### awoo's blog

By awoo, history, 17 months 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

1569E - Playoff Restoration

Idea: BledDest

Tutorial
Solution (BledDest)

1569F - Palindromic Hamiltonian Path

Idea: BledDest

Tutorial
Solution (awoo)  Comments (39)
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 » Fastest editorial I've seen
•  » » Stop disliking him.
 » 17 months ago, # | ← Rev. 2 →   Lesson learned: Don't use segment trees unless required.(My segment tree implementation for D gave TLE on system testing.)
•  » » Interesting. I used segment trees and maps as well, and the highest runtime was ~1 second. My submission 128269298
 » 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
 » 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 $\frac{n!}{(k+1)!}$?
•  » » $A_x^y$ 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.
•  » » » what is the error in the bellow analogy of problem Clet the largest element be x and second largest element be y = x-1; if we fix the first position for x and and fix any position for y after it ,then we have (n-2) remaining element. Thus for this elements we have permutation (n-2)! and for y we can have (n-1) different position which are after x. so in total we have (n-2)!*(n-1) different good permutation when x is in the first position . and when x is in the second position then we have (n-2)!*(n-2) different permutation. So the math goes like this: (n-2)!*(n-3) , (n-2)!*(n-4) and so on.if we do the math : (n-2)!*(n-1) + (n-2)!*(n-2) + .... + (n-2)! = (n-2)!* ( (n-1)*n/2) but surely it isnt give the correct answer.
•  » » » » I thought of a similar line but there's a problem with that. First, there can be multiple occurrences of y as well so we cannot simply write (n-2)! for the remaining. Secondly, there can be multiple occurrences of x as well and in that case, the answer will be n!. I tried handling these 2 cases but it became very complex. The editorial's approach is quite simpler than that.
•  » » » » » In this question every element is distinct right because each number is given has a judge id?
•  » » » » why does the above approach fails ?
•  » » » 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);
•  » » » » 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.
 » 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 (^^)
•  » » 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.
•  » » » 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?
•  » » » » 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).
•  » » By the way, answer can be simplified to n! * k / (k+1). That way you don't need to subtract modulo MOD :)
 » 17 months ago, # | ← Rev. 3 →   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<
•  » » 17 months ago, # ^ | ← Rev. 2 →   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
•  » » » Thanks for your answer I got it. I didn't consider errors about mods.
 » 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"; } 
•  » » Is it worth ?
•  » » 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
•  » » » What about finding count of 'a' and 'b' for each range using segment/fenwick trees ? Isn't that useful ?
 » 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
 » 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
 » Your codes for D/E/F are soo looooong
 » 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
 » Hello, Can anybody please explain what does (1<<(1<
•  » » It means $2^{2^k}-1$. The << (bitwise shift) operator shifts the bits in the left number by the right number of times.
 » 17 months ago, # | ← Rev. 2 →   [deleted]
•  » » Both $m$ and $n$ can be upto $2e5$. So yes, $O(mn)$ will exceed both memory and time limits.
 » Hi I was upsolving the C problem, can someone tell me why this wont pass the second test case? Thanks in advanced! my solution
•  » » If you want to modulo division, you should use the inverse.
•  » »
 » 16 months ago, # | ← Rev. 2 →   There is a typo in the last line of paragraph 5, 1569D. strret->street :D @awoo
 » 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?
•  » » 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