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By awoo, history, 22 months ago, translation, 1511A - Review Site

Idea: BledDest

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1511B - GCD Length

Idea: BledDest

Tutorial
Solution (awoo)

1511C - Yet Another Card Deck

Idea: BledDest

Tutorial
Solution (Neon)

1511D - Min Cost String

Idea: BledDest

Tutorial
Solution (Neon)

1511E - Colorings and Dominoes

Idea: BledDest

Tutorial
Solution (BledDest)

1511F - Chainword

Idea: BledDest

Tutorial
Solution (awoo)

1511G - Chips on a Board

Idea: BledDest

Tutorial
Solution (BledDest)  Comments (73)
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 » solved 1st 3 ques very fast but got struck on 4th ques,btw nice contest :)
 » Can someone please explain the solution that's given in tutorial of problem D thanks. I have done a simple solution in that that prints a + ab + ac + .... + az and so on , but I want to understand how the editorial solution works for this
•  » » Can you explain your solution? I dont get it
•  » » What a solution !
•  » » My solution is the same as you and I quickly solve this problem. Amazing.
 » 22 months ago, # | ← Rev. 2 →   For 1511B — GCD Length it can be another easy approach cout<<(ll)pow(10, a-1)<<" "<<(ll)(pow(10, b-1)+pow(10, c-1))<<"\n";
•  » » Can u please explain this qn. it took lot of time and i didnt got soln and even not understood the editorial also
•  » » » First solve the case with $c = 1$. This is easy, we just need two coprime numbers of a certain length, which we can get in many ways. For example $10^{a-1}$ and $10^{b-1}+1$.Now note that given a solution to the problem $a,b,1$, we can transform it to a solution of $a+c-1, b+c-1, c$ by multiplying by $10^{c-1}$.
•  » » » » thanks , I also got the editorial now
•  » » » 22 months ago, # ^ | ← Rev. 2 →   I think a example can help you to understand Let , a = 4, b = 5, c = 3X be a number of length a & Y be a number of length bX can be obtained by many ways but an optimal value of X can be 10^(a-1) -> as example the X is 1000 (10 ^ (4 — 1))same goes for Y. Y is 10000 (10 ^ (5 — 1))Now here comes the GCD part. GCD of X, Y (1000, 10000) is 1000But g = 1000 is having a length of 4 that is not what we wanted.Same as X & Y, we can have the g. g = 10 ^ (3 — 1) = 100Our GCD has to be 100. We can have this GCD easily by adding it to Y. So, our new Y is Y + g = 10000 + 100 = 10100.Now, the GCD of X, Y = GCD(1000, 10100) will be 100, it is having the length of C. I guess this will help.
•  » » » » nice explanation thanks
 » 22 months ago, # | ← Rev. 2 →   My (faster) solution to G, using binary jumping/lifting/whatever: It's obvious that for range $[L, R]$ we want to consider a game of nim where a token at column $L\leq i\leq R$ corresponds to a pile of size $i-L$. Let $x[i][j]$ denote the xorsum of tokens in the range $[i, i+2^j)$, where the left border is $i$ (i.e, a token at $i+k$ contributes $k$ to the xorsum).If we know $x[i][j]$ for all $i, j$, the xorsum of $[L, R]$ can be computed as follows. (Below, $\oplus$ denotes xor) Initialize the xorsum $G = 0$. For each $k$ in descending order, if $L+2^k \leq R$, do: 1. $G = G \oplus x[i][k]$ 2. If there are an odd number of tokens in the range $[L+2^k, R]$, further do $G = G\oplus 2^k$ 3. $L = L+2^k$ At the end of this process, $G$ is exactly the xorsum we want, so we simply check whether it is $0$ or not.So, we want to be able to compute all the $x[i][j]$ fast — this can be done with a similar process: 1. $x[i] = 0$ for all $1\leq i\leq m$. 2. $x[i][j] = x[i][j-1] \oplus x[i+2^{j-1}][j-1]$ 3. If there are an odd number of tokens in $[i+2^{j-1}, i+2^j)$, further do $x[i][j] = x[i][j] \oplus 2^{j-1}$ Note that we need to be able to check whether some range contains an odd number of tokens, but that's easily done with prefix sums. Time complexity is $\mathcal{O}(mlogm)$ for the binary jumping precalculation, and then each query is answered in $\mathcal{O}(logm)$.My submission: 112843116
•  » » Hi! I didn't really get why you need $x[i][j]$. It's just a xorsum on a segment. Can't you just use prefix xors too to find it?My solution is also $\mathcal{O}(m \log m)$ but uses a different idea and actually needs $\mathcal{O}(m \log m)$ memory. But it seems like yours needs only prefix sums and xorsums.
•  » » » It's not quite the xorsum of a segment, more like the shifted xorsum of a segment. For example, if there are tokens are at columns $1, 3, 6, 14, 104$, I would have $x = (3-2)\oplus (6-2)\oplus (14-2) = 9$.I'm not quite sure how you would do the shifting with prefix xors.
•  » » » » Oh, I'm sorry. I misunderstood your explanation.
•  » » Nice solution!
•  » » Very nice solution!
•  » » Fantastic stuff, I did something similar using a segtree with offline queries.My memory complexity is linear, but total time complexity is $O(m \log^2 m)$.The idea is that you can maintain a xor-lazy segment tree, if you want to add/subtract $2^k$ from all stored elements, you can do that by issuing lazy xor-updates in groups of $2^k$. You can "visit" all subtractions from $1$ to $m$ recursively, starting from low values of $k$. For example, with $m=7$, you can have these additions/subtractions in order: $[-1, -2, -4, +4, +2, -4, +4, +1, -2, -4, +4, +2, -4, +4]$ and you will visit all numbers from $1$ to $7$. Notice that $2^k$ is added/subtracted exactly $2^{k+1}$ times, so there are $O(m \log m)$ segment tree updates.This idea was completely new to me when I first solved this problem.
 » BledDestForces
 » I believe that everyone wonders how top 1 risujiroh use O(n^2) brute force to solve G without being hacked. Can someone explain why this solution 112827572 run so fast:)
•  » » 22 months ago, # ^ | ← Rev. 2 →   That is basically because of two things:1. Using GNU C++17 (64) instead of the 32 bit version.2. The presence of these two lines at the top: #pragma GCC optimize("O3") #pragma GCC target("avx2")  Remove any of those and you get a TLE :) Edit: Why so many downvotes? I don't think I said anything wrong. If you don't believe me then try for yourselves.
•  » » » In this case it seems that you can make it even faster if you submit in C++14 but add the line: #pragma GCC optimize("unroll-loops") See here (it runs in around 4000ms instead of the 4700ms for the original). I'm not sure how exactly this whole thing with pragmas works, this was just an experimental discovery.Another thing that seems to contribute to it passing is the usage of cout << "AB"[x == 0]; Instead of something like cout << (x == 0?"B":"A"); See here. Again, not at all sure why this makes such a difference, but it does.Maybe someone can say why these things are the way they are?
 » Could someone pls tell the combinatorics / dp approach for E ? I saw in the announcements page but is very unclear.
•  » » 22 months ago, # ^ | ← Rev. 2 →   we can do it in the following way,first of all lets calculate what will be the answer if we would have 1 segment with length i that contains only 'o'.its can be done in the following waycnt_odd[i] = 2^(i-2) * cnt_odd[i - 2];dp[i] = dp[i - 1] *2 + cnt_odd[i -1]where cnt_odd[i] it is mean how much segments [0..i] that has sufix wich contains only '0' and has odd len. than for each consecutive segments of 'o' ans+=dp[len of that segment ] * (count of all possible сoloring of all other matrix).we should do it for horizontal and vertical segments. it's may become more clear if u look at code 112952569. Also i can explain better how we calculate cnt_odd or othere things if needed.P.S sorry for my pure english)
•  » » » 22 months ago, # ^ | ← Rev. 2 →   Yeah. Could you pls explain that (calculating count odd ad etc) ? Also how do you merge after calculating dp arrays horizontally and vertically ?
•  » » » » 22 months ago, # ^ | ← Rev. 4 →   first of all lets name strings that has sufix which contain only symbol 'o' of odd len good. cnt_odd = 1. becouse string "o" is good. As well as “**”cnt_odd = 1. becouse there is string "*o"."oo" has even len so its not good.let assume that we have string of len i that has last symbol equal to 'o'. Than to be good (i-1)-th symbol of that string should either be '*' than we dont care about others symbol so there can be all possible coloring 2^(i-1).Otherwise (i-1)-th symbol is 'o',so to be good string [0..i -2] should be good as well.so cnt_odd[i] = 2^(i-1) + cnt_odd[i -2]. Now about merging.We have some horizontal consecutive segments of 'o' and some vertical.Let solve them separately.we know that 1 consecutive segment of len i create dp[i] domino. however it will create dp[i] domino for every of possible coloring of the rest of matrix.For example we have segment "ooo" and "o" than answer will be 6 becouse segment "ooo" create 3 domino if second string colored red and 3 if second string colored blue.So 1 segment add to the answer len * 2^(count_of_white -len).I dont know how to explain why we can do separately for horizontal and vertically but i think that is not hard to undestand if u think a little.has it become clearer,or should i try to explain one more time? Do u also need an explanation how we calculate dp?
•  » » » » » Can you explain how you got the transition function for the dp i.e the relation dp[i] = dp[i - 1] *cnt_odd[i -1]
•  » » » » » » i have mistakenly write not correct formula in the post(in a code its correct). I am realy sorry for that. So correct formula is dp[i] = dp[i -1] * 2 + cnt_odd[i-1]. if we know an answer for segment of len (i — 1).Than let us take a look at a segment of len i.We can see that answer for dp[i] is at least 2-times bigger becouse we will add at least dp[i -1] if we would color i-th element in blue and add at least dp[i -1] if we would color i-th element in a red.However for each segment [0..i-1] that has sufix of only 'o' and its len is odd we will add one more domino. Has it become more clear or should i explain anything more? P.S take a look at code it's may become more clear 112952569
•  » » » » » » » Sorry , I could not understand the explanation . Can you explain this a bit more ( with a simple diagram , if possible ).
 » For problem E, can anyone tell the way to find the OEIS sequence (incase anyone else did by finding contributtion method).
 » 22 months ago, # | ← Rev. 2 →   In problem D how will we get a Euler cycle of length k*k + 1, do we also have to add self loop for that, because without that you cannot get an Euler cycle of k*k + 1, can someone explain, possible taking example of k = 3.
 » How to find eulerian cycle in 1511D - Min Cost String
 » can someone explain e more properly?using dp etc
 » I problem F I had slightly different approach. I tried to compute $DP[L][i][j][k]$ which would mean the number of pairs of chainwords, such that first of them has length $(L + i)$, second $(L + j)$ and last part of longer chainword comes from $k$-th word. We can compute it quite straightforward without using any kind of tree structures. All we have to check is whether some parts of words match. We just see that it is enough to store only $((d * (d - 1)) / 2) * n + d$ interesting states (though we don't have to use symmetry to pass tests).
 » 22 months ago, # | ← Rev. 3 →   Was anyone else confused by the legend of F, thinking it meant count the number chainwords that have resolved ambiguity? This isn't the case tho, right?
 » Is this rare way to solve B...? •  » » Lol I did the same thing, searching for prime numbers that big was a real hassle though :P
•  » » I also had a fun solution while preparing the task. Code#include #define forn(i, n) for (int i = 0; i < int(n); i++) using namespace std; mt19937 rnd(42); int main() { int t; scanf("%d", &t); vector pw(1, 1); forn(i, 9) pw.push_back(pw.back() * 10); forn(_, t){ int a, b, c; scanf("%d%d%d", &a, &b, &c); int z = rnd() % (pw[c] - pw[c - 1]) + pw[c - 1]; int la = (pw[a - 1] + z - 1) / z, ra = (pw[a] - 1) / z; int lb = (pw[b - 1] + z - 1) / z, rb = (pw[b] - 1) / z; while (true){ int x = rnd() % (ra - la + 1) + la; int y = rnd() % (rb - lb + 1) + lb; if (__gcd(x, y) == 1){ printf("%d %d\n", x * z, y * z); break; } } } return 0; } 
•  » » I did this too, but on my alt account.
 » " Now, let's suppose there are two strings i and j such that cnti−cntj≥2. Then, if we somehow reduce the number of occurrences of the string i by 1 and increase the number of occurrences of the string j by 1, the cost will decrease " What does that mean ?
•  » » "strings" here means pair of two characters like aa, ab, ac, etc.
•  » » » Actually the solution of many participants is easier & simpler than the editorial
 » Can someone help me to in problem d (min cost string)? What's the intuition behind the greedy approach. like thisTaken from submission (random)
 » Can someone explain how to produce your logic for questions similar to problem B during competitions? I solved C and D in the second hour but I couldn't solve B after wasting the first hour in virtual participation...
•  » » Let t = gcd(x, y)=> x = Aty = AtSo I choose t = 10^(c — 1), then choose A = 10^(a — 1 — (c — 1)), B = 10^(b — 1 — (c — 1)) + 1It can be proven that gcd(A, B) = 1, so gcd(x, y) = t
 » 22 months ago, # | ← Rev. 2 →   For problem F, I generate the input that requires 141 states. caseabcde bcdef cdefg defga efgab fgabc gabcd a And I believe this to be an upper bound. If there is no string that become other string prefix, the number of states will be at most 49. If there is a string that become other string prefix, the number of states will be at most 161 — 20 = 141
•  » » In the way I modeled the problem this case takes only 99 states, so I must be doing something different.I imagined the graph as a finite state machine where each accepting path is a way to construct a possible chainword. The states are pairs $(u, v)$ where $u$ and $v$ are proper (and possibly empty) suffixes of a word in the dictionary, and $u$ is a prefix of $v$, and they represent a look-ahead of the chainword we are building. The initial and only accepting state is $(\varepsilon, \varepsilon)$. Edges from $(u, v)$ to $(u', v')$ represent a way to add one character to the chainword, they fall in three cases: $u$ and $v$ are non-empty so $u'$ and $v'$ are equal to $u$ and $v$ resp. minus the first character in each. $u = \varepsilon$, so $u'$ is a word in the dictionary (again, minus its first character) to continue the chainword from $\varepsilon$. $u = v = \varepsilon$, so $u'$ and $v'$ represent two words in the dictionary that are prefixes from one another to start building a chainword. I believe this approach is very similar to using a trie but there must be a difference somewhere I am not seeing. Is it that I am merging some states into one?Code
•  » » » The writer's solution contains unnecessary states. For example, ("abcde", ""), ("abcde", "abc"), etc. The number of unnecessary states is 42, and your answer removes these states.
 » Can someone please explain me how the logic of D work?
•  » » It works on the concept Euler circuit. you make every letter up to k as a vertex in a graph and connect every vertex to every other one because we can pair every letter with every other one. after that you print the euler cycle of graph which will general all distinct string of size 2 using k letters. and after that you can just repeat this string. Euler circuit says that you can visit every edge of the graph only once. and every edge contains two letter so you will be visiting every substring of size 2 as the graph is complete graph. This is the concept of the problem. now you can implement it by Euler circuit and graph and all. But as I didn't know how to do that I did it using two loops.
•  » » » Thank you
•  » » » » Your Welcome :D
•  » » » » » Please explain your 2 loop solution?
•  » » » » » » first loop I'm starting with a to less than k and adding that aplhabet to string and in the jth loop I'm starting from i + 1, that is from the next letter which I took in i loop. for example: if in the ith loop I have added b to the string then the jth loop I will start from c and not from a because ba would already have occured in the previous iteration when i was a. for a, b and c first iteration i = 0 string a then in jth loop i'm starting from b and adding ab, ac .. so string will be a + ab + ac => aabacthe i will become 1 I will add b string => aabacbnow j will start from c , because you can see ba has already occured in previous iteration so string will be aabacb + bc and so on
 » Problem G can be solved with MO.
•  » » Can you explain how to shift the bounds (of MO, i.e. l, r) fast enough?
 » 22 months ago, # | ← Rev. 2 →   1511G - Chips on a Board Isn't complexity of solution from editorial should have log N outside of square root?
 » I want to share my solution of 1511G - Chips on a Board.All we need to do is to find XOR of $c_i - L$. For details "why?" look into editorial. I'll describe here how I calculate it. Notice that we can represent each number as two digit number XY in some base $2^k$. Then $c_i$ is some xy and L is XY. But this is the same as XOR all (y-Y) and (x-X) but some of them will turn into (x-X-1) when y < Y because of carry. My idea was to view it as points in table, where X digit is on X axis, and Y digit is on Y axis. Notice, that size of table is $2^{2k}$, so with appropriate k it is O(m) memory. Then we need to XOR all xy-XY. Then, to count how many times there is x among $c_i$ is just segment in column which can be calculated as prefix sums. And, to count how many times there is y among $c_i$ is just segment in row which can be calculated as prefix sums. But here is the problem: how can we tell should we subtract 1 for carry or shouldn't? This is most tricky part of whole solution. It turns out that we subtract 1 when y < Y, but it also can be calculated as prefix sum! Because we need to find out how many of x from column we should subtract 1 and how many we shouldn't but within single column x there are exactly numbers xy where v < Y we should subtract 1, and v >= Y we shouldn't, because carry won't happen. So, we can make two tables of prefix sums: for columns and rows. Each table has $2^k$ in both dimensions, and then for each X, and each Y we calculate prefix sum which is $O(2 \cdot 2^k)$ for each request. Overall complexity then is $O(n+m+q\cdot\sqrt{m})$ 113146022
 » can you explain gcd length question i didnt get it
 » 22 months ago, # | ← Rev. 2 →   Alternative solution for problem 1511G - Chips on a Board Consider each bit $b$ separately and keep a data structure that contains the remainder of the position of the coins in a suffix, modulo $2^{b+1}$. It should handle these operations: range sum on $[2^b, 2^{b+1} - 1]$ (get number of coins with that bit turned on) point update on $0$ (add a new coin) add $1$ to all elements (add a column on the left) So you can use a Fenwick tree with Venice trick for each bit. Total complexity: $O(n \log^2(n))$. 113692737
 » I have an intutive approach but I want to know why he work??I want proof?? suppose we will use the first 5 character.. first: we will put the the characters at even position like that: a_b_c_d_e_a_b_c_d_e_a_b_c_d_e  the the first k chars at even positions we will shift them by 1 and put the shifted chars at odd positions like that a_b_c_d_e _b_c_d_e_a abbccddeea  the second k chars at even positions we will shift them by 2 and put the shifted chars at odd positions like that a_b_c_d_e _c_d_e_a_b acbdcedaeb  you can say that the char at odd position j is equal to the char at position (j-1) plus number of shifts like s[j]=(s[j-1]+sh)%k; sorry for my bad english...can someone prove it?? here's my submision