### awoo's blog

By awoo, history, 5 years ago, translation,

1295A - Display The Number

Tutorial
Solution (BledDest)

1295B - Infinite Prefixes

Idea: Roms

Tutorial

1295C - Obtain The String

Idea: Roms

Tutorial
Solution (Roms)

1295D - Same GCDs

Tutorial

1295E - Permutation Separation

Tutorial

1295F - Good Contest

Idea: Roms

Tutorial
Solution (BledDest)
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 » 5 years ago, # |   +10 Finally, the editorial is out :D
•  » » 5 years ago, # ^ |   +1 O(n^3) solution for F 69782525
•  » » » 5 years ago, # ^ |   0 69879879 lagrang interpolation can be done in O(n). This is also a O(n^3) solution.
 » 5 years ago, # | ← Rev. 3 →   0 in F how are we dividing into segments, little confused after reading tourist solution, also could someone help me what is this another way(link) of solving increasing sequences question.Thanks in advance.
•  » » 5 years ago, # ^ |   0 We use the values of $l_i$ and $r_i + 1$ as left borders of the segments: we sort all of them, get rid of the unique values, and each value now defines the beginning of a new segment. So, for each segment we get, all values from it can be used at the same positions of the non-descending sequence we are trying to construct.
•  » » » 4 weeks ago, # ^ |   0 In fact,problem F is very similar to another problem in APIO 2016,and in that problem, $n\le 500$,it means you have to solve it with $O(n^3)$And the link of that problem on Luogu is https://www.luogu.com.cn/problem/P3643,I think you are supposed to see it.
 » 5 years ago, # |   0 Please,can anyone explain problem C,how you are calculating nxt table since I am not able to visualize it. thank you.
•  » » 5 years ago, # ^ | ← Rev. 2 →   0 If you are at index i in the string and want to calculate the smallest index of jth char between i and |s|, two possibilities occur. 1. s[i] is the jth character 2. s[i] is some other character than the jth characterFor the second case, s[i][j]=s[i+1][j] and for the first case s[i][j] is simply i
•  » » » 5 years ago, # ^ |   0 Think about it this wayFor each position we need the next position for all characters from 'a' — 'z' if they exist. So starting from back to front for the current position we get results from index in front and for the current index we can set next position for the character at index i to be index i, since that's closer.Hope this helps.
•  » » 5 years ago, # ^ | ← Rev. 6 →   0 case -1: if any char in t is not present in s. other wise: store the positions of a,b,c...z occurring in S, in a vector of size 26.now build z from t by taking one char at a time (starting from 0). now suppose you have build some part of z and you had just chosen a character which was at position 'cur' in s.now we want to add next char of t at the end of z — lets say the next char is x. hence we need the position of 'x' in s. if we think greedily we will chose that position which is just after our cur. if there is no x after cur. we can start from beginning and therefore setting cur=0, and number of steps ++.https://mirror.codeforces.com/contest/1295/submission/69747737
•  » » 5 years ago, # ^ |   0 Just maintain an array recording the nearest char C whose index is strictly bigger than the current index enumerated from n to 1
 » 5 years ago, # |   +7 What can be the possible divide and conquer solution for problem E? Thank you
•  » » 5 years ago, # ^ |   0 I think my point-update segment tree solution counts as divide-and-conquer, since it solves the problem for sublengths of the permutation and combines them: https://mirror.codeforces.com/blog/entry/73413#comment-576470
 » 5 years ago, # |   +14 Small correction: The combinatoric identity in F, the denominator of the last fraction is k and not k+1.
•  » » 5 years ago, # ^ |   +6 Will be fixed in a few minutes. Thank you!
 » 5 years ago, # |   +3 Is there any inclusion exclusion based solution to D?
•  » » 5 years ago, # ^ |   +13 Maybe this could help
•  » » 5 years ago, # ^ |   +5 Of course there is :) https://mirror.codeforces.com/contest/1295/submission/69760075
•  » » » 5 years ago, # ^ |   0 Sorry, can you please explain your idea? I understand the general idea but not the specifics. Thanks
•  » » » 4 years ago, # ^ |   0 can you please explain
•  » » 5 years ago, # ^ |   +3
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 can you plz plz explain atleast explain if(nf%2==1){ // wh(count/mul); rem=rem+count/mul; } else{ // wh(count/mul); rem=rem-count/mul; } } this
•  » » 5 years ago, # ^ |   0 There exists a combinatorial proof of Euler Phi Function formula that uses the inclusion-exclusion principle.
•  » » 4 years ago, # ^ | ← Rev. 2 →   +4 my stupid ass was not able to understand even the tutorial..... for all the stupid ones out there my friend Thallium54 have a blog https://thallium.space/题解/tutorial/2020/01/29/CF1295D
•  » » 4 years ago, # ^ |   +1 I also did it with Inclusion-Exclusion https://mirror.codeforces.com/contest/1295/submission/84912597
 » 5 years ago, # |   0 IN PROBLEM D HOW IS 0 GETTING MANAGED BY USING THE EULER TOILENT FUNCTION.EDITORIAL SAYS THAT GCD(0,m')=m'but 0 can be added to a and gcd(a,m) will be same as gcd(a+0,m);please help..Thanks in advance ..
•  » » 5 years ago, # ^ |   0 $x' = x' ' g$ and author shows why 0 is not valid for $x' '$. The case $gcd(a, m) = gcd(a + 0, m)$ is covered when $x' ' = a'$
•  » » » 4 years ago, # ^ |   0 i am not able to understand the tutorial for 1295D can you explain
•  » » » » 4 years ago, # ^ |   0 Let $gcd(a,m) = g$. Also Let $x' = x+a$.We want $gcd(x',m) = g$ which is equivalent to $gcd(\frac{x'}{g},\frac{m}{g}) = 1$So we need find number of $x'' = x'/g$ co-prime with $m' = m/g \in [1,m'-1]$ Since $x+a > 0$ This is the Euler Phi Function.
•  » » » » » 4 years ago, # ^ |   0 thanks finally got it
 » 5 years ago, # |   +6 Please explain problem E in simple words..
•  » » 5 years ago, # ^ |   +12 Which words do you not understand?
•  » » » 5 years ago, # ^ |   0 We are iteration on val or on POS and what we store on segment tree. And Finally this thing -- So what will happen if we increase val by 1? Let's define the position of pk=val as k. For each pos≥k we don't need to move pk to the second set anymore, so we should make t[pos]−=ak. On the other hand, for each pos
•  » » » » 5 years ago, # ^ | ← Rev. 2 →   0 Every node of segment tree stores the minimum cost if we select a segment [0,r] and we want all the elements between [0,r] lie between [0,val]. Initially assume val is -1. Therefore for every segment [0,r] we will have to shift its elements to the other segment (since no value lies between [0,val], i.e., [0,-1]). Now we will iterate over val. Now when value is 0, let the position of 0 in the original array be 'ind'. How to update the answer now?? For every segment [0,r] which contain the index 'ind', we do not need to shift val to the other segment (we shifted it when val was -1,). Thus for every segment [0,r] which contain 'ind' we will add (-a[ind]). And for every segment [0,r] which does not contain 'ind' we need to get it, therefor we will add (a[ind]). Hope you got it. Feel free to ask if you didn't understand anything.
•  » » » » » 5 years ago, # ^ |   0 We build segment tree on the given array and then check ind for every node Right? And why we iterate on Val not on pos.
•  » » » » » » 5 years ago, # ^ |   0 As we are sweeping according to val, which means we already know the answer for val, and we want to know the answer for val+1.
•  » » » » » » » 5 years ago, # ^ |   0 Why are we sweeping from 1 to n+1 ? shouldn't it be from 1 to n ? n+1 doesn't even exist in the permutation.
•  » » » » » » » » 5 years ago, # ^ |   +1 "All elements in the left set smaller than all elements in the right set" means that there is such value val that all elements from the first set less than val and all elements from the second set are more or equal to val. " Since all elements on the left set are less than (not less than equal to )val therefore sweeping till n+1, means left set will contain numbers from [1,n].
 » 5 years ago, # |   0 what is n in the editorial of Problem-B ? i mean what does n mean?
•  » » 5 years ago, # ^ |   0 $n$ is the length of the string.
 » 5 years ago, # |   0 C can also be solved using binary search, pretty easy to understand.69761627
•  » » 5 years ago, # ^ |   0 Hey, I didn't get your code.How you are applying a binary search. Can you explain in detail please.
•  » » » 5 years ago, # ^ |   0 Firstly we store the positions of all the appearance of each characters in string s. Then for each character in string t, we want the first appearance of that character after the position of the last character, which could be done with binary search.
•  » » » » 5 years ago, # ^ |   0 last character of what? string s or t ?
•  » » » » » 5 years ago, # ^ |   0 I mean the last character that has been obtained in string t. Sorry for my bad English.
 » 5 years ago, # | ← Rev. 5 →   +3 F says: we will count all the non-descending sequences but must be non-increasing, since pair (1, 2) is inversion according to taskUPD: sorry, must be descending with possible equal, since descending with possible equal != non-increasing (1, 3, 2) is not-increasing, but not a descending with possible equal
•  » » 5 years ago, # ^ |   0 descending means a[i+1]=a[i]
•  » » » 5 years ago, # ^ |   +3 yes, but we need the array to look something like this: 3 3 2 1 a[i] >= a[i+1] it's not a strict decrease, but not a non-decrease or non-increase
•  » » » » 5 years ago, # ^ | ← Rev. 2 →   0 a[i+1] is less than a[i]----------a[i+1]=a[i]. -------non-descendingIn this problem, we need to count the sequence a[] which satisfied:a[1]<=a[2]<=...<=a[n].And it means the sequcence a[] is a non-descending sequence.
•  » » 5 years ago, # ^ |   +3 Yeah I got confused by that; once I realized the issue, I just reversed the array after reading the input to get the right answer
 » 5 years ago, # |   +4 What programming language is the D solution written is? Is that just pseudocode?
•  » » 5 years ago, # ^ |   +12 Looks like Kotlin.
•  » » » 5 years ago, # ^ |   0 Sure, it is Kotlin!
 » 5 years ago, # | ← Rev. 2 →   +1 Any other solution for problem D? or more detailed explanation for the one mentioned in the tutorial?
•  » » 5 years ago, # ^ | ← Rev. 2 →   +21 1) gcd(a, m) is fixed -> let it be G2) Want to find gcd(a + x, m) = G, 0 <= x < m3) Same as gcd(X, m) = G, a <= X < a + m3.5) gcd(X, m) = gcd(X % m, m) (Think about euclidean algorithm)3.75) When you loop X from a to a + m — 1, X % m includes all possible outcome from 0 to m — 1.4) Same as gcd(Y, m) = G, 0 <= Y < m5) Same as gcd(Y / G, m / G) = 1, 0 <= Y / G < m / G (Floor division)6) So count number of Z such that gcd(Z, m / G) = 1 under m / G. That's the definition of euler's totient function.
•  » » » 5 years ago, # ^ |   0 How could you be sure that (in point 5) after divide all numbers (0 to m-1) by G there can be found φ(n) numbers of value those are coprime to m/G ??
•  » » » » 5 years ago, # ^ |   0 For gcd(a, b) = G, a = k1*G and b = k2*G where gcd(k1, k2) = 1So gcd(a / G, b / G) = 1 is a necessary and sufficient conditionDoes that help?
•  » » » » 5 years ago, # ^ | ← Rev. 3 →   0 What do you mean by $\phi(n)$? If you want number of numbers coprime to $m/G$, then you use $\phi(m/G)$. $\phi(n)$ gives number of numbers coprime to $n$, and smaller than $n$.
•  » » » 3 years ago, # ^ |   0 THANKS
•  » » 5 years ago, # ^ | ← Rev. 2 →   +2 I hope my blog would help a bit.
•  » » » 3 years ago, # ^ |   0 can you please re-upload the blog that explains problem D?It says the blog is no more existing . Thallium54
•  » » » » 3 years ago, # ^ |   0 Fixed
•  » » » » » 3 years ago, # ^ |   0 thanks
•  » » » » » 12 months ago, # ^ |   0 not working again.
 » 5 years ago, # |   +1 Can someone explain editorial for B?
•  » » 5 years ago, # ^ |   +1 To me it was easier to see it like this.Suppose the input was 011101000100, then you build the "balance" array of it, which is what the editorial mentioned: s: 0 1 1 1 0 1 0 0 0 1 0 0 b: 1 0 -1 -2 -1 -2 -1 0 1 0 1 2 The last number in the balance array, $bal(n)$, is going to be a "delta", and you can see that in the next iteration of the input the balance array will be the same, only delta units higher, like this: s: 0 1 1 1 0 1 0 0 0 1 0 0 b1: 1 0 -1 -2 -1 -2 -1 0 1 0 1 2 b2: 3 2 1 0 1 0 1 2 3 2 3 4 b3: 5 4 3 2 3 2 3 4 5 4 5 6 From here you can see that from each position $i$ the numbers you can reach are $bal(i) + \delta \times k$, where $\delta = bal(n)$, for some $k \geq 0$.You also need to be careful of two things. What if the delta is 0? then the balance array is going to repeat itself to infinity, so if the $x$ is in the initial balance, then the answer is -1, and if it's not it's 0. Lastly, how many times will $x$ appear? well for each position $i$, $x$ can be reached 0 or 1 times. If it can be reached more than once, that immediately means delta is 0, so the answer is -1. You can check if $x$ can be reached from an initial balance $bal(i)$ easily using modulo or the formula in the editorial.
•  » » » 5 years ago, # ^ |   0 Thanks for helping!
 » 5 years ago, # |   +4 "So let's make a sweep line on val from 1 to n+1 while trying to maintain all answers for each prefix pos". What does making a sweep line mean?
•  » » 5 years ago, # ^ | ← Rev. 4 →   0 It means we reconceptualize the $val$ dimension as physical time, and find a way to transition our program's state from $val = x$ to $val = x + 1$ (so like a line "sweeping" over that dimension over time). Working it out as in the editorial reveals that we want an array that we can do range additions and range minimums quickly, which can be implemented by a lazy-propagating segment tree.
 » 5 years ago, # |   0 Thanks guys for the editorial. By the way. Happy Lunar New Year !
 » 5 years ago, # |   +13 F says: ** Calculating the number of ways to take K elements from an interval [L,R) in sorted order can be reduced to calculating the number of ways to compose K as the sum of R-L non-negative summands (order matters). **, but must be the number of ways to compose R-L as the sum of k non-negative summands (order matters)
 » 5 years ago, # |   0 can anyone please optimize this code ? thanks in advance. this is my submission
 » 5 years ago, # |   +1 REGARDING QUESTION D can this question be solved by taking prime factors of m and then by inclusion and exclusion of factors of these prime numbers?
 » 5 years ago, # | ← Rev. 3 →   0 Can anyone help me to find out the bug of my solution? Problem: 1295E - Permutation Separation Code/** * “Experience is the name everyone gives to their mistakes.” – Oscar Wilde * * author : prodipdatta7 * created : **/ #include #include #include // #pragma GCC optimize("Ofast") // #pragma GCC target("avx,avx2,fma") // #pragma GCC optimize("unroll-loops") using namespace std ; using namespace __gnu_pbds ; #define ll long long #define ld long double #define ull unsigned long long #define pii pair #define pll pair #define vi vector #define vll vector #define vvi vector> #define debug(x) cerr << #x << " = " << x << '\n' ; #define rep(i,b,e) for(__typeof(e) i = (b) ; i != (e + 1) - 2 * ((b) > (e)) ; i += 1 - 2 * ((b) > (e))) #define all(x) x.begin() , x.end() #define rall(x) x.rbegin() , x.rend() #define sz(x) (int)x.size() #define ff first #define ss second #define pb push_back #define eb emplace_back #define mem(a) memset(a , 0 ,sizeof a) #define memn(a) memset(a , -1 ,sizeof a) #define fread freopen("input.txt","r",stdin) #define fwrite freopen("output.txt","w",stdout) #define TN typename typedef tree , rb_tree_tag, tree_order_statistics_node_update > ordered_set; typedef tree , rb_tree_tag, tree_order_statistics_node_update > ordered_multiset; /* Note : There is a problem with erase() function in ordered_multiset (for less_equal tag). lower_bound() works as upper_bound() and vice-versa. Be careful to use. i) find_by_order(k) : kth smallest element counting from 0 . ii) order_of_key(k) : number of elements strictly smaller than k. */ /*###############-> Input Section <-###############*/ template inline void Int(T &a) { bool minus = false; a = 0; char ch = getchar(); while (true) { if (ch == '-' or (ch >= '0' && ch <= '9')) break; ch = getchar(); } if (ch == '-') minus = true; else a = ch - '0'; while (true) { ch = getchar(); if (ch < '0' || ch > '9') break; a = a * 10 + (ch - '0'); } if (minus)a *= -1 ; } template < TN T, TN T1 > inline void Int(T &a, T1 &b) {Int(a), Int(b) ;} template < TN T, TN T1, TN T2 > inline void Int(T &a, T1 &b, T2 &c) {Int(a, b), Int(c) ;} template < TN T, TN T1, TN T2, TN T3 > inline void Int(T &a, T1 &b, T2 &c, T3 &d) {Int(a, b), Int(c, d) ;} template < TN T, TN T1, TN T2, TN T3, TN T4> inline void Int(T &a, T1 &b, T2 &c, T3 &d, T4 &e) {Int(a, b), Int(c, d, e) ;} /*###############-> Debugger <-###############*/ #define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator _it(_ss); err(_it, args); } void err(istream_iterator it) {cout << endl ;} template void err(istream_iterator it, T a, Args... args) { cerr << *it << " = " << a << ' ' ; err(++it, args...); } /*###############-> Constraints <-###############*/ const int N = (int)2e5 + 5 ; const int maxN = (int)1e6 + 6 ; const ll Mod = (ll)1e9 + 7 ; const int inf = (int)2e9 ; const ll Inf = (ll)1e18 ; const int mod = (int)1e9 + 7 ; inline int add(int a, int b, int mod) {a += b ; return a >= mod ? a - mod : a ;} inline int sub(int a, int b, int mod) {a -= b ; return a < 0 ? a + mod : a ;} inline int mul(int a, int b, int mod) {return (ll)a * b % mod ;} /*...... ! Code starts from here ! ......*/ int a[N], b[N] ; ll L[N], R[N] ; int main() { // ios_base::sync_with_stdio(false) ; // cin.tie(NULL) ; cout.tie(NULL) ; int test = 1 , tc = 0 ; //Int(test) ; while (test--) { int n ; Int(n) ; for(int i = 1 ; i <= n ; ++i){ Int(a[i]) ; } for(int i = 1 ; i <= n ; ++i){ int x ; Int(x) ; b[a[i]] = x ; } set < int > extra ; ll sum = 0 ; for(int i = 1 ; i <= n ; ++i){ while(sz(extra) > 0 and *extra.begin() <= i){ sum -= b[*extra.begin()] ; extra.erase(extra.begin()) ; } if(a[i] > i)sum += b[a[i]], extra.insert(a[i]) ; L[i] = sum ; } extra.clear() ; sum = 0 ; for(int i = n ; i >= 1 ; --i){ while(sz(extra) > 0 and *extra.rbegin() >= i){ sum -= b[*extra.rbegin()] ; extra.erase(*extra.rbegin()) ; } if(i > a[i])sum += b[a[i]], extra.insert(a[i]) ; R[i] = sum ; } ll res = min(b[a[1]], b[a[n]]) ; for(int i = 1 ; i < n ; ++i){ res = min(res, L[i] + R[i + 1]) ; } cout << res << '\n' ; } return 0 ; } ApproachFor every prefix of from 1 to i, I calculate the dollar needed to make this segment fully covered with only elements less or equal to i (stored this result to L[i]). The same process for every suffix of the array and stored the result to R[i]. Then for all possible partitions from 1 to n, I calculate the minimum one.
 » 5 years ago, # |   0 Can anybody help me with the following?"Otherwise for each such j there is at most one k satisfying bal(j) + k.bal(n) = x ?
•  » » 5 years ago, # ^ |   0 We look at the string in iterations. One iteration is one time copy the string s to the end of the whole infinite string.If the balance for the string s bal(n)!=0, then it is so that the balance at a position j is at most once equal to x in all iterations.This is so because the balance at a position j changes from iteration to iteration by bal(n), and because it changes, it can be equal to x only once.That one time is the k-th iteration. To find the answer, we count the positions j where such a k exists.
 » 5 years ago, # |   0 Can anybody help me figure out why my DP solution for E is wrong ? dp[i] means the minimum cost if I choose there is i elements in the first set. Considering array a is a permutation, so there is exactly [1, i] in the first set if it is "good", am I right? I wrote my solution based on that. Here is my submission. 70131818 Thanks!
•  » » 5 years ago, # ^ |   0 Sorry, never mind this, I figured it out already.
•  » » » 5 years ago, # ^ |   0 Can you explain where was the problem? My approach is similar with yours and I have the same wrong answer on 10th test case (69928822).
•  » » » » 5 years ago, # ^ |   +3 We miss many situations, if we split the array into [1, k] and [k+1, n] at first, the "good" array after we move the elements may not as the same length as the original(k in the first and n-k in the second, this is nay not the minimum). That is the point. So naive but right solution must be two loops, one is the pos to split, another is the length(or val).
 » 5 years ago, # |   0 Problem A is almost same with 774C - Maximum Number.
 » 5 years ago, # |   +3 Can anybody help me please: I cannot understand why in problem E we cannot compute L and R arrays as well as in prodipdatta7' solution in such way: Let's consider given permutation P and sequence 1,..., n — denote as N; current sum for L — denote as sum; set S. For i from 1 to n — 1: If k > i such that P(k) = i then: S.insert(i); sum += a(i) (- function a(x): cost of x in P) If P[i] belongs to S then: sum -= a(P[i]); S.erase(P[i]). Such approach fails on 10th test.
•  » » 5 years ago, # ^ |   0 I think that the following example can help:Input:42 4 1 38 6 9 7Correct output: 6Your output: 7The correct solution (total cost = 6):Step 1: Split the initial permutation into [2, 4, 1] and [3];Step 2: Move 4 from the first set to the second set (total cost = 6). The final sets are [2, 1] and [4, 3].Your solution (total cost = 7):Step 1: Split the initial permutation into [2, 4, 1] and [3];Step 2: Move 3 from the second set to the first set (total cost = 7). The final sets are [2, 4, 1, 3] and [].
•  » » » 5 years ago, # ^ |   0 Thank you very much !
•  » » 5 years ago, # ^ |   0 You're making the same mistake as https://mirror.codeforces.com/blog/entry/73467?#comment-578439
•  » » » 5 years ago, # ^ |   0 And you too !
•  » » » » 5 years ago, # ^ |   0 Yeah :')
 » 5 years ago, # |   0 in Problem D why m>1 ? I need explanation :( it's too hard
 » 4 years ago, # |   +5 E can also be solved using Ternary search.Solution link — here
 » 4 years ago, # |   0 C using Binary Search https://mirror.codeforces.com/contest/1295/submission/87246971
 » 4 years ago, # | ← Rev. 2 →   0 Problem D is just Insane. Since I found the editorial and all other comments quite tough to understand, here's my Detailed Beginner friendly approach. As Someone above said, this is what I tend to explain, a bit more clearly Perhaps.Step By Step Approach For D: Let gcd(a,m) = g. So, we can say a=g*k and m=g*l. Now Putting back a and m , gcd(g*k,g*l) = g. Obviously , we can say gcd(k,l)=1. Now looking closely, we need to find all x such that gcd(a+x,m)=g. means that in our answers, x should also be a multiple of g. So let x=y*g. Now, gcd(g*k + g*y , g*l) = g, or gcd(k+y , l) = 1. Now What could be the range of y ? See, we have x=y*g and xl, we can say temp = l*some_number + tmp%l . See, we can also say that gcd(temp%l,l)=1 . (See this step again). Now, we have established that we have a Range (of length l) [k....l.....k+l] to look for which of these have gcd()=1 with l, and for all those numbers z in this range greater l, we can simply replace them by z%l. So in the end, we have our Range as [1...l] only. Now to find all the numbers in [1,l] which are coprime with l, just use the Totient Function in O(sqrt(n) Code.
•  » » 13 days ago, # ^ |   0 Thank You for your explanation
 » 4 years ago, # |   0 I really liked Problem E
 » 8 months ago, # |   0 Explanation for D : https://blatherstrike.blogspot.com/2024/01/1295d-same-gcds.html