### chokudai's blog

By chokudai, history, 2 years ago,

We will hold AtCoder Beginner Contest 180.

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

We are looking forward to your participation!

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 » 2 years ago, # |   +39 Wow that's an early announcement
•  » » 2 years ago, # ^ |   0 yessir
 » 2 years ago, # |   0 How to solve D?
•  » » 2 years ago, # ^ |   +3 The idea is that it will only be optimal to use Kakomon Gym (multiply by A) a few times, after which it will always be better to add instead of multiply. You can simulate the first part while x*a <= x+b, then use division to find out the number of operations of the second type to do.
•  » » » 2 years ago, # ^ |   0 Can you share your code?
•  » » » » 2 years ago, # ^ |   0
•  » » » » » 2 years ago, # ^ |   0 Any Idea why this submission gets WA?https://atcoder.jp/contests/abc180/submissions/17481021
•  » » » » » » 2 years ago, # ^ |   0 Because x*a overflow when x = 10^18 && a>10 something like that
•  » » » » » 2 years ago, # ^ |   0 can you please tell me why you subtract 1 from the answer(int ans) at the end??
•  » » » » » » 2 years ago, # ^ | ← Rev. 2 →   +8 The way the code is implemented has the 2 following cases: 1. x stopped right at y, i.e, x == y; 2. x exceeds y after the last increase by b operation, i.e, x > y.This is reflected in the else block inside the while loop. We add 1 to take the remaining y — (y — x) / b * b into consideration. So this add 1 makes x exceeding y. In both case, we must have 1 fewer operation to make x strictly smaller than y. So we subtract 1.
•  » » » » » 2 years ago, # ^ |   0 can you explain LLONG_MAX/x >= a why this was done in if condition?
•  » » » » » » 2 years ago, # ^ |   0 To prevent overflow. If that condition is not my, then x*a will overflow, which can lead to wrong answers and infinite loops.
•  » » » » » » » 2 years ago, # ^ |   +8 ohk thank you so much got it!!
 » 2 years ago, # |   0 How to solve B(only Euclidian distance part) and D ??
•  » » 2 years ago, # ^ |   +3 Sum up the squares of elements in a long long, then cast to long double and take square root.
•  » » » 2 years ago, # ^ |   +1 Can you please explain how to solve D and how to avoid overflow condition?
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   +1 while(a*x <= x+b && a*x < y) This condition was giving me TLE in one case but when I did this while(a*x <= x+b && a*x < y && a
•  » » » » » 2 years ago, # ^ |   +4 using a*x < y will result in integer overflow for larger cases, to avoid that u should use a < y/x, please see https://www.geeksforgeeks.org/check-integer-overflow-multiplication/
•  » » » 2 years ago, # ^ | ← Rev. 2 →   0 Can you explain why this works please? i.e why casting to long double works for B.
•  » » » » 2 years ago, # ^ |   0 Only cause is range !!
•  » » 2 years ago, # ^ |   0 The easier thing would be to use Python for questions like D, or Java's BigInteger.
•  » » » 2 years ago, # ^ |   0 or you can cast long long to (__int128) while comparing to prevent overflows
•  » » » » 2 years ago, # ^ |   0 I did using logs .. dunno if that was actually a good practice or not
•  » » » 2 years ago, # ^ |   0 you can use double in c++ also and convert to int final ans as i did my submission
 » 2 years ago, # |   +9 Solution for F, anyone?
•  » » 2 years ago, # ^ |   +29 Let DP[n][m][b] be the number of ways to build a graph with n labeled nodes and m unlabeled edges with b being a flag for restricting the size of the maximal connected component.From there at each iteration and to avoid double counting you only look at the components covering the first node.For example if I want a component of size 5 and n = 10 first I find the number of ways I can choose nodes to form this component which is 9 choose 4 (node #1 is already included). Then I find the number of way to build a path (5! / 2) or a cycle (4! / 2) and call the appropriate state.
•  » » » 2 years ago, # ^ |   +13 I didn't get how are you avoiding double counting?
•  » » » » 2 years ago, # ^ |   0 cause if you reverse a chain, the result will be the same as the origin, so there are k!/2 ways to form a chain with size k, or you will count a chain twice. as to circle, you should consider symmetry and shift, so there are k!/2/k = (k-1)!/2 ways to form a circle with size k.
•  » » » » » 2 years ago, # ^ |   +13 He is referring to the over counting of the total number of ways to make the connected components using DP, not the over counting of a particular component itself.
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   +5 sorry, i misunderstood it.for example, with node 1, 2, 3, 4 and 5, choose 1,2 to form a component, and then choose 3,4,5 to form a component. choose 3,4,5 to form a component, and then choose 1,2 to form a component. these 2 ways generate same graph, but if you count it more than once, then double counting happen.if you only look at the components covering the first node, you will only count the first way.
 » 2 years ago, # |   -19 How was E intended to be solved? I just copy pasted a DP code from stack overflow and calculated distances of graph[i][j] using the formula in the question and got AC.
•  » » 2 years ago, # ^ |   0 Travelling Salesman Problem using dynamic programming
•  » » » 2 years ago, # ^ |   -6 That's what I did, I am saying if it was meant to be solved in a simpler way.
•  » » » » 2 years ago, # ^ |   +1 Given that $n \le 17$, I guess the intended solution $O(n^{2} \times 2^{n})$ bitmask dp.
•  » » » » 2 years ago, # ^ |   +2 Well I guess we'll find out when(if) the editorial comes out
•  » » 2 years ago, # ^ |   +3 Well, I just honestly implemented dp for traveling salesman problem... I guess it was the intended solution
•  » » 2 years ago, # ^ | ← Rev. 2 →   +11 I guess the main difference in the problem was to visit at least once, while in standard TSP it is exactly once.So consider if the problem was a general directed graph, then the standard DP solution would not have worked, but since the shortest path between two vertices, in this case, is the same as their direct edge between them, the problem had the exact same solution.
•  » » » 2 years ago, # ^ |   +11 Even for the general graph, we can run floyd warshall first, then the classic TSP. It's a shame I took 50 minutes in this problem.
•  » » » » 2 years ago, # ^ |   0 I never thought they would straight away give classic tsp for a problem at "E". Man .. I took so long to implement floyd warshall version of it .. . Should have tested with the classic version first
•  » » » » 2 years ago, # ^ |   0 Can you elaborate on the Floyd-Warshall + TSP Solution for general graphs, or share any resource to learn from.
•  » » » » » 2 years ago, # ^ | ← Rev. 3 →   +3 This is what I did in my submission in contest,now you would have run through comments that direct edge is always optimal in this problem which i never observed so instead what i did was to first apply floyd warshall on the graph with saving a "next" Matrix in order to trace the path when i move from a node to other because when I would do so unintentionally I would also visit some unvisited cities(maybe or maybe not). Now after doing this i just do normal tsp and updating the mask. However i did see printing some paths and saw that direct edges are the shortest ones but thought they have intentionally given such samples.
•  » » 2 years ago, # ^ |   +11 if n<=20 its 95% of the time bitmask dp
•  » » 2 years ago, # ^ |   0 Can you share the link? I searched for such one, but did not found any, and was not able to implement a bugfree version.
•  » » » 2 years ago, # ^ |   0 https://stackoverflow.com/questions/33527127/dynamic-programming-approach-to-tsp-in-javaI think this is rather lengthy after looking at people having much shorter codes.
•  » » » 2 years ago, # ^ |   +6
•  » » » 2 years ago, # ^ |   0 Found the bug, typed r-c instead of max(0, r-c) in distance function :/
 » 2 years ago, # |   0 Was solution to E was intended to be solved using floyd warshall and then tsp, or did I do some overkill ?
•  » » 2 years ago, # ^ |   0 I don't think you need floyd warshall. Since the distance metric used always gives shortest path between two vertices.
•  » » » 2 years ago, # ^ |   0 So basically it was just classic tsp ?
•  » » » » 2 years ago, # ^ |   0 Yes. That's what I did. https://atcoder.jp/contests/abc180/submissions/17473708
•  » » 2 years ago, # ^ | ← Rev. 2 →   +14 I solved it using dp hint1use $dp(int \; lst, vector \; \; v, bool \; isGoingBack)$$lst$ means last city we visit$v[i]$ shows we visited city $i$ before$isGoingBack$ means : are we visited all and going back to 1 hint2use hash for v in dpmy code : Code#include using namespace std; const int N = 17, M = 272144; int x[N], y[N], z[N]; int dp[M][N]; int dis(int f, int s) { return abs(x[f] - x[s]) + abs(y[f] - y[s]) + max(0, z[s] - z[f]); } int solve(int lst, vector v, bool is_back = false) { if (is_back) { if (lst == 0) return 0; } int h = 0; for (int i = 0; i < v.size(); i++) { h *= 2; h += (int)v[i]; } h *= 2; h += (int)is_back; if (dp[h][lst] != -1) return dp[h][lst]; bool f = true; for (auto x : v) if (!x) f = false; if (lst == 0 && f) { return 0; } if (f) { for (int i = 0; i < v.size(); i++) { v[i] = false; } v[lst] = true; return solve(lst, v, true); } int ans = INT_MAX; for (int i = 0; i < v.size(); i++) { if (v[i]) continue; v[i] = true; int t = solve(i, v, is_back) + dis(lst, i); ans = min(ans, t); v[i] = false; } return dp[h][lst] = ans; } int32_t main(){ int n; cin >> n; for (int i = 0; i < n; i++) { cin >> x[i] >> y[i] >> z[i]; } for (int i = 0; i < M; i++) for (int j = 0; j < N; j++) dp[i][j] = -1; vector v; for (int i = 0; i < n; i++) v.push_back(false); v[0] = true; cout << solve(0, v); } 
 » 2 years ago, # | ← Rev. 2 →   -8 This is my solution to B problem, can anyone tell me why it's WA My codeEdit: After Changing everything to long long, it is working now, even long double didn't work, can anyone explain to me why long long is working and remaining are not.
•  » » 2 years ago, # ^ |   +3 Try long double instead of double, and sqrt instead of pow.
•  » » » 2 years ago, # ^ |   0 Tried now, but still shows WA, can you please help me out and point the mistake in this
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   0 You have to take max of absolute of arr[i] instead of just arr[i] to calculate 3rd distance. This is the same which I did and llost 700 ranks bcoz of thatEDIT: sorry I overlooked that part.
 » 2 years ago, # |   +13 How to solve Problem F? DP?
•  » » 2 years ago, # ^ |   0 Yes
 » 2 years ago, # | ← Rev. 3 →   0 Hi, could someone point out my mistake in BEdit: But x[i] is long long right?
 » 2 years ago, # |   0 Could someone help me.. i dont know why my dp runs incorrectly :< https://paste.ubuntu.com/p/nrYYHqvxw7/
 » 2 years ago, # | ← Rev. 2 →   0 TLE AC Solution after contest 4th problem why anyone?
•  » » 2 years ago, # ^ |   0 Because a * x will overflow for big cases, leading to a negative number and an endless condition
 » 2 years ago, # |   0 How to solve d please explain anyone?? i was thinking it was dp but other's solution just amazed me
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 Solution Just think about a short case and try to work on it. Keep a check on overflows.
 » 2 years ago, # |   0 Why this solution gets TLE while this gets AC? Using int instead of long long should result in wrong answer instead of TLE, then what other reasons behind it?
•  » » 2 years ago, # ^ |   0 For bigger values of n, i * i will overflow and will result in i * i <= n being true.
 » 2 years ago, # |   0 Please can somebody explain why using even unsigned long long is giving WA on D? I tried different values of x,y,a,b and checked but they were all running fine, but during contest ((ull)x*a<2e18) dosen't work but ((double)x*a) worked; why??
•  » » 2 years ago, # ^ |   0 consider x as 1e18 and a as 1e9. It will overflow range of unsigned long long
•  » » » 2 years ago, # ^ |   0 Thanks
 » 2 years ago, # |   0 How to solve E?
•  » » 2 years ago, # ^ |   0 Use dynamic programming with bitmask, dp[i][j] is the min cost of visting all cities specified by state i with the last visiting city as j. For a given dp[i][j], if we already know its result, we use it to make the next hop. For all valid hops from a city that has been included in i, to a city j from 0 to n — 1, update dp[next][j] where next is i | (1 << j). The final answer is dp[(1 << n) — 1][0].
•  » » » 2 years ago, # ^ |   0 Can you give me your AC code? Thank you.
•  » » » » 2 years ago, # ^ |   0
 » 2 years ago, # |   0 I have a problem with B. I am using long long and long double wherever needed and I am still getting compilation error. I think I am getting an error while using "abs(x)". Here's my code, plz help : problem B
•  » » 2 years ago, # ^ |   +3 There are a few problems:The error with abs function is due to ll being defined as unsigned.Then after this, you will be getting a WA because you have your cd as cd=max(cd,x) instead of cd=max(cd,abs(x))After that, you still are getting a WA because you are submitting problem B's solution for problem A :)
 » 2 years ago, # |   0 Hello guys, I need your help on my solution to the third problem. Here's a link to my submission, I kept getting TLE. How could I have optimized my solution? https://atcoder.jp/contests/abc180/submissions/17476685On the second problem, I was flat out getting Wrong Answer. What was wrong also? https://atcoder.jp/contests/abc180/submissions/17474677Thank you very much. PS; ABC 180 was my first CP contest, so also, any general tips for CP will do.
•  » » 2 years ago, # ^ |   0 For C problem,You are running a loop N times which is too slow as N can be upto 10^12.Here you need to run a loop till sqrt(N) and then it will work. Here is my submission:-https://atcoder.jp/contests/abc180/submissions/17445269To understand why this works you can see this tutorial:-https://www.youtube.com/watch?v=ga9N_ey4La4&list=PL2q4fbVm1Ik4liHX78IRslXzUr8z5QxsG&index=2
•  » » » 2 years ago, # ^ |   0 Still getting TLE, but a few more are reading AC https://atcoder.jp/contests/abc180/submissions/17578344
•  » » » » 2 years ago, # ^ |   0 Several things are wrong.Vector should be long long type.U should push j only if i != j which means if i is not the square root.And i and j should be long long type also.
 » 2 years ago, # |   0 A question: How much would be the corresponding rating of problems C, D and E on codeforces?