### maroonrk's blog

By maroonrk, history, 13 months ago, We will hold AtCoder Regular Contest 123.

The point values will be 300-400-600-700-700-1000.

We are looking forward to your participation! Comments (65)
 » Hope for a good contest!
 » Front seat!
 » Ah, It's too hard! :(
 » For the first time I solve three problems in ARC.I'm so happy :)
•  » » 13 months ago, # ^ | ← Rev. 2 →   How did yo solve it? Can you explain your idea?
•  » » » I think my solution is not so good.You can read the official editorial now.
•  » » How did you solve $C$? Please tell.
 » » How to solve C?
•  » » Indeed how to solve $C$. Greedy algorithms like taking largest possible number fail. I felt it was kind of similar to yesterday's A
•  » » Digit dp-esque dp worked for me
•  » » » colud you elaborate ?
•  » » » » First, use the fact that the answer should be maximum 10 (or even lower couldn't bother to prove a tighter bound), and then we should think about how each of $a_i$ contributes to each digit of n. A length k number will contribute k digits of n from the first digit to the kth digit once, so instead of figuring out what the individual $a_i$ could be, we could shift our focus to how many numbers focus on a specific digit of n. Let $b_i$ denote the number of $a_i$ that contributes to the ith lowest digit. Then, it should be clear that $b_i$ is non_increasing similar to a cumulative frequency. Now, we just want to know if it is possible to create n from a sequence of $b_i$ that is where digit dp comes in. The first parameter of the dp is the current position starting from the lowest digit, and the second is $b_i$ at that position and finally, the last parameter is how much we carried over from the last digit.
•  » » » » » Amazing use of digit dp!
•  » » » » » I didn't understand why you use array f in your solution you did not update array f.
•  » » » » » » I mean I did use the array to store the dp results in the line int &res = f[pos][choice][up]? Next time when you comment please read the code carefully and then comment.
•  » » dp digit
•  » » » I have a greedy solution, but I can't prove it :https://atcoder.jp/contests/arc123/submissions/24372393
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me too but i can't solve the num like 91 

# include

using namespace std;

typedef long long ll;

int main() { int t; cin >> t; while (t--) { ll n; cin >> n; int a = { 0 }, pos = 0; while (n) { a[++pos] = n % 10; n /= 10; } int ans = 0; for (int i = 1; i <= pos; i++) { if (a[i] == 0)ans = max(ans, 4),a[i+1]--; ans = max(ans, a[i] / 3 + (a[i] % 3 ? 1 : 0)); } cout << ans << "\n"; }

return 0;


} 

•  » » » How did you learn digit DP? I have tried learning it a couple of times already but have never been able to understand it. What kind of prerequisite/background does one need for really understanding it? I currently can only solve classic DP problems.
•  » » » » I had to learn after I couldn't solve this question, yet I couldn't solve C
•  » » » » I learned it from this blog post Link:. The explanation is great and also has a handful of problems which you can solve to practice digit dp.
•  » » 13 months ago, # ^ | ← Rev. 3 →   for a given k can we do it? if we can answer this then its easy just do k++ until we get one. It will never go too far. Now lets try to answer the first part that for a given k can we do it? Try to think of a dp state. run a loop from right to left digits one by one in n. ( i=1 to len(n)) i=1 corresponding to the rightmost digit.we will pass three variables after every iteration i. lowest and highest carry, and mx_k that is the maximum no. of i-1 digit numbers possible up to this point. and lowest and highest carry should correspond to mx_k.tough to explain the mx_k part. Give a good read and think you will get what I meant to say.my solution
•  » » » lovely explanation, thanks!!
•  » » 13 months ago, # ^ | ← Rev. 2 →   Another solution with dfs: The restriction that no digit can be 0 is troublesome, so I considered to clear out this. We minus n with numbers like 111111, 1111 first, and then the digits can be 0,1,2, which is a greedy problem similar to 1530A. Thus it can be proved that the answer is not greater than 5, and we are able to solve it by dfs. For each number, we only need to fix its length, and minus n with 11..1 with the same length. In the end check whether n (after minus) can be constructed by numbers which digits are 0 or 1 or 2. With a little amount of optimization, the solution was accepted.
•  » » »
•  » » » Another implement https://atcoder.jp/contests/arc123/submissions/24377005
 » how to solve D? there's no editorial for it
•  » » 13 months ago, # ^ | ← Rev. 3 →   If $a_i > a_{i+1}$, it's optimal to choose $b_i = b_{i+1}$, else it's optimal to choose $c_i = c_{i+1}$. Then, the $b_i$ and the $c_i$ can be uniquely determined for a fixed $b_1$ and they have the form $\mid b_1 - k \mid$ ($k$ is different for each $b_i$ and $c_i$). The optimal $b_1$ is the median of the $2n$ values of $k$.
•  » » 13 months ago, # ^ | ← Rev. 2 →   ternary search on the value of $B_1$, knowing that $B_i=B_{i-1}+max(A_i-A_{i-1},0)$ and C as the complement of B
•  » » » Can you share your submission? My ternary search fails for some reason. Submission
•  » » » »
•  » » » » I think you make an overflow in Calc, sum can be up to 1e15*2e5 which is to big
•  » » » » » 13 months ago, # ^ | ← Rev. 2 →   Thanks. The actual bug was that minimum sum can be over 1e18. So I have to take initial value as LLONG_MAX. Also as you pointed out in the ternary search, 1e15 will overflow. So taking low as -1e13 and high as 1e13 now worked for me. Even 1e12 is bad.
•  » » It's the first editorial in E. Seem like the author put it in the wrong place.
•  » » 13 months ago, # ^ | ← Rev. 3 →   One of the (almost) middle elements can be 0After fix we do simple greedy to left and rightarc123/submissions/24365978
 » Err......Starting from C, the problems are too hard for me:(I have to say, contests in AtCoder are becoming more and more challanging. Is that because the problem setters are stronger and stronger or I'm weaker and weaker?
•  » » I think AtCoder is turning harder and harder.
•  » » All the latest ARCs are mostly hard, you also need to get used to the ARC problems, on my first ARCs I always had a bad performance and the problems seem quite difficult, however after some contests, you will find the problems more accessible (excluding all F's, and some E's, and the other problems are still hard, but at least you can elaborate more on how to solve them). They are really fun and you can learn and enjoy a lot. I am really grateful to all atcoder writers, coordinators, and staff, the contests are great.
 » Why number of contestant is so small than number of contestant at any begginer contest !!
 » The approach of problem D is similar with 1406D - Three Sequences, 1442A - Extreme Subtraction.
 » How large could the answer be for problem D? I thought the answer could go up to $10^{18}$ which was a terrible mistake as I got 7 WA because of it and changing the limit to LLONG_MAX just solved the issue.
 » 13 months ago, # | ← Rev. 5 →   For A the approach I followed is,Let mean of consecutive differences in the array be dd;So I iterated on differences from (dd — epsilon) to (dd + epsilon) (epsilon can be anything I considered 2) and checked all possible arrays for a particular difference and took the array with minimum operations. It got accepted but I cannot prove it. Can anyone help me with the proof?Here's my submission Task AThanks in advance :)
 » SpoilerARC likes floor_sum
•  » » Spoilerwhile I hate :(
 » can anyone help me with understanding the editorial of problem A.i am having difficulty in understanding the answer part "Let k be the number of times we add 2 to X. Then, it is necessary that k≧0 and X+2k≧0, which can be represented as k≧k0 where k0=max(0,⌈(−X)/2⌉) If we fix k≧k0 , we need to do X+2k additions of −1, for a total of X+3k operations. In order to minimize this number, we should minimize k, which means the answer is X+3k0. The time complexity of our approach is Θ(1)." Here what's the need of taking k0... why do we need to do x+2k additions of -1, and how did the total operations became as x+3k...and should i have prerequisite of any topic to understand this editorial!!
•  » » the sequence [a,b,c] is an arithmetic one if a-b == c-b so 2b — a — c must equal 0 let x = 2b — a — c and we will make x equal to 0 by two operations : 1. increase x by 2 (same to add 1 to b in the original problem) 2. decrease x by 1 (same to add 1 to a or c ) now there are 2 cases : x > 0 then the only way to make it 0 is decreasing it by one x times (so we need x operations). x < 0 then we will keep adding 2 until x becomes positive then back to the case 1 this case requires Ceil( x / 2 ) operations.
•  » » » You didn't explain $x+3k$ logic.
•  » » » » 13 months ago, # ^ | ← Rev. 3 →   Let's explain two examples here- $a={4 ,8 ,10}$ $x = 2*a-a-a = 2.$So we need to increase either $a$ or $a$ total of $2$. Here $x$ is the answer cause both odd and even number can be formed with $a$ and $a$.But when $a={10 ,3 ,5}$ $x = 2*a-a-a = 6-15=-9.$$As$ $x$ is odd we can not increase $a$ to any odd number (as it's a multiple of $2$).So we need to increase $min(a,a)$ and $a$ by $1$ after adding the $x/2$.For the above example array would become $a={10,7,5}$$after$ adding $x/2$ to the $a$ and Finally the answer array would be $a={10,8,6}$ code...  cin>>a>>b>>c; x=2*b-a-c; if(x>=0){ cout<
•  » » » thanks for the explaination..it was really helpful
 » How to solve B?
•  » » 13 months ago, # ^ | ← Rev. 2 →   First sort all the three input arrays(A,B,C) and then count for valid triplets such that the upper bound of first element from the first array A uniquely exists in second array B and upper bound of that element from second array B uniquely exists in third array C. Codesort( a.begin(), a.end() ); sort( b.begin(), b.end() ); sort( c.begin(), c.end() ); int count = 0; for(int i = 0; i < n; i++){ auto itr1 = upper_bound( b.begin(), b.end(), a[i] ); auto itr2 = upper_bound( c.begin(), c.end(), *itr1 ); if ( itr1 == b.end() || itr2 == c.end() ){ break; } else{ count += 1; b.erase(itr1); c.erase(itr2); } } cout << count << "\n"; 
•  » » » 13 months ago, # ^ | ← Rev. 2 →   thanks nice explanation :)
•  » » » » 13 months ago, # ^ | ← Rev. 4 →   It's surely TLE.How can you expect erase function to work in $log(n)$ complexity. I used ordered set to erase the elements whose complexity is $log(n)$ unlike vectors. code... #include #include #include using namespace __gnu_pbds; using namespace std; #define ordered_multiset tree, rb_tree_tag,tree_order_statistics_node_update> #define ordered_set tree, rb_tree_tag,tree_order_statistics_node_update> #define int long long int void Malena(){ int n,x;cin>>n; ordered_multiset os,os1,os2; for(int i=0;i>x,os.insert(x); } for(int i=0;i>x,os1.insert(x); } for(int i=0;i>x,os2.insert(x); } auto Upper_Bound1=[&](int element)->int{ int pos=os1.order_of_key(element+1); if(pos==os1.size()) return -1; else{ return *(os1.find_by_order(pos)); } }; auto Upper_Bound2=[&](int element)->int{ int pos=os2.order_of_key(element+1); if(pos==os2.size()) return -1; else{ return *(os2.find_by_order(pos)); } }; int cnt=0; for(auto ele:os){ int a=Upper_Bound1(ele); int b=Upper_Bound2(a); if(a!=-1 and b!=-1){ cnt++; os1.erase(os1.find_by_order(os1.order_of_key(a))); os2.erase(os2.find_by_order(os2.order_of_key(b))); } } cout<>t; while(t--){ Malena(); } return 0; } 
•  » » » » » 13 months ago, # ^ | ← Rev. 2 →   I mean all self-balancing binary search trees, such as the std::(multi)set (red-black tree) or avl trees, allow deletion in $O(log(n))$ complexity due to its self balancing nature. For this specific problem, I used std::multiset which worked under 150ms.
•  » » » » » 13 months ago, # ^ | ← Rev. 4 →   yeah you are absolutely correct , i also got tle but then i use unordered_map
•  » » 13 months ago, # ^ | ← Rev. 2 →   I used 3 pointers approach.Sort all the three arrays, start from the beginning. If x[i] < x[j], then there are two possibilities i.e x[j] < x[k] then we get the valid triplet so increment the answer and all the pointers by 1, otherwise if x[j] > x[k], then we'll have to pickup a bigger element from the 3rd array so we increment k by 1. Similarly if x[i] > x[j], we'll have to pickup a bigger element from the 2nd array so we increment j by 1.Time Complexity of this method will be O(n*logn + n) = O(n*log(n)) as we are doing this in a single iteration after sorting all three arrays. Codefor u in range(int(input())): n = int(input()) x = map(int, input().split()) y = map(int, input().split()) z = map(int, input().split()) x = sorted(x) y = sorted(y) z = sorted(z) i = 0 j = 0 k = 0 ans = 0 while(i
 » Could someone find my mistake in Problem B. I sorted all the input array A,B and C and then finding index in C using upper bound. If found then I store it in variable named k and start searching again for other possible index starting from k index. 13 test cases were passed but other failed. Please help My code link
 » Is the question level as same in ARC as DIV2 codeforces.Is it true that ARC(a
 » 13 months ago, # | ← Rev. 2 →   Atcoder solution B REWhy I am getting runtime error can anyone explain? Spoiler... `Your code here... #include using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); long int n; cin>>n; long int a[n],b[n],c[n]; for(int i=0;i>a[i]; for(int i=0;i>b[i]; for(int i=0;i>c[i]; sort(a,a+n); sort(b,b+n); sort(c,c+n); vectorv2,v3; long int j=0; for(int i=0;ia[j]){ v2.push_back(b[i]); //b[i]=0; j++; } } j=0; for(int i=0;iv2[j]){ v3.push_back(c[i]); //c[i]=0; j++; } } /*for(int i=0;i
 » Thanks a lot for a detailed editorial like this.
 » A very good contest! But now Atcoder is being maintained so that I can't see the editorial for problem E, can anyone explain how to solve problem E? Thanks a lot!
 » Will you provide the code for F?
 » 13 months ago, # | ← Rev. 2 →   In Problem C Can Anyone Explain A number can be written as the sum of K good numbers if and only if it can be written as the sum of the following: An integer between K and 3K; The sum of at most K good numbers, multiplied by 10. The above statement in Editorial. It would be a Great Help