Codeforces Round #696 Editorial

№	Пользователь	Рейтинг
1	jiangly	3640
2	Benq	3593
3	tourist	3572
4	orzdevinwang	3561
5	cnnfls_csy	3539
6	ecnerwala	3534
7	Radewoosh	3532
8	gyh20	3447
9	Rebelz	3409
10	Geothermal	3408

№	Пользователь	Вклад
1	maomao90	174
2	awoo	164
3	adamant	163
4	TheScrasse	159
5	nor	158
6	maroonrk	156
7	-is-this-fft-	151
8	SecondThread	147
9	orz	146
10	pajenegod	145

Спасибо за участие в раунде! Я надеюсь, что вам понравились задачи.

1474A - Загадка из будущего

Hint 1

Hint 2

Editorial

Solution

1474B - Различные делители

Hint 1

Hint 2

Hint 3

Editorial

Solution

1474C - Уничтожение массива

Hint 1

Hint 2

Hint 3

Hint 4

Editorial

Solution

1474D - Уборка

Hint 1

Hint 2

Hint 3

Hint 4

Editorial

Solution

1474E - Что это?

Hint 1

Hint 2

Hint 3

Hint 4

Editorial

Solution

1474F - 1 2 3 4 ...

Bonus

Editorial

Solution

Удачи в будущих контестах!

If I don't use the superability,it will be b[1]=a[1] b[2]=a[2]-a[1] b[3]=a[3]-a[2]+a[1] ... b[n]=a[n]-a[n-1]+a[n-2]-... For odd numbers : b[n]=a[n]-a[n-1]+a[n-2]-...-a[2]+a[1] For even numbers : b[n]=a[n]-a[n-1]+a[n-2]-...+a[2]-a[1] 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...

//#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector") #include <bits/stdc++.h> #include<set> #include <array> 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<ll>& arr) { for (ll i = 0; i < si(arr); i++) cin >> arr[i]; } void read_graph(vector<vector<ll>>& g, ll m) { while (m--) { ll x, y; cin >> x >> y; x--, y--; g[x].pb(y); g[y].pb(x); } } template <class S, S (*op)(S, S), S (*e)(), class F, S (*mapping)(F, S), F (*composition)(F, F), F (*id)()> struct lazy_segtree { public: lazy_segtree() : lazy_segtree(0) {} lazy_segtree(int n) : lazy_segtree(std::vector<S>(n, e())) {} lazy_segtree(const std::vector<S>& v) : _n(int(v.size())) { log = ceil(log2(_n)); size = 1 << log; d = std::vector<S>(2 * size, e()); lz = std::vector<F>(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[1]; } 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 <bool (*g)(S)> int max_right(int l) { return max_right(l, [](S x) { return g(x); }); } template <class G> 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 <bool (*g)(S)> int min_left(int r) { return min_left(r, [](S x) { return g(x); }); } template <class G> 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<S> d; std::vector<F> 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<ll> arr; ll calc() { if (si(arr) == 2) { return (bool)(arr[0] == arr[1]); } vector<ll> narr(n - 1); narr[0] = arr[1] - arr[0]; 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<S> 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<S, op, e, F, mapping, composition, id> seg1(seg11); lazy_segtree<S, op, e, F, mapping, composition, id> 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(); }

char a[(int)1e5+5],b[(int)1e5+5]; int ai,bi; int main(){ int t; cin>>t; vector<string>sol; 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<n; ++i){ cin>>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<<x<<endl; } }

Комментарии (122)

Показать архивные | Написать комментарий?

Mestu

3 года назад, # |

+18

thanks for fast editorial

→ Ответить

Heptadecagon

+10

Hope I'll become Candidate master today without new year's magic ;)

slim-shady

3 года назад, # ^ |

+25

You already are ;)

FatihSolak

You can see unofficial results with CF predictor

breakthejinx

Man!!! That was really quick! Nice hints and clear insight.

It_Wasnt_Me

If negative numbers allowed in problem C, Is this problem still can be solved ?

Flower08

← 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}

ItsEZ

Thanks!

Java

+40

Good contest, despite O(N^2LogN) not passing in java for C

Portugas_D_Ace

+14

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.

rkguptagkp462

← 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.

sarthak2007

+12

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.

DammitDwight

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.

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.

pratikb2805

There doesn't seem to be any hacking round, so When is the rating gonna be updated?

MubtasimShahriar

Wait for 1-5 hours.

vdmedragon

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.

HelloWorld

Use iterators to avoid that!

Dan4Life

you can just use

m.erase(m.find(element));

it looks elegant.

kkite_gk

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

Thallium54

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.

1000101

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)?

Thallium54 kkite_gk

Sadly, no one can prove it because it is simply wrong. See my comment above.

somyamehta_24

Can you please tell why my submission is not accepted like I am also doing using similar approach

https://codeforces.com/contest/1474/submission/104876110 [My submission]

Sorry, no idea. Try checking if all the indices are correct.

← Rev. 4 →

Some people were lucky and their solution got accepted otherwise it is giving wrong answer on

1 2 2 3 1 3 2 1 2 1 2 2 2

answer should be no but your and mine approach gives Yes

Just figured that out too.

Letscode_sundar

Can you please explain how to choose x for C problem ? i'm stuck at that.

nipul1

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.

why

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.

tonyb7

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.

arman.t

I think I rewrote it somewhat cleaner in 105191466.

aTHiNkEr113

Video Editorial for

problem A : https://www.youtube.com/watch?v=2u6zr-tdEF4&t=8s problem B : https://www.youtube.com/watch?v=ACbXqdm2uHU

0verHaul

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.

th_rahulthorat

Can you help me with my Java submission for C. Its giving TLE even though its O(n^2 logn) 104873241

adedalic

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 $$$O(n)$$$.

broiampro

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...

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.

NewLul

"It is guaranteed that the total sum of n over all test cases doesn't exceed 1000."

thanks

kevin_007

+16

This editorial is awesome as it totally focuses on concepts instead of spoon-feeding.

CannibalCorpse

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 ?

rgnerdplayer

Start from a[0], it necessarily takes a[0] operations. Apply those and now a[1] is fixed and a1~n-1 is a similar problem.

AjaySabarish

Why start from a[0] ? why not from a[n-1] ?

Uh, obviously it is equivalent?

aryanc403

+55

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[0][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

IgorI

+34

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.

sh_maestro

+17

Great problems. The hints in the editorial are excellent as well. Thanks for a very interesting problem set.

TANGRA

Hey, can someone explain why am I getting time limit exceed for problem C : https://codeforces.com/contest/1474/submission/104834766

I 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

dhruv7888

Can someone tell me Time complexity of my code , I think it is O(N^2) approx. https://codeforces.com/contest/1474/submission/104834839

galen_colin

+27

Here are my video solutions of this round for A-E, which I'm really glad I didn't do a screencast for

searleser97

+13

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: $$$1$$$ $$$p$$$ $$$p^2$$$ $$$p^3$$$ ?

EnterYourName

$$$p_1.p_2 < p_1^3$$$ 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?

mtw

+21

https://en.wikipedia.org/wiki/Bertrand%27s_postulate

Bertrand's postulate, proven in 1852, states that there is always a prime number between k and 2k, so in particular $$$p_{n+1} < 2p_n$$$.

rananjay23

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 $$$p<p_2<p^2$$$ , since smallest prime is 2. Hence $$$p_1*p_2$$$ is better answer.

You asked very good question , I never thought of it during the contest.

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 $$$p_n . p_x < p_n^3$$$.

← 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.

Using Bertrand Postulate we can say that there must exist prime $$$p_2$$$ such that $$$p^2<p_2<p^3$$$.

I don't know if this is called proof by contradiction. It'd be correct if you change the order as $$$p_2 < p^2 < p^3$$$ according to Bertrand Postulate.

← Rev. 5 →

In Bertrand Postulate , take $$$n=p$$$ , also $$$p≥2$$$ , thus there must exist prime $$$p_2$$$ such that $$$n<p_2<2n≤p^2$$$. So how my argument wrong ?

Thus we have better answer $$$p_1*p_2$$$ but that's contradiction since best answer was $$$p^3$$$ . Hence our assumption that $$$p^3$$$ will be best answer was wrong . Thus answer must be of form $$$p_1*p_2$$$.

proof by contradiction . In this to disprove something we first assume it's correct and then arrive at some contradiction . Here i assumed $$$p^3$$$ is correct and then we arrived at contradiction .

In Bertrand Postulate , take $$$n=p^2$$$ , also $$$p\geq2$$$ , thus there must exist prime $$$p_2$$$ such that $$$n<p_2<2n≤p^3$$$.

I get what you're saying. But $$$p^2$$$ has to be a prime number in order to use Bertrand's postulate.

Thank You, I didn't read about this before.

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

read here :

A less restrictive formulation is: for every $$$n>1$$$ there is always at least one prime p such that $$$n<p<2*n$$$

nitr0gen

← 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 $$$p<p_2<p^2$$$ and thus $$$p_1*p_2$$$ is smaller.

DreamPop1984

+26

This "Hint by Hint" format is nice. Keep it up.

achint_2.0

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

Shinpi

Testcase:

2

8  2  4  5

Correct Answer:NO

Your OUTPUT:

YES

13

5 8

4 4

2 2

akhil.arza

You can’t use the same number multiple times

Yes,that's why above person's solution is incorrect.

Thank you so much

rsFalse

Problem B can be solved with help of regexes, as N is not big. Try it yourself.

Spoiler 1

Spoiler 2

Spoiler 3

Solution 104855574

theunseenblade

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

Bojack_human

If anyone is stuck at C https://youtu.be/Y_53vVG7RMw

hehepig

I still have a question about prob D, why we just talk about the swap between 'a[i]' & 'a[i+1]' ?

wsyhb

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?

Sindy

I think there is n^2 solution then

shivtom

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

timreizin

Is fails on test case:

1 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 ?

Josh_josh

brilliant idea for problem D

brownTeddy

Is +460 in first codeforces contest good? What is the metric for good and bad for codeforces contests?

https://codeforces.com/blog/entry/77890

AatifRashid

This format of multiple hints and then answering them is great

nesco

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?

Zahid_Hasan_Sahin

Why I'm getting RE in Case 6 in problem C, Help me (O_O) :

104870245

abhiaps

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.

Indigosky

What's wrong with my D? I'm confused. 104868453

I'm not good at English..Emm...

If I don't use the superability,it will be ``` b[1]=a[1] b[2]=a[2]-a[1] b[3]=a[3]-a[2]+a[1] ... b[n]=a[n]-a[n-1]+a[n-2]-...

For odd numbers : b[n]=a[n]-a[n-1]+a[n-2]-...-a[2]+a[1] For even numbers : b[n]=a[n]-a[n-1]+a[n-2]-...+a[2]-a[1]

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...

upd: Accepted

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).

__hermit__

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 <bits/stdc++.h>
#include<set>
#include <array>
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()) &mdash; (a).begin())
#define maxi(a)    ( max_element((a).begin(), (a).end()) &mdash; (a).begin())
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) &mdash; (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) &mdash; (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 &mdash; 1) / b;
}
void read(vector<ll>& arr)
{
    for (ll i = 0; i < si(arr); i++)
        cin >> arr[i];

}

void read_graph(vector<vector<ll>>& g, ll m)
{
    while (m--)
    {
        ll x, y; cin >> x >> y;
        x--, y--;
        g[x].pb(y);
        g[y].pb(x);
    }
}
template <class S,
          S (*op)(S, S),
          S (*e)(),
          class F,
          S (*mapping)(F, S),
          F (*composition)(F, F),
          F (*id)()>
struct lazy_segtree {
  public:
    lazy_segtree() : lazy_segtree(0) {}
    lazy_segtree(int n) : lazy_segtree(std::vector<S>(n, e())) {}
    lazy_segtree(const std::vector<S>& v) : _n(int(v.size())) {
        log =  ceil(log2(_n));
        size = 1 << log;
        d = std::vector<S>(2 * size, e());
        lz = std::vector<F>(size, id());
        for (int i = 0; i < _n; i++) d[size + i] = v[i];
        for (int i = size &mdash; 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[1]; }

    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 <bool (*g)(S)> int max_right(int l) {
        return max_right(l, [](S x) { return g(x); });
    }
    template <class G> 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 <bool (*g)(S)> int min_left(int r) {
        return min_left(r, [](S x) { return g(x); });
    }
    template <class G> 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<S> d;
    std::vector<F> 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<ll> arr;
ll calc()
{
    if (si(arr) == 2)
    {
        return (bool)(arr[0] == arr[1]);
    }

    vector<ll> narr(n - 1);
    narr[0] = arr[1] - arr[0];
    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<S> 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<S, op, e, F, mapping, composition, id> seg1(seg11);
    lazy_segtree<S, op, e, F, mapping, composition, id> 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

Appple

Can anyone one help me with D: https://codeforces.com/contest/1474/submission/104881006

Try this test case:

1 1 3 2 3

demonstrator

104885466 why i am getting tle

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.

FukuNozi

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)

REALBOY

I advise everyone to watch this video, hope would be helpful!

coder_Aditya_404

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.

Maxis

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 ?

Vedkribhu

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)

combinatorialist

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

Sameer_Mishra100

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

hope this helps...

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

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;

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?

here co is cout<<

char a='1',b='2';
co a+b<<nl;///99
co (char)(a+b)<<nl;///c
co a-'0'+b-'0'<<nl;///3

when you add two char it automatically becomes int until you convert it into char.

frogcatcher

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.

Into_Your_Arms

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 ?

ShAiDSk

11 месяцев назад, # |

Problem B:

cout << min(1ll * p[0] * p[1], 1ll * p[0] * p[0] * p[0]) << "\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

B_Consistent

5 месяцев назад, # |

The main point I missed out on was that I was just concentrating on the effect of swap which is obviously on the right side of pair but totally forgot about the condition that alternating sum should be >=0 Even then I think it would have been nightmare to implement it in easier manner bcoz even to implement right side effect, i couldn't think of easier implementation

Блог пользователя IgorI