### IgorI's blog

By IgorI, history, 3 years ago, Thank you for participating in the contest! I hope you enjoyed the tasks. Here is an editorial with hints.

1474A - Puzzle From the Future

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Editorial
Solution

1474B - Different Divisors

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Solution

1474C - Array Destruction

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Editorial

1474D - Cleaning

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Solution

1474E - What Is It?

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Editorial
Solution

1474F - 1 2 3 4 ...

Bonus
Editorial
Solution

Good luck in your future contests! Tutorial of Codeforces Round 696 (Div. 2) Comments (121)
| Write comment?
 » thanks for fast editorial
 » Hope I'll become Candidate master today without new year's magic ;)
•  » » You already are ;)
•  » » You can see unofficial results with CF predictor
 » Man!!! That was really quick! Nice hints and clear insight.
 » If negative numbers allowed in problem C, Is this problem still can be solved ?
•  » » 3 years ago, # ^ | ← Rev. 2 →   Sorry I underestimated the fact that, the X will become variable in case of negatives. Ex : 6 4 1 -2 Here we can make X = 5, {4, 1} -> {6, — 2}
 » Thanks!
 » Good contest, despite O(N^2LogN) not passing in java for C
 » 3 years ago, # | ← Rev. 2 →   I am not understanding why i am getting tle verdict in third problem. I think i am doing it in O(n2⋅logn) but i am wrong maybe . Here is a link to my submission https://codeforces.com/contest/1474/submission/104839937 I would be extremely glad if someone can point out the error.
•  » » 3 years ago, # ^ | ← Rev. 6 →   for me also the multiset solution was not passing then I did it using the count array as the numbers were restricted to 10**6. So, I modified it to O(N^2) solution. Upd: sorry did not knew that count operation in multiset takes linear time.
•  » » 3 years ago, # ^ | ← Rev. 2 →   I think the issue is in the part after you calculate as where you do s.erase(dusra). Instead you should do s.erase(it).
•  » » » Thanks it solved the provlem. I meant to write it only there i don't know what i was thinking then.
•  » » » I don't understand why 104806538 is getting TLE in problem C. I think it is n2*logn and should pass. It would be very helpful if someone can point out the issue.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   as rkguptagkp462 pointed out count opertaion takes linear time. Upd: my bad i thought multiset was used instead of set
•  » » » » » that is for multiset no? count in std::set take logn time only.Although, i tried this as well 104853778.
 » There doesn't seem to be any hacking round, so When is the rating gonna be updated?
•  » » Wait for 1-5 hours.
 » Very well written tutorial. While solving C using multiset; one needs to remember that erasing an element by value will delete all elements having same value. And it costs you time to debug it and to read online documents.
•  » » Use iterators to avoid that!
•  » » you can just use m.erase(m.find(element));
•  » » » it looks elegant.
 » 3 years ago, # | ← Rev. 2 →   Problem 1474D - Cleaning. I supposed that, if we don't use the ability, the answer is "yes" if and only if the sum of numbers on odd positions is equal to the sum of numbers on even positions, and every number is not greater than the sum of it's neighbors. But the solution fails: 104836378. Is my idea wrong, or is it because of some mistake in the implementation? I wasn't able to find any couterexample(and the test on which it fails is unvisible) nor any formal proof. Thanks in advance.UPD: I have found the counterexample. Very weird my solution passed all those millions of tests. 1 8 2 2 4 3 1 2 1 1 
•  » » 3 years ago, # ^ | ← Rev. 2 →   I used the same idea and it worked. You can check my submission.UPD: the counterexample isn't complex but I just couldn't find one during the contest.
•  » » » Thank you, it turns out I used bool instead of int for the prefix sums array, that was the mistake.
•  » » » » Do you have any formal proof of this solution(I also had the same intuition while trying to upsolve it but sadly i cannot prove it)?
•  » » » » » Sadly, no one can prove it because it is simply wrong. See my comment above.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   Can you please tell why my submission is not accepted like I am also doing using similar approachhttps://codeforces.com/contest/1474/submission/104876110 [My submission]
•  » » » » » Sorry, no idea. Try checking if all the indices are correct.
•  » » » » » » 3 years ago, # ^ | ← Rev. 4 →   Some people were lucky and their solution got accepted otherwise it is giving wrong answer on 1131 2 2 3 1 3 2 1 2 1 2 2 2answer should be no but your and mine approach gives Yes
•  » » » » » » » Just figured that out too.
•  » » » Can you please explain how to choose x for C problem ? i'm stuck at that.
•  » » » » From the problem statement it can be observed that x will decrease with each operation. So it is necessary for us to take maximum element as one of the constituent of x. whyIf we choose both elements lesser then the maximum elements as constituent of x, then in that case after first operation we are going to have new x as lesser then the maximum value of the remaining elements in the array. In that case it is clear that eventually we will not be able to get solution using that x. So we must combine each of the element with the maximum to get a new potential value of x. and check if it is feasible to go with that value of x.
 » I don't really understand the explanation for D. Can somebody explain/give an example for how p_i and s_i are calculated, and what the check mentioned in the last paragraph is? A link to a solution that implements this method would be just as good as well.
•  » » I think I rewrote it somewhat cleaner in 105191466.
 » Video Editorial for
•  » » I think you should first focus on improving your problem solving skills rather than wasting time in making youtube videos.p.s. NO intention of offending you but just an advice, rest on you .
 » My Java Solution for Problem C: 104839108 Isn't it O(n^2logn) ? I used to HashMap.
•  » » Can you help me with my Java submission for C. Its giving TLE even though its O(n^2 logn) 104873241
•  » » » No, it's $O(n^3)$: for each starting $x$ ($2n$ variants) you $O(n)$ times call $contains()$ and $remove()$ from $ArrayList$ $l$, where at least one call of $remove()$ is removing the first element and works in $O(n)$.
 » in 1 sec per test: will our code of O(n^2) passes? if 1<=t<=1000 and 1<=n<=1000 because i read only 10^7 operations will pass... but here (n^2)*t will be around 10^9 right.. any clarification on this will be appreciated...
•  » » 3 years ago, # ^ | ← Rev. 2 →   It is guaranteed that the total sum of n over all test cases doesn't exceed 1000. This means: Whether how many test cases there are, n will never exceed 1000. You can remove 't' from your complexity here. Suppose t=1000. Then the value of n in each test case will be 1 only.
•  » » "It is guaranteed that the total sum of n over all test cases doesn't exceed 1000."
•  » » » thanks
 » 3 years ago, # | ← Rev. 2 →   This editorial is awesome as it totally focuses on concepts instead of spoon-feeding.
 » It's hard for me to understand why there is only one way to make all numbers equal to 0 in problem D. I don't get how proof works. Can someone explain ?
•  » » Start from a, it necessarily takes a operations. Apply those and now a is fixed and a1~n-1 is a similar problem.
•  » » » Why start from a ? why not from a[n-1] ?
•  » » » » Uh, obviously it is equivalent?
 » 3 years ago, # | ← Rev. 4 →   My soln of F is similar to the model soln mentioned in the editorial of 1295F - Good Contest.Firstly in linear time, we can compute the length of LIS.Let's fix a minimum element of LIS say X and length by L. Then we know that LIS is $X,X+1,..,X+L-1$. We will loop over all possible Xs.dp[X-1]=1.dp[i][j] is no of LIS starting at X and end at j using first $d_1,d_2,...,d_i$This soln works in $O(n*(max(Pi)-min(Pi))$.One can refer to this Bruteforce soln for more details about DP states.Soln in one line — dp[i] is a piecewise polynomial function of degree atmax i in j. Let's denote B be the set of all values in prefix sum of the given array in increasing order.Important thing to note here is there exsists a polynomial P(x) of degree less than or equal to i such $P(B_l)=dp[i][B_l],P(B_l+1)=dp[i][B_l+1],...,P(B_{l+1}-1)=dp[i][B_{l+1}-1]$.So instead of maintaining $O(B_{l+1}-Bl)$ states we can just maintain i+1 values and interpolate this polynomial whenever required. Also, X belongs to B.For each polynomial, we maintain O(n) values. Interpolation takes O(n). There are $O(n*n)$ polynomials for each $X$.This soln works in $O(n^4)$.AC Submission
 » 3 years ago, # | ← Rev. 2 →   I just added solutions in C++ to all problems.There are some problems with spoilers now, so hints are published without any spoilers (actually it is the reason editorial wasn't pusblished just after the contest).Anyway, thanks again for taking part in my contest and discussing its problems.UPD: Now hints should be ok
•  » » The Problems were beautiful.
•  » » Great problems. The hints in the editorial are excellent as well. Thanks for a very interesting problem set.
 » 3 years ago, # | ← Rev. 2 →   Hey, can someone explain why am I getting time limit exceed for problem C : https://codeforces.com/contest/1474/submission/104834766I used a O(n^2log(n)) solution, normally it should pass..I used an unerdored map in the last seconds and got accepted in 982 ms.Thanks in advance
 » Can someone tell me Time complexity of my code , I think it is O(N^2) approx. https://codeforces.com/contest/1474/submission/104834839
 » Here are my video solutions of this round for A-E, which I'm really glad I didn't do a screencast for
 » 3 years ago, # | ← Rev. 2 →   Can someone explain why we don't need to check the case where $a = p^3$ in problem B ?In other words why can't a solution be of the form: $1$ $p$ $p^2$ $p^3$ ?
•  » » 3 years ago, # ^ | ← Rev. 2 →   $p_1.p_2 < p_1^3$ where $p_1 > d, p_2 \geq p_1 + d.$ Since, we're asked to find the minimum integer $x$. Maybe, It's also because prime numbers are dense at the start.
•  » » » what if $p_1^2 < p_2$ ? or that just never happens due to the limits of this problem ?$p_1^2$ being greater or equal than $p_1 + d$$p_2$ being next prime greater or equal than $p_1 + d$any formal or convincing proof that this never happens?
•  » » » » 3 years ago, # ^ | ← Rev. 4 →
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   Bertrand's postulate, proven in 1852, states that there is always a prime number between k and 2k, so in particular $p_{n+1} < 2p_n$.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   As pointed in above comments , we can prove using Bertrand Postulate .More formally suppose you say answer is $p^3$ and thus it's factors are $1,p,p^2,p^3$ . Difference between each of them should be greater than $d$. Thus $p_1≤p$ . Using Bertrand Postulate we can say that there must exist prime $p_2$ such that $p •  » » » » » No, I don't think$p_1^2 < p_2$since$2p_1 < p_1^2, \forall p > 2$.We can further postulate that, there also exist a prime,$p_x$that is less than$4p_n$with Bertrand's Postulate since,$p_{n + 1}$may not be greater than or equal to$p_n + d$.Thus, We can prove that$p_n . p_x < p_n^3$. •  » » » » » » 3 years ago, # ^ | ← Rev. 3 → I wrote$p < p_2$.It's called proof by contradiction. So i assumed that$p^3$is the best answer but then we arrived at contradiction that a better answer would exist in that case. •  » » » » » » » 3 years ago, # ^ | ← Rev. 2 → Using Bertrand Postulate we can say that there must exist prime$p_2$such that$p^2
•  » » » » » » » » 3 years ago, # ^ | ← Rev. 5 →   In Bertrand Postulate , take $n=p$ , also $p≥2$ , thus there must exist prime $p_2$ such that $n •  » » » » » » » » » 3 years ago, # ^ | ← Rev. 3 → In Bertrand Postulate , take$n=p^2$, also$p\geq2$, thus there must exist prime$p_2$such that$n
•  » » » » » » » » » 3 years ago, # ^ | ← Rev. 2 →   It can be any integer . I get what you're saying. But $p^2$ has to be a prime number in order to use Bertrand's postulate A less restrictive formulation is: for every $n>1$ there is always at least one prime p such that $n •  » » » » » 3 years ago, # ^ | ← Rev. 7 → When you said p ^ 2 < p2 < p ^ 3, didnt you actually prove p1 * p2 is a worse answer? Because suppose p1 = p so then p1 * p2 > p * p ^ 2, because p2 > p ^ 2. So optimal would be p ^ 3. •  » » » » » » Thanks for pointing it out , I have corrected it now. Actually i needed to write$p
 » This "Hint by Hint" format is nice. Keep it up.
 » I would be very thankful if someone can find the bug in my code https://codeforces.com/contest/1474/submission/104847038 I'm unable to figure it out
•  » » 3 years ago, # ^ | ← Rev. 2 →   Testcase: 2 8 2 4 5 Correct Answer:NO Your OUTPUT: YES 13 5 8 4 4 2 2 
•  » » » You can’t use the same number multiple times
•  » » » » Yes,that's why above person's solution is incorrect.
•  » » » » » Thank you so much
 » Problem B can be solved with help of regexes, as N is not big. Try it yourself. Spoiler 1You can create a string which contains 0 on prime indexes, 1 otherwise. Spoiler 2You can use quantifiers with minimal range, i.e {X,} for preserving minimal distance d. Spoiler 3Even generating an array of primes you can use similar trick.Solution 104855574
 » The time constraints for C are absolutely atrocious, I tried 3 different n^2 log n implementation and I wasn't able to make any of them pass
 » If anyone is stuck at C https://youtu.be/Y_53vVG7RMw
 » I still have a question about prob D, why we just talk about the swap between 'a[i]' & 'a[i+1]' ?
•  » » Before the start of cleaning, you can select two neighboring piles and swap them.
•  » » » Oppps, what an idot i am... :(by the way, what if there's no this "neighboring" limit?
•  » » » » I think there is n^2 solution then
 » My solution is getting WA on test 2 and I don't understand why. Can someone please give me a test case where this fails?https://codeforces.com/contest/1474/submission/104864709
•  » » Is fails on test case:151 1 3 3 2
 » I got a few doubts in n* +max(a) solution of c if (cnt[a[j]] < 0 || cnt[x - a[j]] < 0) { cnt[a[j]] = 0; cnt[x - a[j]] = 0; break; } when the cnt of a[j] will be less than zero ? as for in while loop we check that cnt[a[j]]>0 also if i comment it out, then why it gives wrong answer since in reset(a) after ops, we are anyway setting cnt[a[i]] to zero ?
 » brilliant idea for problem D
 » Is +460 in first codeforces contest good? What is the metric for good and bad for codeforces contests?
•  » »
 » This format of multiple hints and then answering them is great
 » 3 years ago, # | ← Rev. 2 →   Problem D, Now we have 0 stones in the 1-st pile and a2−a1 stones in the 2-nd pile. I understood this.We can apply the idea above and get that we should have a2−a1 stones in the 3-rd pile. How do we get a2-a1 for 3-rd pile?
 » Why I'm getting RE in Case 6 in problem C, Help me (O_O) :104870245
•  » » upper limit for a[i] is 10^6 but you have made the size of array f as 10^5 so f[a[i]]++ will end up with run time error(as 10^6>10^5 so its array size issue).
•  » » » Tnx Man :). Got Ac . Verdict Time : 998 :( 104875146
•  » » » » For each test case, you create a new array $f$ of size $10^6$. Don't know why you are sad — you are lucky to get AC on Java with it.
 » 3 years ago, # | ← Rev. 2 →   What's wrong with my D? I'm confused. 104868453I'm not good at English..Emm...If I don't use the superability,it will be  b=a b=a-a b=a-a+a ... b[n]=a[n]-a[n-1]+a[n-2]-...For odd numbers : b[n]=a[n]-a[n-1]+a[n-2]-...-a+a For even numbers : b[n]=a[n]-a[n-1]+a[n-2]-...+a-aif we swap(a[pos],a[pos]+1]) b[pos] will -a[pos]+a[pos+1]for i>=pos+1 i%2=(pos+1)%2: b[i] will -2*a[pos+1]+2*a[pos] i%2=(pos+2)%2: b[i] will +2*a[pos+1]-2*a[pos]I want b[n]->0 and all i in [1,n] b[i]>=0 I calculate suffix(suf) and prefix(pre)  Oh,I can't make it any clearer,I'm poor in English...upd: Accepted
•  » » If I don't use the superability,it will be b=a b=a-a b=a-a+a ... b[n]=a[n]-a[n-1]+a[n-2]-... For odd numbers : b[n]=a[n]-a[n-1]+a[n-2]-...-a+a For even numbers : b[n]=a[n]-a[n-1]+a[n-2]-...+a-a if we swap(a[pos],a[pos]+1]) b[pos] will -a[pos]+a[pos+1] for i>=pos+1 i%2=(pos+1)%2: b[i] will -2*a[pos+1]+2*a[pos] i%2=(pos+2)%2: b[i] will +2*a[pos+1]-2*a[pos] I want b[n]->0 and all i in [1,n] b[i]>=0 I calculate suffix(suf) and prefix(pre) Oh,I can't make it any clearer,I'm poor in English... 
 » 3 years ago, # | ← Rev. 3 →   C was just awesome.I solved it using the idea that at each operation we have to select two numbers among which one is always the maximum number among the remaining non selected numbers and the other number can be any number among non selected numbers such that sum of both numbers is equal to x(and then update x and repeat the process) If we can not do it for any selected x then answer will be No. And initial x can be (maximum of all numbers+ any other number from array).
 » 3 years ago, # | ← Rev. 3 →   Hello , Thanks for this awesome contest and awesome solution for Problem D.However I tried to solve D using segment trees and I am able to get AC on it. What I am doing is creating two array for odd and even position and trying every swap by making appropriate updates and then rolling back to original states. I could'nt see anything wrong in my solution. Can someone help me figure it out ? My code//#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector") #include #include #include using namespace std; #define ll long long #define endl '\n' #define mod 1000000007 #define pb push_back #define ff first #define ss second #define con continue #define ub upper_bound #define lb lower_bound #define si(x) int(x.size()) #define sum_all(a) ( accumulate ((a).begin(), (a).end(), 0ll)) #define mine(a) (*min_element((a).begin(), (a).end())) #define maxe(a) (*max_element((a).begin(), (a).end())) #define mini(a) ( min_element((a).begin(), (a).end()) — (a).begin()) #define maxi(a) ( max_element((a).begin(), (a).end()) — (a).begin()) #define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) — (a).begin()) #define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) — (a).begin()) const double pi = 2 * acos(0.0); const int dx[] = { -1, 0, 1, 0 }; const int dy[] = { 0, -1, 0, 1 }; const int dx8[] = { -1, 0, 1, 0,1,1,-1,-1 }; const int dy8[] = { 0, -1, 0, 1,1,-1,1,-1 }; ll min(ll a, ll b) { if (a < b)return a; return b; } ll max(ll a, ll b) { if (a > b)return a; return b; } ll ceil1(ll a, ll b) { return(a + b — 1) / b; } void read(vector& arr) { for (ll i = 0; i < si(arr); i++) cin >> arr[i]; } void read_graph(vector>& g, ll m) { while (m--) { ll x, y; cin >> x >> y; x--, y--; g[x].pb(y); g[y].pb(x); } } template struct lazy_segtree { public: lazy_segtree() : lazy_segtree(0) {} lazy_segtree(int n) : lazy_segtree(std::vector(n, e())) {} lazy_segtree(const std::vector& v) : _n(int(v.size())) { log = ceil(log2(_n)); size = 1 << log; d = std::vector(2 * size, e()); lz = std::vector(size, id()); for (int i = 0; i < _n; i++) d[size + i] = v[i]; for (int i = size — 1; i >= 1; i--) { update(i); } } void set(int p, S x) { assert(0 <= p && p < _n); p += size; for (int i = log; i >= 1; i--) push(p >> i); d[p] = x; for (int i = 1; i <= log; i++) update(p >> i); } S get(int p) { assert(0 <= p && p < _n); p += size; for (int i = log; i >= 1; i--) push(p >> i); return d[p]; } S prod(int l, int r) { assert(0 <= l && l <= r && r <= _n); if (l == r) return e(); l += size; r += size; for (int i = log; i >= 1; i--) { if (((l >> i) << i) != l) push(l >> i); if (((r >> i) << i) != r) push(r >> i); } S sml = e(), smr = e(); while (l < r) { if (l & 1) sml = op(sml, d[l++]); if (r & 1) smr = op(d[--r], smr); l >>= 1; r >>= 1; } return op(sml, smr); } S all_prod() { return d; } void apply(int p, F f) { assert(0 <= p && p < _n); p += size; for (int i = log; i >= 1; i--) push(p >> i); d[p] = mapping(f, d[p]); for (int i = 1; i <= log; i++) update(p >> i); } void apply(int l, int r, F f) { assert(0 <= l && l <= r && r <= _n); if (l == r) return; l += size; r += size; for (int i = log; i >= 1; i--) { if (((l >> i) << i) != l) push(l >> i); if (((r >> i) << i) != r) push((r - 1) >> i); } { int l2 = l, r2 = r; while (l < r) { if (l & 1) all_apply(l++, f); if (r & 1) all_apply(--r, f); l >>= 1; r >>= 1; } l = l2; r = r2; } for (int i = 1; i <= log; i++) { if (((l >> i) << i) != l) update(l >> i); if (((r >> i) << i) != r) update((r - 1) >> i); } } template int max_right(int l) { return max_right(l, [](S x) { return g(x); }); } template int max_right(int l, G g) { assert(0 <= l && l <= _n); assert(g(e())); if (l == _n) return _n; l += size; for (int i = log; i >= 1; i--) push(l >> i); S sm = e(); do { while (l % 2 == 0) l >>= 1; if (!g(op(sm, d[l]))) { while (l < size) { push(l); l = (2 * l); if (g(op(sm, d[l]))) { sm = op(sm, d[l]); l++; } } return l - size; } sm = op(sm, d[l]); l++; } while ((l & -l) != l); return _n; } template int min_left(int r) { return min_left(r, [](S x) { return g(x); }); } template int min_left(int r, G g) { assert(0 <= r && r <= _n); assert(g(e())); if (r == 0) return 0; r += size; for (int i = log; i >= 1; i--) push((r - 1) >> i); S sm = e(); do { r--; while (r > 1 && (r % 2)) r >>= 1; if (!g(op(d[r], sm))) { while (r < size) { push(r); r = (2 * r + 1); if (g(op(d[r], sm))) { sm = op(d[r], sm); r--; } } return r + 1 - size; } sm = op(d[r], sm); } while ((r & -r) != r); return 0; } private: int _n, size, log; std::vector d; std::vector lz; void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); } void all_apply(int k, F f) { d[k] = mapping(f, d[k]); if (k < size) lz[k] = composition(f, lz[k]); } void push(int k) { all_apply(2 * k, lz[k]); all_apply(2 * k + 1, lz[k]); lz[k] = id(); } }; struct S { ll sum, len, m1; }; struct F { ll a, b; }; S op(S l, S r) { return S{ l.sum + r.sum,l.len + r.len ,min(l.m1,r.m1) }; } S e() { return S{ 0,0,(ll)(1e18) }; } S mapping(F l, S r) { ll a = r.sum * l.a + r.len * l.b; return S{ a,r.len,r.m1 * l.a + l.b }; } F composition(F l, F r) { ll a, b; a = l.a * r.a; b = l.a * r.b + l.b; return F{ a,b }; } F id() { return F{ 1,0 }; } ll n; vector arr; ll calc() { if (si(arr) == 2) { return (bool)(arr == arr); } vector narr(n - 1); narr = arr - arr; for (ll i = 1; i < n - 1; i++) { narr[i] = arr[i + 1] - narr[i - 1]; } ll n1 = ceil1(si(narr), 2); ll n2 = si(narr) - n1; vector seg11, seg22; for (ll i = 0; i < n1 + n2; i++) { if (i % 2 == 0) seg11.pb(S{ narr[i],1,narr[i] }); else seg22.pb(S{ narr[i],1,narr[i] }); } lazy_segtree seg1(seg11); lazy_segtree seg2(seg22); // no swap ll m1 = min(seg1.prod(0, n1).m1, seg2.prod(0, n2 ).m1); ll last; if (n2 == n1) { last = seg2.prod(n2 - 1, n2 ).m1; } else last = seg1.prod(n1 - 1, n1 ).m1; if (m1 == 0 and last == 0) return true; for (ll i = 0, j = 0, k = 0; i < n - 1; i++) { // apply if (i % 2 == 0) { ll val = 2 * (arr[i] - arr[i + 1]); if(j<=n1-1) seg1.apply(j, n1 , F{ 1,val }); if(k<=n2-1) seg2.apply(k, n2 , F{ 1,-val }); } else { ll val = 2 * (arr[i] - arr[i + 1]); if(j<=n1-1) seg1.apply(j, n1 , F{ 1,-val }); if(k<=n2-1) seg2.apply(k, n2 , F{ 1,val }); } // query ll m1 = min(seg1.prod(0, n1 ).m1, seg2.prod(0, n2 ).m1); ll last; if (n2 == n1) { last = seg2.prod(n2 - 1, n2 ).m1; } else last = seg1.prod(n1 - 1, n1 ).m1; if (m1 == 0 and last==0) return true; // rollback if (i % 2 == 0) { ll val = 2 * (arr[i] - arr[i + 1]); if(j<=n1-1) seg1.apply(j, n1 , F{ 1,-val }); if(k<=n2-1) seg2.apply(k, n2 , F{ 1,val }); j++; } else { ll val = 2 * (arr[i] - arr[i + 1]); if(j<=n1-1) seg1.apply(j, n1 , F{ 1,val }); if(k<=n2-1) seg2.apply(k, n2 , F{ 1,-val }); k++; } } return false; } void solve() { cin >> n; arr.clear(); arr.resize(n); for (ll i = 0; i < n; i++) cin >> arr[i]; bool ans = calc(); reverse(arr.begin(), arr.end()); ans = ans || calc(); string res = "YES"; if (!ans) res = "NO"; cout << res << endl; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif int t; cin >> t; while (t--) solve(); } Update: Figured out my bug, I skipped the fact that I had to update previous element also. xd
 » Can anyone one help me with D: https://codeforces.com/contest/1474/submission/104881006
•  » » Try this test case:151 1 3 2 3
 » 104885466 why i am getting tle
•  » » 3 years ago, # ^ | ← Rev. 2 →   Because if your use find(ms.begin(), ms.end(), val) it is linear in time, instead you should use ms.find(val), which is O(log n). I submitted your code 104886371 with it and got AC.
 » for some reason i feel tired of constructive problem.i dont hate them but while doing them i just feel so weird and solution often suprised very hard (like an unexpectd slap)
 » I advise everyone to watch this video, hope would be helpful!
 » Can we solve problem D if we are allowed to swap any two piles once? It could be a variation for this problem. Suggesions are much appreciated.
 » If someone pls could help ! their solution for C without set ..isn't it O(n³) ? if not then why coz the three loops in worst case are gonna iterate and complete to their limit ?
 » Can anyone tell what is wrong in my D? WA on test 2: 105030694
 » Still stuck in problem D, can anyone help me? :| thanks. 105049987
•  » » my latest WA submission #105067344
•  » » Oh, I find out why. in case 0 if (Smark==n+1 && s[n]==0) in case 2 (a[i]-s[i-1]>=0 && a[i+1]-p[i+2]>=0 && a[i]-s[i-1]==a[i+1]-p[i+2] && s[i-1]>=0 && p[i+2]>=0)
 » 3 years ago, # | ← Rev. 2 →   somebody please explain why this is giving TLE for problem 1. https://codeforces.com/contest/1474/submission/104916660 again for problem 1, the following code prints all digits as 1, i cant see why it is happening https://codeforces.com/problemset/submission/1474/105175403
•  » » hey combinatorialist, well I don't know the reason , but in your code this is because of you are using strings and append in strings like a=a+'1'; and b=b+digit; my suggestion is if you use the char array like this char a[(int)1e5+5],b[(int)1e5+5]; int ai,bi; int main(){ int t; cin>>t; vectorsol; while(t--){ memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); ai=0,bi=0; int prevsum=0; char digit; int n; cin>>n; for (int i=0; i>digit; b[bi++]=digit; if(b[i]+1!=prevsum){ a[ai++]='1'; prevsum=b[i]+1; continue; } if(b[i]!=prevsum){ a[ai++]='0'; prevsum=b[i]; } } sol.push_back(a); } for(auto x:sol){ cout<
•  » » » ok.....can you also look at my second submission, it is printing every digit as 1, i cant understand why. https://codeforces.com/problemset/submission/1474/105175403
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   bro your code is correct! You just make 1 to '1' and 2 to '2' in if(sum-b[i]==2) and else if(sum-b[i]==1) as it is character so it's sum is 97 ans you are comparing it with 2;
•  » » » » » 3 years ago, # ^ | ← Rev. 2 →   ok, it worked...but i am still confused. when i write int sum=a[i]+b[i], this means that int sum='1'+'1' which should mean int sum="11", and then the string "11" will be converted to int? basically i am getting confused in adding two chars and then converting into int, can you explain or give me a resource from where i can read this?
•  » » » » » » 3 years ago, # ^ | ← Rev. 2 →   here co is cout<< char a='1',b='2'; co a+b<
 » 3 years ago, # | ← Rev. 2 →   Hints for problems is such a great idea! I believe that hints+editorial is way better because some people may need just a bit of help to upsolve. Setters, please, make hints+editorial.
 » 3 years ago, # | ← Rev. 2 →   My $O(N^2)$ java sol for problem C is getting TLE. But such similar solution of others getting passed. Can anyone tell why it's getting TLE. I thought a lot but unable to figure it out. Anyone please.
 » For problem D,in case of no superability, why should we start from 0 ? why not from position n-1 ?
 » Problem B:cout << min(1ll * p * p, 1ll * p * p * p) << "\n"; Instead of we can do this:  int ans = 1; for (auto &it : primes) ans *= it; cout << ans << nln; It will also gives you Accepted vardict.To know more see my submission: https://codeforces.com/contest/1474/submission/207939116