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### chokudai's blog

By chokudai, 9 months ago, We will hold AtCoder Beginner Contest 185.

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

We are looking forward to your participation! Comments (115)
 » 5 min left to start !
 » 9 months ago, # | ← Rev. 3 →   40 seconds left.....
 » 9 months ago, # | ← Rev. 2 →   .
 » Am I the only one who felt C and F were too classical problems? :/
•  » » no
•  » » » 9 months ago, # ^ | ← Rev. 2 →   Then, I think they should not include these kind of problems because they are just basic.
•  » » » » Then, I think they should not include these kind of problems because they are just basic. Then you need to learn the basics.
•  » » » » » 9 months ago, # ^ | ← Rev. 2 →   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.
•  » » » » » » I mean what's the point if they are only giving problems whose solutions are already known to people? 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?
•  » » » » » » » 9 months ago, # ^ | ← Rev. 2 →   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.
•  » » » » » » » » » I can agree with you for E. But for C and F no way man!
•  » » 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.
•  » » It's literally this same one.
•  » » 9 months ago, # ^ | ← Rev. 2 →   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?
•  » » 9 months ago, # ^ | ← Rev. 3 →   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) : transition 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 .
•  » » 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. ll solve(ll ar1[],ll ar2[],ll i,ll j,ll n,ll m) { if(i == n) { return m-j; } if(j == m) { return n-i; } if(dp[i][j] != -1ll) return dp[i][j]; if(ar1[i] == ar2[j]) { return dp[i][j] = solve(ar1,ar2,i+1,j+1,n,m); } return dp[i][j] = min({solve(ar1,ar2,i+1,j,n,m),solve(ar1,ar2,i,j+1,n,m),solve(ar1,ar2,i+1,j+1,n,m)}) + 1; } 
•  » » » Thanks for explanation !!
 » How to solve C ??I solved D but not C
•  » » 9 months ago, # ^ | ← Rev. 3 →   oh I tried the stars and bars technique. a simple ncr(n+r-1,r) gives the answer. basically n here is 12 and r is l-12.
•  » » can we see the test cases where we got wrong ans
•  » » Just calculate nCr(n-1,r-1).
•  » » 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/18756020Appreciate the help.
•  » » » » overflow. Try reducing the gcds first
•  » » » » naturell Check this outll nCr(ll n , ll r) { if(n(n-r);k--) { ans *= k; if(p<=r) { ans/=p; p++; } } return ans;}
•  » » C using recursion + memoization :https://atcoder.jp/contests/abc185/submissions/18763610
 » 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. CodeProblem 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 dp(x,y) = dp(x-1,y) + dp(x-1,y-1); // you cut a piece of 1 from prefix or you don't cut The base cases if(x==y) dp(x,y) = 1; // all pieces of 1 if(x
•  » » C is also l - 1 choose 11 because we have l - 1 choices and have to select 11.
•  » » Main problem in E is to get your head arround what the problem statement asks for.
•  » » 9 months ago, # ^ | ← Rev. 2 →   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 cut
•  » » For Problem D, I binary searched over [0, 10^9] to find the maximum stamp width, so as to get the minimum possible operations; but this solution got 'WA' for 14 test cases. Can't find the possible error, any help would be really appreciated.
 » somebody tell me how to solve B. as i m beginner .
•  » » 9 months ago, # ^ | ← Rev. 2 →   Just do what the problem says.
•  » » » 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
•  » » For every cafe calculate time taken from the last cafe and subtract if from n. If n<=0 at any stage we know there's not enough mah left. Also after cafe add (end time-start time) for charging. Also consider special cases like first cafe and last cafe. coden,m,t = map(int,input().split()) bank = n prev = 0 flag = 1 for i in range(m): a,b = map(int,input().split()) bank -= (a-prev) if bank<=0: flag = 0 bank += (b-a) bank = min(n,bank) prev = b bank -= (t - b) if flag: if bank<=0: print("No") else: print("Yes") else: print("No")
 » I solved A,B,D,E,F but was not able to solve C. Any hint would be appreciated.
•  » » Combinatorics
•  » » » » 9 months ago, # ^ | ← Rev. 2 →
•  » » » » 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 :)
•  » » » » » »
•  » » » »
•  » » » » Use __int128 — https://atcoder.jp/contests/abc185/submissions/18738450
•  » » » » Check this https://cp-algorithms.com/combinatorics/binomial-coefficients.html and use long long as data-type in order to overcome overflow.
•  » » » » 9 months ago, # ^ | ← Rev. 2 →   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
•  » » replace n with n-12 in this Reasonwe need at least 1 so we put 1 in all 12 parts
•  » » 9 months ago, # ^ | ← Rev. 2 →   Finding_Infinity For C, I simply calculated the ways I can select 11 positions out of L-1 positions.For example :-L = 141 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14'|' represent a possible point on the rod where I can cut it. As per questions we have to cut it at 11 positions (given L >= 12). so now the solution reduces to (L-1)C(11) [nCr is binomial coeff.] i.e. 13C11 = 78.L-1 because the number of positions will one less than the total lengthLink to my solutionIt's a pretty straight forward solution. Please ask questions if its still not clear to you.
•  » » ans is l-1C11. We have total l-1 positions to place bar in which we have to choose only 11 of them.
•  » » 9 months ago, # ^ | ← Rev. 3 →   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. Spoilerint 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.
»

This is my code for problem B but it was wrong for few test cases can anybody help me out

# include <bits/stdc++.h>

using namespace std; int main(){ #ifndef ONLINE_JUDGE freopen("input.text", "r", stdin); freopen("output.text", "w", stdout); #endif int n,m,t; cin>>n>>m>>t; int item = 0; for ( int i = 1; i<=m*2; i++){ int ele; cin>>ele; if ( i%2 == 0){ n = n + ( ele — item)*2; } else{ n = n — ( ele — item)*2; } if ( n<0){ n = 0; } item = ele;

    }
n = n-(t-ele)*2;
if ( n <= 0){
cout<<"No";

}
else{
cout<<"Yes";
}
return 0;


}

 » Can someone help find the mistake ?
•  » » 9 months ago, # ^ | ← Rev. 12 →   is if(S<=s && E>=e)return tree[id]; not if(S>=s && E<=e)return tree[id]; here
•  » » » 9 months ago, # ^ | ← Rev. 2 →   Thanks!
 » 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 belowcommon_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 : 5 5 1 0 9 2 4 1 2 8 7 4 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 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.
 » 9 months ago, # | ← Rev. 2 →   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?
•  » » it is a basic problem of segment tree , xor on segment tree if you ever use segment tree then its ok and if not then go to codeforces edu and watch some video of segment tree in segment tree section after watching it there is a problem of sum on segment tree if you can understand that then you can easily solve that problem, in that problem you have to do some update and queries. and by creating a segment tree you can solve it in O(nlogn). you can see the solution here : https://atcoder.jp/contests/abc185/submissions/18787613 and if it not clear then send a message.
 » 9 months ago, # | ← Rev. 2 →   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 $dp[m][n]$.
•  » » 9 months ago, # ^ | ← Rev. 2 →   Check this test: 4 3 1 1 2 3 2 3 1 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 LinkI 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!
 » 9 months ago, # | ← Rev. 3 →   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!
 »
 » Hey guys I made a screencast of this contest and also explained the solutions of all the problems except E. You can find that here :https://www.youtube.com/watch?v=ViFutg-bZJM
 » D was a very lame problem.
 » Would anyone be able to find the problem in my code for the segment tree in F? My implementationIt 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 tree with the update query? It's being updated at - tree[ind] = val in one of the updateTree cases. Moreover, val ,that I passed was equal to arr[x-1]^y Shouldn't that be enough? Thanks for looking into it...
•  » » » » Consider two update queries for the same index x. You pass arr[x-1]^y, but then never update arr[x-1], so in the next update query with this index, you will XOR with the initial value itself.
•  » » » » » That's awesome. Thanks man!
 » If anyone is interested, I streamed my virtual participation & explained the solutions after, you can find the upload here: https://youtu.be/PLIm4jYYaHk
 » 9 months ago, # | ← Rev. 3 →   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.
 » 9 months ago, # | ← Rev. 2 →   Hi, Why in problem E — sequence matching, the answer = abs(n-m) + min(n,m) — lcs(n,m) isn't correct? lcs(n,m) — longest common sequence of a and b. For example: a = (1, 3, 2, 4), b = (1 5 2 6 4 3). ans = 2 + 4 — 3 = 3. lcs(n,m) = (1 2 4).