@JR OIer attended on "Spectral: :Cup 2026 Round 2 (Codeforce Round 1100, Div. 1 + Div. 2) Editual"

Thank you for participating in our round!

Prepared by:

Editoral by: reirigan

Let

Let

We want to choose the value of

#IncludeWe're talking namespace std;

typedef long long longll;

#definine debug(x) cout < #x < "= "< x < "\n";

#define vdebug(a) cout < < #a < < " = " ; for (auto x: a) cout < < < < " ; cout < < " \ " ;

mt19937 rang (chrono:steady clock:now() .time since epoch() .count());

Intuid(int a, int b)(a, b) (rang);}

ll uld(ll a, ll b)(a, b) (rang);}

void Solve(){

I don't know.

I'm sorry.

vector a (n);

For (int & x: a) cin > x;

Int mx = *max election (a.begin(), a.end());

Int mn = *min election (a.begin(), a.end());

Cout < (mx - mn + 1) / 2 < "n";

♪ I'm sorry ♪

Intmain(){

OS:: sync with stdio (false);

cin.tie (0);

Int t;

cin> > t;

(b) While (t-);

♪ I'm sorry ♪

Idea by: Intellegent

Prepared by:

Editoral by: reirigan

For one of the indices, you will gain equal to

It is never bad to put the larger value in

We should swim any

#IncludeWe're talking namespace std;

typedef long long longll;

#definine debug(x) cout < #x < "= "< x < "\n";

#define vdebug(a) cout < < #a < < " = " ; for (auto x: a) cout < < < < " ; cout < < " \ " ;

mt19937 rang (chrono:steady clock:now() .time since epoch() .count());

Intuid(int a, int b)(a, b) (rang);}

ll uld(ll a, ll b)(a, b) (rang);}

void Solve(){

I don't know.

I'm sorry.

vector a (n), b (n);

For (int & x: a) cin > x;

For (int & x: b) cin > x;

ll sum = 0;

for (int i = 0; i < n; i ++)

sum + = max (a [i], b[i];

♪ Ans = 0;

for (int i = 0; i < n; i ++)

Ans = max (ans, sum + min (a[i], b[i]));

What's going on?

♪ I'm sorry ♪

Intmain(){

OS:: sync with stdio (false);

cin.tie (0);

Int t;

cin> > t;

(b) While (t-);

♪ I'm sorry ♪

2229C1 - We Be Flipping (Easy Version) 2229C2 - We Be Flipping (Hard Version)

Idea by: Intellegent

Prepared by:

Editoral by: reirigan

You can make this elegance without placing any later elegations by applying an operation to it directly.

Let

Read the points. In fact, it is always possible to make all elections before

To make all of the elections before

#IncludeWe're talking namespace std;

typedef long long longll;

#definine debug(x) cout < #x < "= "< x < "\n";

#define vdebug(a) cout < < #a < < " = " ; for (auto x: a) cout < < < < " ; cout < < " \ " ;

mt19937 rang (chrono:steady clock:now() .time since epoch() .count());

Intuid(int a, int b)(a, b) (rang);}

ll uld(ll a, ll b)(a, b) (rang);}

void Solve(){

I don't know.

I'm sorry.

vector a (n);

For (int & x: a) cin > x;

Int par = 0;

vector I don't know.

For (int i = n - 1; i > = 0; i-){

If (par = 1)

a[i] = -a[i];

If (a[i] > 0){

Ans.push back(i);

Par ^ 1;

♪ I'm sorry ♪

♪ I'm sorry ♪

Cout < ns. size() < "n";

for (int i = 0; i < ans. size(; i ++)

cout < < ans[i] + 1 < "\n" [i = ans.size() - 1];

♪ I'm sorry ♪

Intmain(){

OS:: sync with stdio (false);

cin.tie (0);

Int t;

cin> > t;

(b) While (t-);

♪ I'm sorry ♪

#IncludeWe're talking namespace std;

typedef long long longll;

#definine debug(x) cout < #x < "= "< x < "\n";

#define vdebug(a) cout < < #a < < " = " ; for (auto x: a) cout < < < < " ; cout < < " \ " ;

mt19937 rang (chrono:steady clock:now() .time since epoch() .count());

Intuid(int a, int b)(a, b) (rang);}

ll uld(ll a, ll b)(a, b) (rang);}

void Solve(){

I don't know.

I'm sorry.

vector a (n);

For (int & x: a) cin > x;

vector pre(n), suf(n + 1);

[0] = abs (a [0];

for (int i = 1; i < n; i ++)

Pre[i] = pre[i-1] + abs (a[i];

suf[n - 1] = a [n - 1];

For (int i = n - 2; i > = 0; i--)

suf[i] = suf[i + 1] + a[i];

ll best = suf [0];

Int idx = -1;

for (int i = 1; i < n; i++){

If (a[i] > 0){

ll score = pre[i - 1] + suf [i + 1] - a[i];

If (score > best){

Best = score;

idx = i;

♪ I'm sorry ♪

♪ I'm sorry ♪

♪ I'm sorry ♪

If (idx = -1)

I'm sorry.

I don't know.

♪ I'm sorry ♪

vector I don't know.

For (int i = idx-1; i > = 0; i--){

If (ans.size() & 1)

a[i] = -a[i];

If (a[i] > 0)

Ans.push back(i);

♪ I'm sorry ♪

Ans.push back(idx);

Cout < ns. size() < "n";

for (int i = 0; i < ans. size(; i ++)

cout < < ans[i] + 1 < "\n" [i = ans.size() - 1];

♪ I'm sorry ♪

Intmain(){

OS:: sync with stdio (false);

cin.tie (0);

Int t;

cin> > t;

(b) While (t-);

♪ I'm sorry ♪

Idea by: Intellegent

Prepared by:

Editoral by: reirigan

Try solving the problem on a binary basis; that is, all values of

Supper you want to check if it is possible to achieve a final value of at least

This allows you to believe in the final answer.

- Yes, sir.

Now, it is impossible to address

#IncludeWe're talking namespace std;

typedef long long longll;

#definine debug(x) cout < #x < "= "< x < "\n";

#define vdebug(a) cout < < #a < < " = " ; for (auto x: a) cout < < < < " ; cout < < " \ " ;

mt19937 rang (chrono:steady clock:now() .time since epoch() .count());

Intuid(int a, int b)(a, b) (rang);}

ll uld(ll a, ll b)(a, b) (rang);}

void Solve(){

I don't know.

I'm sorry.

vector a (n), b (n);

For (int & x: a) cin > x;

For (int & x: b) cin > x;

Int l = 0, r = 1e9 + 5;

While (l< r){

Int Mid = (l + r) / 2;

Int one = 0, zero = 0;

Not prev = -1;

for (int i = 0; i < n; i+){

Int type = 0;

If (a[i] > = mid)

type++;

If (b[i] > = mid)

type++;

If (type = 1)

I'm sorry.

If (type = 2)

One++;

Prev = 1;

♪ I'm sorry ♪

If (type = 0){

If (prev! = 0)

zero++;

Prev = 0;

♪ I'm sorry ♪

♪ I'm sorry ♪

If (one > zero)

l = mid + 1;

I'm sorry, else.

r = mid;

♪ I'm sorry ♪

I'm sorry, I'm sorry.

♪ I'm sorry ♪

Intmain(){

OS:: sync with stdio (false);

cin.tie (0);

Int t;

cin> > t;

(b) While (t-);

♪ I'm sorry ♪

Idea by: Intellegent

Prepared by:

Educational by:

Try rooting at

Read the points, through this editoral I will consider

Lets try to define the following for each node.

& max For this to happen there must be some leave outside the subtree than each element of the subtree. If

We can solve this problem with dynamic programming where

It should be note that transfers to

#IncludeWe're talking namespace std;

typedef long long longll;

#definine debug(x) cout < #x < "= "< x < "\n";

#define vdebug(a) cout < < #a < < " = " ; for (auto x: a) cout < < < < " ; cout < < " \ " ;

mt19937 rang (chrono:steady clock:now() .time since epoch() .count());

Intuid(int a, int b)(a, b) (rang);}

ll uld(ll a, ll b)(a, b) (rang);}

= 998244353;

timedate / template was not available from https://codeforce.com/profile/SharpEdd

Other Organiser

Static constexpr bool is big mod =mod > numeric limites:: max();

Using S = general t;

Using L = general t;

S x;

Modenum(): x(0){

modnum(ll x){

x %= status cast(mod);

x+=mod;}

x = x;

♪ I'm sorry ♪

modnum pow (ll n) const {

Modnum rees = 1;

Modnum Cur = *this;

While (n > 0) {

If (n & 1) res* = cur;

Cur*=cur;

n = 2;

♪ I'm sorry ♪

I'm sorry.

♪ I'm sorry ♪

modnum inv() const {return(*this). pow(mod-2;}

{(contmodnum&a){

x + = a.x;

If (x > = mod) x - = mod;

What's going on?

♪ I'm sorry ♪

{(contmodnum&a){

If (x < a.x) x + = mod;

x - = a.x;

What's going on?

♪ I'm sorry ♪

== sync, corrected by elderman ==

x = status cast(x) *.x %mod;

What's going on?

♪ I'm sorry ♪

=(contmodnum & a)

{return Modenum(a)+=b;}

{return Modnum(a) - =b;}

*

}

== sync, corrected by elderman == @elder man

== sync, corrected by elderman ==

{cont MODnum & b) {return a. x < b.x;}

{os < a.x; return os;}

{ll x; is > x; a = modnum (x); return is;}

};

We're going to have to go to the hospital.;

void Solve(){

I don't know.

I'm sorry.

vector> g (n);

for (int i = 0; i < n - 1; i ++){

Int u, v;

cin > u > v;

u-;

v-;

g [u].push back(v);

g[v].push back(u);

♪ I'm sorry ♪

If (g[n-1]. size()=1){

What's going on?

I don't know.

♪ I'm sorry ♪

vector mx(n);

Autu dfs = [ ] (auto self, int c, int p) - >void{

For (int x: g[c]{

If (x = p)

I'm sorry.

Self (self, x, c);

mx[c] = max({mx[c], x, mx[x]});

♪ I'm sorry ♪

};

dfs (dfs, n - 1, 1);

Set> s;

for (int i = 0; i < n - 1; i ++)

s.insert(i);

vector I'm sorry, Cur;

Auto dfs2 = [ /] (auto self, int c, int p) - >void{

s.erace(c);

cur.push back(c);

For (int x: g[c]{

If (x = p)

I'm sorry.

Self (self, x, c);

♪ I'm sorry ♪

};

vector OK (n); / can change to n

For (int x: g[n-1]){

dfs2 (dfs2, x, n - 1);

For (int u: cur){

If (u > (*s.begin()))

Ok[u] = true;

♪ I'm sorry ♪

For (int u: cur)

s.insert (u);

Cur.clar();

♪ I'm sorry ♪

[n-1] = true;

Int idx = -1;

for (int i = 0; i < n; i+){

If (g[i]. size() = 1)

idx = i;

♪ I'm sorry ♪

vector dp(n), pre(n);

dp [idx] = 1;

= 1;

for (int i = idx + 1; i < n - 1; i ++){

Int l = mx[i] + 1;

If (l< i){

dp[i] = pre[i-1];

If (l > 0)

dp[i] - = pre[l-1];

♪ I'm sorry ♪

pre[i] = dp[i] + pre[i-1];

♪ I'm sorry ♪

mint ans = 0;

for (int i = 0; i < n; i+){

If (ok[i])

ns + = dp[i];

♪ I'm sorry ♪

What's going on?

♪ I'm sorry ♪

Intmain(){

OS:: sync with stdio (false);

cin.tie (0);

Int t;

cin> > t;

(b) While (t-);

♪ I'm sorry ♪

Idea by: myst-6

Prepared by:

Editoral by: sammyuri

Which element of

Now what does the problem return to you?

To maximise the minute value of

We can make the bitmask dp in the form dp [mask] = number of groups, leftover. There are

We can use this greeny algorithm to protect that

#Include We're talking namespace std;

typedef long long longll;

#define all(x) x.begin(), x.end()

#define vecin name(len); for (auto & name) cin > > ;

#define vecoout(v) for (auto : v) cout < < < " ; cout < endl;

void Solve() {

Not n, k; cin > n > k;

(aa, n); sort (all (aaa));

No, no, no, no.

ll extra = aa.back(); aa.pop back();

ll L = 0, R = extra * (ll)n;

& While

ll Mid = (L + R + 1) / 2;

vector> dp (1 < n, {0, )

for (int i = 1; i < (1< n; i ++){

for (int j = 0; (1< j) < = i; j ++) {

If! (i & (1< j)) continue;

Int res = i ^ (1 < j);

Pair Cur = dp[res];

Cur.second+=aa[j];

If (cur.second > = mid)

Other Organiser

dp[i] = max (dp[i], cur);

♪ I'm sorry ♪

♪ I'm sorry ♪

If (dp[ (1 < n) - 1]. first > = k)

L = mid;

I'm sorry, else.

R = mid-1;

♪ I'm sorry ♪

Not < L + extra < < endl;

♪ I'm sorry ♪

inmain(){

OS base:: sync with stdio (false);

cin.tie (NULL);

Inttt = 1;

cin > tt;

(b) while (tt...) Solve();

Return 0;

♪ I'm sorry ♪

Idea by:

Prepared by:

Editoral by: sammyuri

When is it to stay at a house for more than

Supper you are staying at a house for more than

Supper you are staying at a house for more than

First, until that any house you stay at for longer than

Also, if you are going to stay at a house for longer than

Finely, if you are going to switch directs in an alternative solution, you should go all the way past house

We know that for each house, the average answer on that house at time

For the sake of case, we can maintain which house is the prefix maxim on both sides, and for the better case case, we can use force to keep track of the best conservation if we respond on house

Supper you are staying at a house for more than

In fact, we can show that whenever you reach a prefix maxim, the next house you will move to is either the next High Prefix maxim on that side of

#Include We're talking namespace std;

typedef long long longll;

#define all(x) x.begin(), x.end()

#define vecin name(len); for (auto & name) cin > > ;

#define vecoout(v) for (auto : v) cout < < < " ; cout < endl;

Standard Line

I don't know.

{(cont Line& o)Const {return k < o.k;}

{ll x) const {return p < x;}

ll eval(ll x)

};

Standard LineContainer: multiset> {

= LLONG MAX;

ll div(ll a,ll b)

return a / b -((a ^ b) < 0 & a %b; }

Bool test

If (y = end()) return x->p = inf, 0;

If (x->k = = y->k) x-mp = x-m > y-m? inf:-inf;

=div (y->m-x-m, x->k-y->k);

Return x->p = y->p;

♪ I'm sorry ♪

void add(ll k,ll m)

auto z = insert(k, m, ), y = z++, x = y;

While (isect(y,z)) z = erace(z);

If (x! = begin() & & subject(-, y) isect(x, y = example(y));

While ((y = x)! = begin() & (-x)->p > = y->p)

isect(x, example(y));

♪ I'm sorry ♪

ll query(ll x){

Asert(!impty());

Auto l = *lower born (x);

Return l.eval(x);

♪ I'm sorry ♪

};

vector h;

vector I'm sorry, pref;

llrange sum(int L, int R){

Return pref[R + 1] - pref[L];

♪ I'm sorry ♪

void Solve() {

Int n, t, x; cin > n > t > x;

x-;

h.resize(n); pref.resize(n+1);

Pref[0] = 0;

for (int i = 0; i < n; i ++){

cin > h[i];

pref[i + 1] = pref[i] + h[i];

♪ I'm sorry ♪

vecin(d, n-1);

for (int i = x - 2; i > = 0; i --) {

d [i] = max(d[i], d[i + 1] + 1);

♪ I'm sorry ♪

for (int i = x; i < n - 2; i ++) {

d [i + 1] = max (d [i + 1], d [i] + 1);

♪ I'm sorry ♪

ll ans = h[x]*t;

Int L = x, R = x;

Line Curleft = h[x], 0,

Int left best=x, right best=x;

LineContainer left, right?

left.add(h[x], 0); right.add(h[x], 0);

Que.>add left, add right;

While (L > 0 ||R< n-1) {

Int left time=(L=0)? 2e9 : d [L-1];

Int right time = (R = n - 1) 2e9: d[R];

> t) break;

If (left time< = right time)

L-;

If (h[L] > Curleft.k) {

/ best = max sum on day left time-1 on house L+1

= left best-L;

Instart dist=x-L;

While (add right. size() & add right. front(). first < = left time - start dist) {

Right.add(add right.front().second.first, add right.front().second.second.second);

Add right.pop();

♪ I'm sorry ♪

ll best = max

+ renge sum (L + 1, left best-1),

Right. querry (left time -start dist)+range sum (L+1, x-1)

(b) ;

/ Cout < L < " < h [L] < " < best < endl;

Ans = max(ans, best + (t - left time + 1) *h [L];

left best=L;

Curleft = {h[L], best - (left time - 1) *h [L],

dd left.push({left time + start dist, {h[L], best - (left time-1 + start dist) *h[L] + range sum(L+1, x)});

♪ I'm sorry ♪

♪ Orse ♪

R ++;

If (h[R] > control.k) {

/ best = max sum on day right time-1 on house R-1

= R-right best;

Instart dist=R-x;

While (add left. size() & add left. front(). first < = right time-start dist) {

Left.add (add left.front().second.first, add left.front().second.second);

Add lift.pop();

♪ I'm sorry ♪

ll best = max

Right time - pref dist)+range sum (right best+1, R-1),

Left. Query (right time - start dist)+range sum (x + 1, R-1)

(b) ;

/ cout < R < " < h [R] < " < best < endl;

Ans = max(ans, best + (t - right time + 1) *h [R];

Right best=R;

== sync, corrected by elderman == @elder man

Add right.push({right time + start dist, {h[R], best - (right time-1 + start dist) *h [R] + range sum(x, R-1}});

♪ I'm sorry ♪

♪ I'm sorry ♪

♪ I'm sorry ♪

What's up?

♪ I'm sorry ♪

inmain(){

OS base:: sync with stdio (false);

cin.tie (NULL);

Inttt = 1;

cin > tt;

(b) while (tt...) Solve();

Return 0;

♪ I'm sorry ♪

#Include We're talking namespace std;

We'll = long long long;

= vector;

= vector;

We're using pl = pair;

timedate = vector;

timedate =0;

template

Oh, yeah. = 2'020'000'000'000'000'000'000'000;

#definine FOR2(i, a) for (int i = 0; i < int(a); +i)

#definine FOR3 (i, a, b) for (int i = int(a); i < int(b); +i)

#definine FOR2 R(i, a) for (int i = int(a)-1; i > = 0; --i)

#definine overload4 (a, b, c, d, e, e)

#definine overload3 (a, b, c, d, ...) d

#definine FOR(...) overload4( VA ARGS , 4, FOR3, FOR2, 1) ( VA ARGS )

#definine FOR R(...) overload3( VA ARGS , 3, FOR2 R, 1) ( VA ARGS )

#define all(x).begin(),(x).end()

#definine track(a, x) for (auto& a: x)

#Define eb place back

# Define mp make

# Definine fi first

#Define us second #

timedate Inline bool chmax (T &a, const S & b){

T c = max(a, b);

Bool changed = (c! = a);

a = c;

I'm sorry.

♪ I'm sorry ♪

timedate void rd (T & x){cin > x;}

timedate Void rd(vc) & x){for (auto & d: x)rd(d);}

void read()}

timedate void read (H & h, T & ...t){rd(h), read(t...);}

void pent(){cout< 'n';}

timedate Tail

void print (head & head, tail &...tail) {

I don't know.

If (sizeof... (Tail)) cut < ' ;

print(forward)(tail)...

♪ I'm sorry ♪

#definine INT(...) int VA ARGS; read

#definine LL(...)ll VA ARGS; read( VA ARGS )

#definine VEC (type, name, size) name(size); read(name)

void Solve() {

INT(N); LL(t); INT(X);

X-;

VEC (ll, A, N);

VEC (ll, D, N-1);

vl T(N);

T[X] = 1;

FOR(i, X, N-1){

T[i+1] = max (T[i]+1, D[i];

If (i = = X & D [i] = 1) T [i + 1] = 1;

♪ I'm sorry ♪

FOR R(i, X){

T[i] = max (T[i+1]+1, D[i];

(i) = X-1 & D[i] = 1) T[i] = 1;

♪ I'm sorry ♪

vi (N), lh (N);

vc What are you doing?

stk.eb (mp(N, 2e9);

FOR R(i), {

While (stk.back().se < = A[i]) stk.pop back();

n[i] = stk.back().fi;

stk.eb (mp(i, A[i)]);

♪ I'm sorry ♪

stk.clar();

stk.eb (mp(-1, 2e9);

& FOR(i), {

While (stk.back().se < = A[i]) stk.pop back();

lh[i] = stk.back().fi;

stk.eb (mp(i, A[i)]);

♪ I'm sorry ♪

ll dp [N];

& FOR(i), {

dp[i]=-infoty;

♪ I'm sorry ♪

For(i), N) dp[i] = 0;

vc vals; FOR(i), N) vals.eb (mp(T[i], i));

sort (vals);

♪ Ans = 0;

vl ps; ps.eb (0);

FOR(i), N) ps.eb(ps.back()+A[i]);

Trav(xx, vals){

Int i = xx.se;

If (dp [i] < 0) continue;

Chmax(ans, (tot-T[i]+1) *A[i] + dp[i];

vi ops = n[i], lh[i]};

Trav(j, ops){

If (j < 0 j > = N) continue;

If (abs(j-i) > T[j] - T[i] {

I'm sorry.

♪ I'm sorry ♪

Chmax(dp[j], dp[i]+ (T[j] - T[i] - abs(j-i) + 1) *A[i] + ps[max(i, j)] - ps[min(i, j) + 1];

♪ I'm sorry ♪

♪ I'm sorry ♪

(e) print(s);

♪ I'm sorry ♪

Signed Main()

OS:: sync with stdio (false);

cin.tie (nulrptr);

INT (T);

FOR(t), {t]

Solve();

♪ I'm sorry ♪

Return 0;

♪ I'm sorry ♪

Idea by: Intellegent

Prepared by:

Educational by:

Imagine the string has no

Try to generalise this greeny for when there are

Read the points.

We would like to find a situation of

Say we have a match where at some point we didn't choose the early match, say between

Now lets try to promote this when there are

It fails because we properly assume the value of a

To fix this problem, we can assume our greeny to consider both parents at the same time.

The prof for this greeny is simple to the naïve greeny in the no

Now for accounting, we can run dynamic programming on this greeny. Where?

This is the final time.

#IncludeWe're talking namespace std;

typedef long long longll;

#definine debug(x) cout < #x < "= "< x < "\n";

#define vdebug(a) cout < < #a < < " = " ; for (auto x: a) cout < < < < " ; cout < < " \ " ;

mt19937 rang (chrono:steady clock:now() .time since epoch() .count());

Intuid(int a, int b)(a, b) (rang);}

ll uld(ll a, ll b)(a, b) (rang);}

= 998244353;

timedate / template was not available from https://codeforce.com/profile/SharpEdd

Other Organiser

Static constexpr bool is big mod =mod > numeric limites:: max();

Using S = general t;

Using L = general t;

S x;

Modenum(): x(0){

modnum(ll x){

x %= status cast(mod);

x+=mod;}

x = x;

♪ I'm sorry ♪

modnum pow (ll n) const {

Modnum rees = 1;

Modnum Cur = *this;

While (n > 0) {

If (n & 1) res* = cur;

Cur*=cur;

n = 2;

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I'm sorry.

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modnum inv() const {return(*this). pow(mod-2;}

{(contmodnum&a){

x + = a.x;

If (x > = mod) x - = mod;

What's going on?

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{(contmodnum&a){

If (x < a.x) x + = mod;

x - = a.x;

What's going on?

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== sync, corrected by elderman ==

x = status cast(x) *.x %mod;

What's going on?

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=(contmodnum & a)

{return Modenum(a)+=b;}

{return Modnum(a) - =b;}

*

}

== sync, corrected by elderman == @elder man

== sync, corrected by elderman ==

{cont MODnum & b) {return a. x < b.x;}

{os < a.x; return os;}

{ll x; is > x; a = modnum (x); return is;}

};

We're going to have to go to the hospital.;

void Solve(){

I don't know.

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I don't know.

cin > s;

s = '0' + s;

n+;

vector Pre(n);

= s[0] = = '1';

for (int i = 1; i < n; i++){

Pre[i] = pre[i-1] + (s[i] = = '1');

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vector,2> to (n + 1);

For (int i = 0; i < = n; i ++){

To [i][0][0] = n;

To [i] [0][1] = n;

To [i] [1][0] = n;

To [i] [1][1] = n;

Int cnt = 0;

Bool wowee = false;

For (int j = i + 1; j < n; j +){

If(cnt & 1) &!wowe){

If (s[j] = = '1')

cnt++;

If (s[j] = = '?')

= true;

I'm sorry.

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If (s[j] = = '1')

cnt++;

If (s[j] = = '?')

= true;

Int par = pre[j] & 1;

If (s[j] = = '?')

To [i] [0] [par] = min (to [i] [0] [par], j);

To [i] [1] [par] = min (to [i] [1] [par], j);

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Int og = s[j] - '0';

To [i] [omg] [par] = min (to [i] [omg] [par], j);

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vector Good (n + 1);

Int cnt = 0;

Int wowee = false;

For (int i = n - 1; i > = 0; i-){

good[i] = (cnt %2 = 0) || wowee;

If (s[i]=='1')

cnt++;

If (s[i] = = '?')

= true;

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mint ans = 0;

vector>dp(n + 1, vector(n + 1)

dp [0][n] = 1;

vector> Ord;

For (int i = 0; i < = n; i ++){

for (int j = i; j < = n; j ++){

Ord.push back(i,j});

If (j! = i)

Ord.push back({j, i});

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For (auto [i, j]: ord){

For (int nxt = 0; nxt < 2; nxt++){

Int even =min (to [i][nxt][0], to [j][nxt][0];

Int odd = min (to [i] [nxt [1], to [j] [nxt [1]);

dp[even][odd]+=dp[i][j];

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If (good[i] || good[j])

ns + = dp[i] [j];

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What's going on?

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Intmain(){

OS:: sync with stdio (false);

cin.tie (0);

Int t;

cin> > t;

(b) While (t-);

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Idea by:

Prepared by:

Editoral by: sammyuri

It's not convex, montone, FFT, controid, or any other fancy trick like that. Reformulate the future for scure so that it's in terms of nodes in the subtree from root, not the path to the root.

How many DPs do you compute for the node exclusively care about?

[i] be the maximme score of a subset of nodes of nodeline of nodei, assuring the self-made root isi or an objector of i, who uses literally nodes.

Let's get together.

Supplose we are at a node

Observe that we only need to compute

All that remains is to add the condition of node

Then, for a nonde

We can make it one!

The only thing to watch out for is that communication of the original node

#Include We're talking namespace std;

typedef long long longll;

#define all(x) x.begin(), x.end()

#define vecin name(len); for (auto & name) cin > > ;

#define vecoout(v) for (auto : v) cout < < < " ; cout < endl;

Const int MAXN = 4005;

ll dp down [MAXN] [MAXN], dp down take [MAXN] [MAXN], dp up [MAXN] [MAXN];

ll pref merge [MAXN] [MAXN], suff merge [MAXN] [MAXN];

vector Adj [MAXN];

vector [MAXN];

♪ Light [MAXN];

Int sz [MAXN];

Inn n, k;

void dfs1 (int node, in parent)

sz [node] = 1;

For (auto a: adj [node])

If (a! = parent)

Children [node].push back(a);

dfs1 (a, node);

For (int i = min (sz [node] - 1, k); i > = 0; i-)

for (int j = 1; j < = sz [a] & i + j < = k; j +)

dp down [node] [i + j] = max (dp down [node] [i + j], dp down [node] [i] + dp down take[a] [j];

sz [node]+=sz[a];

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For (int i = min(k, sz [node]; i > 0; i-)

dp down take [node][i] = max (dp down [node][i], dp down [node][i-1] + (ll)i * weight [node];

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void dfs2 (int node){

Int deg = children [node]. size();

If (deg = 1) {

for (int i = 0; i < = k; i ++)

dp up [children [node][0] [i]=dp up[node][i];

{else if (deg > = 2) {

Int p = sz [children [node][0];

for (int i = 0; i < = k; i ++)

= dp down take [children [node][0]][i],

suff merge[deg][i]=dp up[node][i];

for (int c = 1; c < dig; c ++){

for (int i = 0; i < = k; i ++)

= 0;

Int child = children [node] [c];

p + = sz [child];

for (int i = min (k, p); i > = 0; i-)

For (int j = min (i, sz [child]; j > = 0 & & i - j < = p - sz [child]; j --)

pref merge[c][i] = max (pref merge[c][i], pref merge[c-1][i-j]+dp down take[child][j]]);

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For (int c = dig-1; c > = 0; c-) {

for (int i = 0; i < = k; i ++)

suff merge[c][i] = 0;

Int child = children [node] [c];

p - = sz [child];

for (int i = k; i > = max (0, k - p - 1); i-)

For (int j = min (i, sz [child]; j > = 0; j --)

suff merge[c][i] = max (suff merge[c][i], suff merge[c + 1] [i-j] + dp down take[child] [j]]);

for (int i = k; i > = max (0, k - sz [child] - 1); i-)

For (int j = min (i, p); j > = 0; j-)

dp up[child][i] = max(dp up[child][i], suff merge[c + 1] [i-j]+(c > 0]? pref merge[c-1][j]:0);

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For (auto a: children [node])

for (int i = min(k, n - sz[a]; i > = 1; i-)

dp up[a] [i] = max (dp up [a] [i], dp up[a] [i-1] + (ll)i * weight [node]]);

dfs2(a);

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void Solve() {

cin > n > k;

for (int i = 0; i < n; i ++){

for (int j = 0; j < = k; j ++)

dp down[i] [j]=dp down take[i] [j]=dp up[i] [j]=0;

Adj [i].clar(); children[i].clar();

cin> weight[i];

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For (int i = 0; i < n - 1; i ++){

Int a, b; cin > a > b;

a; b -;

Adj [a].push back(b);

Adj [b]. Push back(a);

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dfs1 (0, 1);

dfs2(0);

for (int i = 0; i < n; i ++){

ll best = 0;

for (int j = 0; j < min(k, sz[i]; j ++)

Best = max (best, dp down[i] [j] + dp up[i] [k - 1 - j] + (ll)k * weight[i];

Cout < best < " ;

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I don't know.

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inmain(){

OS base:: sync with stdio (false);

cin.tie (NULL);

Inttt = 1;

cin > tt;

(b) while (tt...) Solve();

Return 0;

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#Include We're talking namespace std;

typedef long long longll;

#define all(x) x.begin(), x.end()

#define vecin name(len); for (auto & name) cin > > ;

#define vecoout(v) for (auto : v) cout < < < " ; cout < endl;

Const int MAXN = 4005;

Inn n, k;

I don't know.

vector Adj [MAXN], children [2* MAXN];

ll beauty [2* MAXN];

Int sz [2* MAXN];

ll dp down [2* MAXN] [MAXN], dp down take [2* MAXN] [MAXN];

ll dp up [2* MAXN] [MAXN];

{lll *target, *ll *bb, int tsize, int *bsize, int *

for (int i = tsize; i > max (0, tsize -topmost); i --)

for (int j = 0; j < = min (asize, bsize), i); j ++

= max (target[i], asize > =bsize? aa[i-j] + bb[j]: bb[i-j] + aa[j];

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void finalise (ll *target, ll time, ll b, int tsize)

for (int i = tsize; i > 0; i-)

+ (ll)i*b);

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void dfs1 (int node, in parent)

vector cc;

For (auto a: adj [node])

If (a! = parent)

dfs1 (a, node);

cc.push back(a);

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Incur = node;

While (cc. size() > 2) {

Children [cur] = {cc.back(), spare};

cc.pop back();

Beauty [spare] = beauty [node];

Cur = spare ++;

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Children [cur] = cc;

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void dfs2 (int node){

sz [node] = (node < n);

For (auto a: children [node])

dfs2(a);

sz [node]+=sz[a];

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If (children [node]. size()=1) {

for (int i = 0; i < = k; i ++)

dp down [node][i] = dp down take[node][0] [i];

{Other if (children [node]. size()=2)}

Merge (dp down [node], dp down take [children [node] [0], dp down take [children [node] [1]), Min (k, sz [node] - (node < n)), sz [children [node] [node], sz [children [node][1], sz [node] - (node < n)]);

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Finalise (dp down take[node], dp down [node], node < n ? beauty [node]: 0, min(k, sz [node]]);

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void dfs3 (int node){

If (children [node]. size()=1) {

(dp up [children [node][0]], dp up [node], pretty [node], min (k, n - sz [children [node][0]]));

{Other if (children [node]. size()=2)}

Int aa = children [node][0], b b = children [node][1];

(dp up [aa], dp up [node], dp down take [bb], Min (n - sz [aa], k), Min (n - sz [node], k), Min (sz [bb], k), Min (sz [aa] + 2, k));

(dp up [bb], dp up [node], dp down take [aa], Min (n - sz [bb], k), Min (n - sz [node], k), Min (sz [aa], k), Min (sz [bb] + 2, k));

If (aa < n)

Finalise (dp up [aa], dp up [aa], pretty [node], min (n - sz [aa], k));

If (bb < n)

Finalise (dp up [bb], dp up [bb], pretty [node], min (n - sz [bb], k));

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For (auto a: children [node])

dfs3 (a);

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void Solve() {

cin > n > k;

for (int i = 0; i < n; i ++)

Adj [i].clar(), children[i].clar();

For (int i = n; i < 2 * n; i ++)

Children[i].clar();

spare = n;

for (int i = 0; i < n; i ++)

cin > beauty[i];

For (int i = 0; i < n - 1; i ++){

Int a, b; cin > a > b;

a; b -;

Adj [a].push back(b);

Adj [b]. Push back(a);

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dfs1 (0, 1);

for (int i = 0; i < spare; i ++)

for (int j = 0; j < = k; j ++)

dp down[i] [j]=dp up[i] [j]=dp down take[i] [j]=0;

dfs2(0);

dfs3 (0);

for (int i = 0; i < n; i ++){

ll best = 0;

for (int j = 0; j < min(k, sz[i]; j ++)

best = max (best, dp down[i][j] + dp up[i][k - 1 - j] + (ll)k * pretty[i];

Cout < best < " ;

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I don't know.

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inmain(){

OS base:: sync with stdio (false);

cin.tie (NULL);

Inttt = 1;

cin > tt;

(b) while (tt...) Solve();

Return 0;

♪ I'm sorry ♪