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Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 22.11.2024 13:56:08 (j1).
Десктопная версия, переключиться на мобильную.
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When I reach the last page of registration process (where there are "About you" and "Rules" sections), and press "Register" button, it does nothing but remaining at the same page.
What do you think?, is it a general problem, or just for me (and my friend from the same country, Syria, and we use a vpn connection to enter this site)?.
Thanks in advance.
I had a similar problem when I tried logging in to the website using Firefox. Opening it on Chromium solved the issue.
Thank you!
Didn't work with me :(
Clashes with the Atcoder round.
How to solve for n=20? (the third question). https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fd5e
Hint: I used this alg: https://mirror.codeforces.com/blog/entry/45223
Can you please elaborate on this approach? During Contest I realised that 3^N can be optimised 2^N via SOS DP but i Couldn't come up with approach. It would be great if you could share your approach.
Let
dpA[mask]be sum of A's happiness for shifts represented bymask. Then for everymasksuch thatdpA[mask] >= H, we need to computeF[mask1], wheremask1is obtained frommaskby flipping its bits, andF[mask1]is the number of supersets ofmask1whose sums on B are >= H. This is equivalent to find all the subsets ofmaskwhose sums on B are <=totalB - H....code
meet in midddle and ordered set will also work fine......
code
Although SOS DP solution seems to be better, but I used Meet in the Middle Technique to solve this problem.
The approach was to solve for both the halves of the array recursively by brute force O(3^(n/2)) and store all the possible pairs of total happiness values for A and B for both the halves separately and sort them according to any one of the person. I did it for A.
Now, how to compute the number of possible pairs from second halves ??
Suppose (x,y) is the first pair for (A,B), you need to find the number of pairs from the second halves (X,Y) such that x+X >= h and y+Y >= h;
So basically find the number of pairs (X,Y) such that (X>=(h-x),Y>=(h-y)). At this stage, I first did coordinate-compression to limit the values. (Think Why)
To calculate the number of pairs, I made a classical segment tree and accordingly kept on updating it by visiting the pair in first half in the sorted order.
Complexity -> O(3^(n/2)*lg(3^(n/2))
At least in terms of time complexity, Meet in the Middle is better.
How to solve (b)the equation?
You can solve it using a greedy algorithm, make use of suffix sum. Solution
Can you please elaborate the solution?I didn't understand the editorial
anyone with rank 150-200 got a call from a google recruiter in this year rounds ??
Seems like there's no way getting a call from a recruiter without being in top ~20-30 (if possible). Not even sure whether recruiters even look at Kickstart standings.
I have seen many cases where people got call even without participating in Kickstart.
I mean if you want to get a call from Google recruiter as a result of your performance in Kickstart. Yeah, sure there are many cases of call without involving Kickstart
Thankyou so much for the information, will work hard to get a better rank. Btw, how can we get a call from recruiters through ways other than Kickstart. If you know anything related to it can you share !!
Hey man I know many getting call. Even after a rank of 213 in Kickstart ayushghd got a call from Google.
May be see this video and your doubts will be clear. Link
And also you should try connecting to recruiters or someone working in Google and knows cp so that (s)he can give you a referral.
Hey CMaster,i got a call for onsites at Google office
I did SOS DP on 3rd Question but I am getting the wrong answer can anyone point out what am I doing wrong here. Solution
You havent understood the question correctly. Either this or your logic is wrong. It says that atleast one of them has to work everyday,ie, both can also work on any day.
See the spoiler if needed. (Based on explanation by @tuananh238)
In Book Reading my code gave wrong answer for Test Set 2. When I replaced the ans variable datatype to long long int from int it got AC.
But according to me int should also work the maximum value of answer will be 1166750. Can anybody help me finding error in my code.
include<bits/stdc++.h>
define ll long long int
using namespace std; ll ct[100005];
int main(){ int t; cin>>t; for(int x = 1 ; x <= t ; x++) { int n,m,q; cin>>n>>m>>q;
for(int i = 1 ; i <= n ; i++) { for(int j = 1 ; j <= sqrt(i); j++) { if(i%j == 0) { int first = i/j; ct[first]++; if(first * first != i) { ct[i/first]++; } } } } for(int i = 1 ; i <= m ; i++) { int val; cin>>val; for(int j = 1 ; j <= sqrt(val); j++) { if(val%j == 0) { int first = val/j; ct[first]--; if(first * first != val) { ct[val/first]--; } } } } int ans = 0; for(int i = 1 ; i <= q ; i++) { int val; cin>>val; ans += ct[val]; } cout<<"Case #"<<x<<": "<<ans<<endl; for(int i = 1 ; i <= 100005 ; i++) ct[i] = 0;} }
Consider the case:
Ohk, you are right. Thanks for the help.
I myself felt so bad about the same mistake too.I was done with all 3 in 1.5 hrs, but after the system test results were out, i realized i shouldn't have been greedy taking int type(to avoid any possible TLE)....
Can someone help me with the 2nd question? I just cannot figure out whats wrong in my approach:-
1.Find total sum
2.Find count of set bits in n numbers at each bit place.
3.Subtract the values (those bits where count(1s) >= count(0s) say ith bit has this property. then subtract (2*count(1s) — n)*(1<<i) )
4.Now add the values (those bits where count(0s)>counts(1s)) — starting from msb
5.Keep updating the value of k
Here's my code
You have an overflow in your code. Just use numbers with 51 bits only(bn=50 in your code instead of 55) and your code will pass.
Can someone help me with 3rd problem? I want to know what's wrong with my solution. I used meet in the middle algorithm and then used merge sort tree. I got TLE with bigger cases. Code
Hey, this is my first time participating in Google's Kickstart. I know my code is probably inefficient but it works beautifully on my local IDE (IntelliJ). However, when I upload it to the kickstart website, it gives me a runtime error. Can someone please help me? TY :D
(This is for Round G, Question A)
Here is the code: http://txt.do/1kcfs
Look at their FAQ, you may have to name the main class Solution.
Hey guys. Stuck on the 3rd question with the following logic: Meet in the middle and then using trie to count no. of numbers less than the needed value.
I think the logic is right but just not able to figure out the bug. Code is well written and self explanatory. Here's my code.
Please help
Screencast of me winning the round: https://youtu.be/bc3DpNRZ0lk.
(sorry for posting it so late)