We will hold AtCoder Beginner Contest 185.
- Contest URL: https://atcoder.jp/contests/abc185
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20201213T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: beet, gazelle, kobae964, kort0n, kyopro_friends, QCFium, satashun
- Rated range: ~ 1999
The point values will be 100-200-300-400-500-600.
We are looking forward to your participation!
5 min left to start !
40 seconds left.....
Am I the only one who felt C and F were too classical problems? :/
no
Then, I think they should not include these kind of problems because they are just basic.
Then you need to learn the basics.
I mean what's the point if they are only giving problems whose solutions are already known to people? (They don't even need to think) They are pretty standard problems.
Do you know all the solutions? If not then it is a good way to learn about the standard problems. It is already mentioned that this is a beginner contest and it is a great opportunity to apply those standard techniques. Then what is the problem?
I'm not talking about all problems but three of them are(C,E,F) and I understand that this is a beginner contest but does that mean that you will give questions which are one google search far from the solution? And if yes, then there is no point of giving such contest because one can practice that stuff at Leetcode, GFG etc. also.
E was a very good standard problem with a twist. Really a very good problem.
C was good at its place. It needs basic DP or Combinatorics knowledge.
You can say F was too standard and I also agree with that.
Not sure of C but F was for sure
C was a standard combinatorics problem
E and F were also classical common problems !!
F was basically editing the merge function in my segment tree template
Or maybe copying the exact same code from gfg ;_;
How solve F ??
use segment tree with xor as merge operation of two segments.
classic segment tree problem
You can use Fenwick tree for range based queries. It's standard for range sum.
Using binary indexed tree for each bit independently, you can check if the number of bits in the range from x to y are odd or not.
Code
How to do E?
Problem E can be solved using DP . Following is transition formula (based on if we want to pair index n with m or we want to discard index n or we want to discard index m) :
dp[n][m] = min({solve(n-1,m-1)+((long long)(a[n]!=b[m])),solve(n,m-1)+1,solve(n-1,m)+1});
where n is index of array a and m is index of array b . Base case is when n is equal to 0 or m is equal to 0 . If n=0 then we need to discard prefix of length m in array b .
submission .
just calculate the edit distance between A and B.
It simply reduces to the classical Edit Distance problem. Try to prove it yourself.
I use dynamic programming for solving E. Starting from the left if both elements are equal then you don't need to do anything otherwise do one of the three:
1. remove element from first array.
2. remove element from second array.
3. don't remove any just count 1 as penalty for element being not equal.
If reach at the end of any array remove remaining elements from the other array.
Thanks for explanation !!
How to solve C ??
I solved D but not C
nCr(n-1,11)
Used this formula, but passed only 10/ 16 Test Cases:/
Where am I going wrong? https://atcoder.jp/contests/abc185/submissions/18756020
Appreciate the help.
naturell
ll nCr(ll n , ll r) {
if(n<r)return 0; if(n==r)return 1; ll ans=1; for(ll k=n,p=1;k>(n-r);k--) { ans *= k; if(p<=r) { ans/=p; p++; } } return ans;}
Does anybody know how to solve the E question? I tried longest common subsequence approach but got WA.
Here's an unofficial editorial .
Problem A : Just take the min of given 4 values
Problem B : Just take the difference of consecutive numbers and alternatively subtract and add. Code
Problem C : One of the simple solution is DP.
Let dp(x,y) is number of ways to cut x, into y pieces then our required result is dp(l,12);
The recurrence is
The base cases
Code
Problem D : Sort the given array A, if the first element is not 1, or last is not n, append them.
Code
Problem E : This was just a simple variaton of standard LCS.
Code
Problem F : That's just a standard segment tree with point update queries. Code
C is also
l - 1choose11because we havel - 1choices and have to select11.Main problem in E is to get your head arround what the problem statement asks for.
it should be obvious to many, but if someone feel confused by,
dp(x,y) = dp(x-1,y) + dp(x-1,y-1);you cut a piece of 1 from prefix or you don't cutread this, just ordered the expression as in text
dp(x,y) = dp(x-1,y-1)+ dp(x-1,y) ;you cut a piece of 1 from prefix or you don't cutsomebody tell me how to solve B. as i m beginner .
Just do what the problem says.
... and check after each step if c<=0.
and see that capacity cannot be more than the maximum capacity
I solved A,B,D,E,F but was not able to solve C. Any hint would be appreciated.
Combinatorics
I know answer is (n-1)c11 but n! getting overflow..please help.
Check this
There is a dynamic programming approach to find the values of $$$n \choose r$$$. The recursive formula is: $$$\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$$$
can i get the code..please sir
It's pretty common. Just google it :)
Calculate nCr using dp.
Use __int128 — https://atcoder.jp/contests/abc185/submissions/18738450
One loop, No __int128: https://atcoder.jp/contests/abc185/submissions/18740485
You can use dp of 12*200 states where dp[i][j] means ith cut at jth position .
submission
it is just simple dp, you can either make a cut at the position you are at or not make a cut, when u have 11 cuts just check if it has made 12 valid cuts and there you go!
You can type the star and bars formular from head or implement some more or less simple dp. I tried to google the formular but without success then implemented the dp.
Number of ways to choose 11 from (L-1). Simple combinatorics.
Basic combinatoric problem just be careful about overflows you need to divide numbers while you are able to have integer result when calculating nCr
ans is l-1C11. We have total l-1 positions to place bar in which we have to choose only 11 of them.
If there is a rod of length L, we have 0,1,2,3,....,N-1,N positions on rod. So in total we have (n+1) points, but we can't make a cut at the extreme ends, so we just have to choose 11 points from the remaining (n-1) points, so answer is simply (n-1)Cr .
In order to avoid the overflow, I divided the numerator and denominator with their gcd in every iteration.
int ans = 1; int deno = 1; fo(i, 1, 11) deno *= i; rfo(i, n, n - 10) { ans *= i; int g = __gcd(ans, deno); ans /= g; deno /= g; }Infact I tried dp but my transitions were wrong
Can we solve problem F using SQRT decomposition? , although I solved using seg tree, still can we solve it?
With segment tree it is O(nlogn), with SQRT aka Mo it is O(n sqrt(n)), which is more, but still could fit the timebox.
F — Segment tree Implementation
Can someone help find the mistake ?
is
not
here
Approach to problem F: goto geeksforgeeks, copy the solution , edith for input and submit. Wooooow, you solve a problem with 600 point. Fuck.........
In problem B, the statement was not clear to me although I solved it after looking at the examples. It said that "Determine whether he can return home without the battery charge dropping to 0 on the way.", so it meant that we have to determine battery charge dropping to 0 when he was on his way home after visiting M cafes?
There are two possible interpretions: Drop to 0 at any point of time, or in the end. So that is what the examples for, we have to interpret them if not sure.
How to solve E using LCS? My logic is given below
common_elements = lcs(arr, brr, n, m); [calculate number of common elements using LCS]
y = min(n, m) — common_elements; [number of elements in the array which are not equal arr[i] != brr[i]]
x = max(n, m) — min(n, m); [numbers you have to delete to keep the length of both arrays same]
mySolution
For the test case :
According to the question, X would be 0 and Y would be 3. Your code would return 2 whereas the answer should be 3 or am I wrong in understanding this...?
I couldn't find the bug in my implementation of segment-tree for F. Please help. Code
I think you have not read the question properly. For every 1 type query, you have to replace a[x-1] with a[x-1]^y, not with y
I did so inside my update function. Inside the else if block Where I updated the tree node as
t[node]=combine(t[node],val);
finally found the bug. In update you have called left node both times.It should be right in second call
thanks a lot brother.
My solutions to this contest, along with detailed explanations of the solutions, can be found here :)
is there any way to get editorial of this contest as i am not able to solve 3 question and i want a editorial for it is there a way.
Check this.
I have written an unofficial English editorial.you can find it here..
thnks bro
can someone explain segment tree in PF?
Can anybode see what is wrong with my code?
My idea is to first swap the two array to make array $$$a$$$ has less or the same amount of numbers than array $$$b$$$. I observe that it is optimal to not deleting any number from array $$$a$$$, since if $$$|A'|<|A|$$$, we can add a number to the end of each of array $$$a$$$ and $$$b$$$ and $$$x$$$ will decrease by $$$2$$$ while $$$y$$$ will only increase by at most $$$1$$$. Then, I use DP, for which $$$dp[i][j]$$$ is the minimum value of $$$x+y$$$ when we only consider the first $$$i$$$ value of array $$$b$$$ and the first $$$j$$$ value of array $$$a$$$. The answer is $$$dp[m][n]$$$.
Check this test:
It's better to delete $$$A_1$$$, $$$A_2$$$ and $$$B_3$$$ with cost = 3. Your code prints 4.
Got it. Thanks!
I also have a similar kind of problem. Could you please help.
Code Link
I understood from your test case, that what I was doing wrong. But, can we not iterate over all the "subsequence sizes", and fix it ?
Just my feedback I think this round was way easier than general. Actually the difficulty of all A-F in this round should actually be A-D, and E, F need to be a bit challenging, requiring more thinking for beginners to be challenged. D was also quite easier more like B. Thank you!
Can someone help me with this solution of D https://atcoder.jp/contests/abc185/submissions/18768487 ?
How to approach D problem? Help me, please!
The idea is that we want to use the biggest k possible, because then we need to use the stamp the minimum number of times.
On the other hand, k must not be bigger than the smallest consecutive segments of blue cells, because if it is bigger, then we cannot use the stamp on that segment.
So we choose k as the number of cells in the smallest such segment, and then count foreach segment how often we need to use the stamp.
I got It. Thanks a lot, Mr. SpookyWooky ... God bless you!
Video editorial
Text editorial: English
Text editorial: Chinese
Would anyone be able to find the problem in my code for the segment tree in F? My implementation
It is being accepted on samples but failing in randomized tests. Thanks
You forgot to update arr after 1st type of operation. Just add
arr[x-1] ^= y;after updateTree and you'll get AC.Shouldn't I just be updating my segment tree array
treewith the update query? It's being updated at -tree[ind] = valin one of the updateTree cases. Moreover,val,that I passed was equal toarr[x-1]^yShouldn't that be enough? Thanks for looking into it...Consider two update queries for the same index
x. You passarr[x-1]^y, but then never updatearr[x-1], so in the next update query with this index, you will XOR with the initial value itself.If anyone is interested, I streamed my virtual participation & explained the solutions after, you can find the upload here: https://youtu.be/PLIm4jYYaHk
Why can I post the editorial? Is it a bug or a feature?
I was curious about the "post editorial" button and tried to write one, but it somehow appeared where the official editorial should be.
Should I delete it?
Upd: It seems that all users with a rating more than 2000 have the access.