### 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! Comments (79)
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 » Wow that's an early announcement
•  » » yessir
 » How to solve D?
•  » » 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.
•  » » » Can you share your code?
•  » » » »
•  » » » » » Any Idea why this submission gets WA?https://atcoder.jp/contests/abc180/submissions/17481021
•  » » » » » » Because x*a overflow when x = 10^18 && a>10 something like that
•  » » » » » can you please tell me why you subtract 1 from the answer(int ans) at the end??
•  » » » » » » 2 years ago, # ^ | ← Rev. 2 →   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.
•  » » » » » can you explain LLONG_MAX/x >= a why this was done in if condition?
•  » » » » » » To prevent overflow. If that condition is not my, then x*a will overflow, which can lead to wrong answers and infinite loops.
•  » » » » » » » ohk thank you so much got it!!
 » How to solve B(only Euclidian distance part) and D ??
•  » » Sum up the squares of elements in a long long, then cast to long double and take square root.
•  » » » Can you please explain how to solve D and how to avoid overflow condition?
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   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
•  » » » » » 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 →   Can you explain why this works please? i.e why casting to long double works for B.
•  » » » » Only cause is range !!
•  » » The easier thing would be to use Python for questions like D, or Java's BigInteger.
•  » » » or you can cast long long to (__int128) while comparing to prevent overflows
•  » » » » I did using logs .. dunno if that was actually a good practice or not
•  » » » you can use double in c++ also and convert to int final ans as i did my submission
 » Solution for F, anyone?
•  » » 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.
•  » » » I didn't get how are you avoiding double counting?
•  » » » » 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.
•  » » » » » 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 →   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.
 » 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.
•  » » Travelling Salesman Problem using dynamic programming
•  » » » That's what I did, I am saying if it was meant to be solved in a simpler way.
•  » » » » Given that $n \le 17$, I guess the intended solution $O(n^{2} \times 2^{n})$ bitmask dp.
•  » » » » Well I guess we'll find out when(if) the editorial comes out
•  » » Well, I just honestly implemented dp for traveling salesman problem... I guess it was the intended solution
•  » » 2 years ago, # ^ | ← Rev. 2 →   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.
•  » » » 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.
•  » » » » 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
•  » » » » Can you elaborate on the Floyd-Warshall + TSP Solution for general graphs, or share any resource to learn from.
•  » » » » » 2 years ago, # ^ | ← Rev. 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.
•  » » if n<=20 its 95% of the time bitmask dp
•  » » Can you share the link? I searched for such one, but did not found any, and was not able to implement a bugfree version.
•  » » » 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.
•  » » »
•  » » » Found the bug, typed r-c instead of max(0, r-c) in distance function :/
 » Was solution to E was intended to be solved using floyd warshall and then tsp, or did I do some overkill ?
•  » » I don't think you need floyd warshall. Since the distance metric used always gives shortest path between two vertices.
•  » » » So basically it was just classic tsp ?
•  » » » » Yes. That's what I did. https://atcoder.jp/contests/abc180/submissions/17473708
•  » » 2 years ago, # ^ | ← Rev. 2 →   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 = true; cout << solve(0, v); } 
 » 2 years ago, # | ← Rev. 2 →   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.
•  » » Try long double instead of double, and sqrt instead of pow.
•  » » » Tried now, but still shows WA, can you please help me out and point the mistake in this
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   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.
 » How to solve Problem F? DP?
•  » » Yes
 » 2 years ago, # | ← Rev. 3 →   Hi, could someone point out my mistake in BEdit: But x[i] is long long right?
 » Could someone help me.. i dont know why my dp runs incorrectly :< https://paste.ubuntu.com/p/nrYYHqvxw7/
 » 2 years ago, # | ← Rev. 2 →   TLE AC Solution after contest 4th problem why anyone?
•  » » Because a * x will overflow for big cases, leading to a negative number and an endless condition
 » 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 →   Solution Just think about a short case and try to work on it. Keep a check on overflows.
 » 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?
•  » » For bigger values of n, i * i will overflow and will result in i * i <= n being true.
 » 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??
•  » » consider x as 1e18 and a as 1e9. It will overflow range of unsigned long long
•  » » » Thanks
 » How to solve E?
•  » » 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].
•  » » » Can you give me your AC code? Thank you.
•  » » » »
 » 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
•  » » 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 :)
 » 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.
•  » » 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
•  » » » Still getting TLE, but a few more are reading AC https://atcoder.jp/contests/abc180/submissions/17578344
•  » » » » 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.
 » A question: How much would be the corresponding rating of problems C, D and E on codeforces?