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By awoo, history, 3 years ago, translation,

1569A - Balanced Substring

Idea: BledDest

Tutorial
Solution (awoo)

1569B - Chess Tournament

Idea: BledDest

Tutorial
Solution (Neon)

1569C - Jury Meeting

Idea: BledDest

Tutorial
Solution (Neon)

1569D - Inconvenient Pairs

Idea: BledDest

Tutorial
Solution (adedalic)

1569E - Playoff Restoration

Idea: BledDest

Tutorial
Solution (BledDest)

1569F - Palindromic Hamiltonian Path

Idea: BledDest

Tutorial
Solution (awoo)
• +108

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 » 3 years ago, # |   -48 Fastest editorial I've seen
•  » » 3 years ago, # ^ |   -59 Stop disliking him.
 » 3 years ago, # | ← Rev. 2 →   +14 Lesson learned: Don't use segment trees unless required.(My segment tree implementation for D gave TLE on system testing.)
•  » » 3 years ago, # ^ |   0 Interesting. I used segment trees and maps as well, and the highest runtime was ~1 second. My submission 128269298
 » 3 years ago, # |   -7 code and explanation for problem D SpoilerJust separate the points into two parts one where the x of the point matches with any of the x of vertical roads and other part where the y of the point matches with the y of any horizontal road. these two parts may or may not be disjoint as it is also possible that point's x as well as y matches with any vertical and horizontal road then this point will be the part of both parts.Now solve the answers for both the parts separately ->for 1st part Sort it using a comparater which sorts first according to y and then according to x. Now just iterate one by one on the points since the points are sorted using the above comparater so the points will be divided into groups which lie on the same horizontal line(same y) and this group is also sorted according to x also. Now we can see that the vertical roads divides the x axis into some segments. (x0,x1) (x1,x2) (x2,x3) ........for calculating the answer we will maintain one array f of size (n-1) where ith element represents the number of points that lie in the ith segment and whose y is also less than the current y(on which we are currently at in the iteration).f[i]=number of points which lie in the segment (x[i],x[i+1]) end points not included.for finding the segment-number of a point (a,b) we can easily find it using binary search(using lower_bound in c++)now let us assume we are at a point (a,b) so to find out the pairs that this point will make with other points that are already visited in the iteration we can make use of the array f.first we find the segment to which the point (a,b) lies and then the bad pairs which it makes is equal to f[segment-number]. because f[segment-number] stores the number of points that lie in that segment in the previous horizontal line (yb) then we need to add the point (a,b) to its respective segment(increment value of f[segment-number]) so that it can contribute the points which lie on the horizontal line above it.what we can do is we can maintain a cnt variable which stores the number of points which lie on the same horizontal line and in the same segment of x axis.When we move to next segment then we can just update the value of f[prev-segment-number]+=cnt.for 2nd part-Sort it using a comparater which sorts first according to x and then according to y. It's exactly same the difference is just the change of axis now we will iterate vertical line by line and in increasing order of x.Now the y axis is divided into segments.Link to my code- https://codeforces.com/contest/1569/submission/128325617
 » 3 years ago, # |   -7 I am facing some difficulty in understanding the equation in Problem C: $A_n^{n-k-1} = \frac{n!}{(k+1)!}$ What is $A$ here and how did we get $\frac{n!}{(k+1)!}$?
•  » » 3 years ago, # ^ |   +18 $A_x^y$ is the number of ordered ways to choose exactly $y$ different objects from $x$. So, it's like a binomial coefficient ${x}\choose{y}$, but with also considering the order in which we choose the object. Hence the formula for $A_x^y$ is ${{x}\choose{y}} \cdot y! = \frac{x!}{(x-y)!}$: there are ${x}\choose{y}$ ways to choose exactly $y$ objects out of $x$, and $y!$ ways to order them.
•  » » » 2 years ago, # ^ |   0 why to select n-k-1 from n we know which elements are n-k-1( n-k-1 elements are those which are not equal to ax or ax-1,where ax is the maximum element);
•  » » » » 2 years ago, # ^ |   0 It is like considering the positions; instead of the elements. There are in total n positions. We wish to pick n-k-1 positions among them for the n-k-1 elements. There will be k+1 gaps after that. The last gap is fixed for the max element. The rest of the k elements goes k! ways into the rest of k gaps. I was also confused at first. Hope this helps.
•  » » 3 years ago, # ^ |   0 nPr formula is n!/(n-r)! where nPr represents permutating r numbers in n places. It is basically n p (n-k-1).
•  » » » 2 years ago, # ^ |   0 is it true? nPr= rAn ? bcoz n should be greater than r.
 » 3 years ago, # |   -7 Problem C : if (cmx == 1) ans = (ans — sub + MOD) % MOD; I don't know why we need plus MOD. I thought we just need : ans = ( ans — sub ) % MOD Thanks for someone explain this (^^)
•  » » 3 years ago, # ^ |   0 It's because ans and sub are both %MOD, and if (ans%MOD — sub%MOD) < 0, so ( ans — sub ) % MOD will be < 0 too, that's because the % of negative numbers is also negative.for example, if MOD = 1e9+7, sub = 5 and ans = 1e9 + 8 (but in your code ans = 1, because you do ans%= MOD): ans = (1 — 5)%MOD = -4%MOD = -4But if you add MOD this never you be negative, because sub < MOD.I hope you understood and sorry for my bad english, I'm not fluent.
•  » » » 3 years ago, # ^ |   0 I get that you are doing this because ans-sub can be negative but how does MOD+(ans-sub) give the right answer when ans-sub is negative. Is this a property of MOD?
•  » » » » 3 years ago, # ^ |   0 The "taking value modulo MOD" operation basically tells you the distance between your number and the previous number that divides MOD. So in case of negative numbers (in range (-MOD; 0)) you need to find the distance between -MOD and your number. If you shift both values by adding MOD it becomes the distance between 0 and (your number + MOD) which is of course (your number + MOD).
•  » » 3 years ago, # ^ |   -7 By the way, answer can be simplified to n! * k / (k+1). That way you don't need to subtract modulo MOD :)
 » 3 years ago, # | ← Rev. 3 →   -7 In problem C my approach giving the wrong answer on test 2 but I couldn't understand why it is giving this error. My SolutionIf the maximum number in the array exists more than once we can simply print N!. Because they will go together to end of the discussion, However, if it exists only once there should be maximum-1 in front of the maximum. So we can find unwanted situations and we can subtract it from N!. We can replace other numbers except maximum and maximum-1. we can do this in:Permutation(N!,number_of_maximum + number_of_maximum-1) different ways.After we have replaced others, we can put the numbers with values maximums and maximum-1. If we don't have any maximum-1 in front of maximum it is an unwanted situation. Hence, we can replace them in:(number_of_maximum-1)! different ways.Because we can put the maximum in the end and we can permutate others. Codevoid solve(){ int n,maxi=0,fact=1; cin>>n; vector v(n); map mp; for(int i=0;i>v[i]; mp[v[i]]++; maxi = max(maxi,v[i]); } for(int i=1;i<=n;i++){ fact *= i; fact %= mod; } if(mp[maxi] != 1){ cout<
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 It's because of this int x = (fact(n) / fact(mp[maxi] + mp[maxi-1])) % mod;fact(n) is divisible by fact(mp[maxi] + mp[maxi-1] if you don't use mod, but with mod it doesn't, for example:16 is divisible by 8 but if i use mod 3 (16%3 = 1 and 8%3 = 2): 1 isn't divisible by 2 To fix this we can use module inverse, you can see it here or hereI hope you understood and sorry for my bad english.upt: I don't know if the rest of your solution is correct
•  » » » 3 years ago, # ^ |   0 Thanks for your answer I got it. I didn't consider errors about mods.
 » 3 years ago, # |   0 For problem A you can use two pointers. Spoiler int t; cin >> t; while (t--) { int n; string s; cin >> n >> s; int cur_a = 0, cur_b = 0; int l = 0, r = n - 1; string rel = ""; if (count(all(s), 'a') > 0 && count(all(s), 'b') > 0) { string k = s.substr(l, r - l + 1); int ca = count(all(k), 'a'), cb = count(all(k), 'b'); while (ca != cb) { if (ca > cb) { if (s[l] == 'a') { l++; } else { r--; } } else if (ca < cb) { if (s[l] == 'b') { l++; } else { r--; } } k = s.substr(l, r - l + 1); ca = count(all(k), 'a'), cb = count(all(k), 'b'); } cout << l + 1 << " " << r + 1; } else { cout << "-1 -1"; } cout << "\n"; } 
•  » » 3 years ago, # ^ |   +23 Is it worth ?
•  » » 3 years ago, # ^ |   0 There are 4 methods for this question. 1. O(n) — simple traversal as described in editorial. 2. O(n^2) — brute force — generating all subarrays 3. O(n) — using two pointers. 4. O(n) — using traversal + map used to store sum till particular index
•  » » » 3 years ago, # ^ |   0 What about finding count of 'a' and 'b' for each range using segment/fenwick trees ? Isn't that useful ?
 » 3 years ago, # |   0 I have a different implementation for E , I am traversing on 1 match per function call by finding the people who haven't lost yet . It was giving TLE until I figured out ( yes figured out , because I didn't Google it ) how to traverse on Only set bits of a number in a given range ( range of bits ) . My submission
 » 3 years ago, # |   +17 I think there exists a method for question D that is a little faster than the tutorial. The method used in the tutorial is to enumerate points, but I think it would be faster to enumerate edges. We divide the points into two groups, on x and on y, and discard the ones that satisfy both, since they cannot be composed. This way a pointer can be used to model the points that are between the two edges. Here the practice of my idea link
 » 3 years ago, # |   -10 Your codes for D/E/F are soo looooong
 » 3 years ago, # |   +1 Easy D approach -Count no of points between each adjacent vertical lines. Let this count be n. Total no of pairs will be n*(n-1)/2. Subtract those pairs which lie on same horizontal line. This can be easily done using map.Do the same for each adjacent horizontal lines128428875
 » 3 years ago, # |   0 Hello, Can anybody please explain what does (1<<(1<
•  » » 3 years ago, # ^ |   0 It means $2^{2^k}-1$. The << (bitwise shift) operator shifts the bits in the left number by the right number of times.
 » 3 years ago, # | ← Rev. 2 →   0 [deleted]
•  » » 3 years ago, # ^ |   0 Both $m$ and $n$ can be upto $2e5$. So yes, $O(mn)$ will exceed both memory and time limits.
 » 3 years ago, # |   0 Hi I was upsolving the C problem, can someone tell me why this wont pass the second test case? Thanks in advanced! my solution
•  » » 3 years ago, # ^ |   0 If you want to modulo division, you should use the inverse.
•  » » 3 years ago, # ^ |   0
 » 3 years ago, # |   0 Can someone help me with my solution for problem C. So I came up with the same idea as the solution that a nice permutation must have at least 1 pair of i, j that i < j, a[i] is the largest number of the array and a[j] = a[i] — 1. I will call the largest number M. So using that, I will choose 2 position for M and M-1 , so there will be n*(n-1)/2 ways to place M and M-1. Suppose there are k number equal to M-1, there will be k*n*(n-1)/2 such ways. The remaining element can be placed anywhere so there will be (n-2)! ways. So in total there will be k*n*(n-1)/2*(n-2)! pair of nice permutation. But this formula gives the wrong answer for the 4th example test. I haven't known what is wrong with my solution yet so I hope u guys will help me with it. Thanks in advance!
•  » » 12 months ago, # ^ |   0 I'm struggling with the same sitaution have you figured out what's wrong with the approach @Dyln
•  » » » 12 months ago, # ^ | ← Rev. 2 →   0 So when we use $C_n^2$ and multiply it with $(n-2)!$ it will count some cases multiple times because the positions of the second max elements are not fixed (ie. when the array is 1 1 2). Therefore, we have to fixed $k+1$ elements first and the max element will be placed anywhere but the last position, so our formula will be $A_n^{n-k-1}k!k$, which will simplify to $\frac{n!}{k+1}k$.
•  » » » » 12 months ago, # ^ |   0 Hey its still doubtful as by doing nC2 we ensure that the second (atleast) one is behind the max one. So how can it count multiple cases
•  » » » » » 12 months ago, # ^ |   0 So let take a look at the array i mentioned: 1 1 2. First we take the first 1 (1st index) and the max value, then pick two random positions for them, let assume they are 1 and 2, so our permutation will be 3 1 2. But when we consider the second 1 (2nd index) and the max value, if we take position 1 and 3 then our permutation will also be 3 1 2. This permutation is counted twice because we didnt fix the position of the 1s. Therefore, the original formula is wrong although intuitively, it seems right XD.
 » 3 years ago, # | ← Rev. 2 →   0 There is a typo in the last line of paragraph 5, 1569D. strret->street :D @awoo
 » 3 years ago, # |   0 In C, this line of code if (cmx == 1) ans = (ans - sub + MOD) % MOD; Why do we use mod like this: (.. + MOD % MOD)? We already calculated 'ans' and 'sub' under MOD, why do we use it again in the if statement?
•  » » 3 years ago, # ^ |   0 you need to +MOD because ans-sub may be a negative number , you need to %MOD because if ans-sub is non-negative , the former +MOD will give a value greater than ( or equal to ) MOD so you have to modulo again