented. Increase duration a bit (maybe like make it 2 hours ? ) Make problem >= F, more in the reach of actual rated range participants. For me personally today, I was done till E in 25 mins and then I stared at screen for rest whole contest. But yes, I understand its my fault and F was actually not that out of reach, but still I think it was pretty hard. (That can be seen by yellow rating to F, G and orange rating to H at kenkoo)
•  » » 2 years ago, # ^ |   +3 Is this going to be the new standard format? If so, sounds fun!
 » 2 years ago, # |   0 Hii, I tried to solve D via method 1, and then method 2, I failed to see why first one didn't work, could u help me out please.
•  » » 2 years ago, # ^ |   0 Try the following trivial testcase with your incorrect code and you should easily see what's wrong with it, the correct output is 2 11: Testcase5 1 1 2 10 1 2 3 3 
 » 2 years ago, # |   0 x^n≡y(mod P) ⇔r^an≡r^b(mod P) ⇔an≡b(mod P−1) How is the second step to the third step transformed? Who can explain? Thank you very much
•  » » 2 years ago, # ^ |   0 this problem is G
•  » » » 2 years ago, # ^ |   0 Because for every i(1<=i<=P-1),r^i is distinct ,so we can know for every r^i there is exactly one such i.