### chokudai's blog

By chokudai, history, 5 days 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 (97)
 » Wow that's an early announcement
•  » » yessir
•  » » I think I'll have to get that contest reminder app.
•  » » » thats y we have google calendar.
 » 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??
•  » » » » » » 4 days 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.
•  » » » » » » » oh! thanks, got it!!
•  » » » » » 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!!
•  » » » 4 days ago, # ^ | ← Rev. 4 →   idk why i thought of dp here may be coz i thought inorder to maintain maximum exp you gotta maintain minmimum str
•  » » » Can you prove why that strategy is optimal?
•  » » 4 days ago, # ^ | ← Rev. 2 →   Just brute force over all the steps,and calculate the maximum. By steps I mean, steps until you reach target by visiting 1st place some k times and visiting 2nd place to complete the remaining EXP. Check for overflows while dong so.Solution Sorry for my English.
•  » » » 4 days ago, # ^ | ← Rev. 2 →   My submission Can someone suggest any modification to my code for task D,I am still getting WAs in 2 test cases.
 » 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?
•  » » » » 4 days 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/
•  » » » » » Did the same mistake
•  » » » 3 days 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.
•  » » » » 3 days 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
•  » » It is a standard question of Dp with bitmasking.
•  » » Well, I just honestly implemented dp for traveling salesman problem... I guess it was the intended solution
•  » » 4 days 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.
•  » » » » » 4 days 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
•  » » 4 days 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); } 
 » 4 days 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
•  » » » » 4 days 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.
•  » » » » » Hey, but I did that right, as you can see, while taking the input of the numbers, if the number is negative, I have written -arr[i].
 » How to solve Problem F? DP?
•  » » Yes
 » 4 days ago, # | ← Rev. 2 →   how to solve C? UPDATE:got it.
•  » » find divisors of number in root(N) time, add it to a list ,sort the list and print it.
•  » » So, first, you have to find the numbers which are divisible by the given number N. For this, we can run a for loop from 1 to N, but you will notice that we shorten up this process, when you get a number i, which is divisible by N, you will also get a number N / i, which is also a divisor of N.when you need to essentially go up to sqrt(n) numbers, as you will cover everything divisor by that time. (Note that a square number wont 2 divisor at sqrt(n)) for(int i = 1; i * i <= n; i++) { if(n % i == 0) { cout << i << '\n'; if(i * i != n) cout << n / i << '\n'; } } But, now you need to print it in the ascending order, just take a vector, and push_back all the (N / i) numbers, print it in the reverse order. vector div; for(int i = 1; i * i <= n; i++) { if(n % i == 0) { cout << i << '\n'; if(i * i != n) div.push_back(n / i); } } for(int i = div.size()- 1; i >= 0; i--) cout << div[i] << '\n'; 
 » 4 days ago, # | ← Rev. 3 →   Hi, could someone point out my mistake in BEdit: But x[i] is long long right?
•  » » x[i]*x[i] may overflow type cast to long
 » Could someone help me.. i dont know why my dp runs incorrectly :< https://paste.ubuntu.com/p/nrYYHqvxw7/
 » 4 days ago, # | ← Rev. 2 →   TLE AC Solution after contest 4th problem why anyone?
 » How to solve C
•  » »
•  » » Count the number of divisors of a number.Here
 » How to solve d please explain anyone?? i was thinking it was dp but other's solution just amazed me
•  » » 4 days 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.
•  » » » Great! Thanks.
 » Hi, can you explain to me why i got WA in problem B? Both sample cases work;n = int(input()) x = list(map(int, input().split())) y = list(map(abs, x)) print(sum(y)) #1 res = [a ** 2 for a in x] j = sum(res) #2 print(math.sqrt(j)) print(max(x)) #3
•  » » print(max(y)) instead of print(max(x)) #3
•  » » » Thank you, it works!
 » 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.
•  » » » »
 » t = int(input()) li = list(map(int, input().split()))manhattan_distance = 0 for i in range(t): manhattan_distance = manhattan_distance + abs(li[i])print(manhattan_distance)import math euclidian_distance = 0 for i in range(t): euclidian_distance = euclidian_distance + (li[i]*li[i])print(math.sqrt(euclidian_distance))chebyshev_distance = max(li) print(chebyshev_distance)Would anyone debug it please?
•  » » li[i]*li[i] may overflow .type cast it into long.
 » 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 :)