Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin >> t;
for(int i = 0; i < t; i++)
{
int x, y;
cin >> x >> y;
int ans = max(x, y) * 2 - 1;
if(x == y) ans++;
cout << ans << endl;
}
}
```

Idea: adedalic

**Tutorial**

Tutorial is loading...

**Solution (adedalic)**

```
fun main() {
repeat(readLine()!!.toInt()) {
val n = readLine()!!.toInt()
val a = readLine()!!.split(' ').map { it.toLong() }
val k = maxOf(a.max()!!, (a.sum() + n - 2) / (n - 1))
println(k * (n - 1) - a.sum())
}
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (pikmike)**

```
def calc(s, x, y):
bal, cnt = 0, 0
for c in s:
if c == y:
if bal > 0:
bal -= 1
cnt += 1
elif c == x:
bal += 1
return cnt
for _ in range(int(input())):
s = input()
print(calc(s, '(', ')') + calc(s, '[', ']'))
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
int add(int x, int y)
{
x += y;
while(x >= MOD) x -= MOD;
while(x < 0) x += MOD;
return x;
}
int mul(int x, int y)
{
return (x * 1ll * y) % MOD;
}
int binpow(int x, int y)
{
int ans = 1;
while(y > 0)
{
if(y % 2 == 1)
ans = mul(ans, x);
x = mul(x, x);
y /= 2;
}
return ans;
}
int divide(int x, int y)
{
return mul(x, binpow(y, MOD - 2));
}
int main()
{
int n;
cin >> n;
vector<int> fib(n + 1);
fib[0] = 0;
fib[1] = 1;
for(int i = 2; i <= n; i++)
fib[i] = add(fib[i - 1], fib[i - 2]);
cout << divide(fib[n], binpow(2, n)) << endl;
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (pikmike)**

```
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
struct seg{
int l, r;
};
int main() {
int n, m, k;
cin >> n >> m >> k;
vector<seg> a(m);
forn(i, m){
cin >> a[i].l >> a[i].r;
--a[i].l;
}
sort(a.begin(), a.end(), [](const seg &a, const seg &b){
return a.l + a.r < b.l + b.r;
});
vector<int> su(m + 1);
forn(i, n - k + 1){
int cur = 0;
for (int j = m - 1; j >= 0; --j){
cur += max(0, min(i + k, a[j].r) - max(i, a[j].l));
su[j] = max(su[j], cur);
}
}
int ans = su[0];
forn(i, n - k + 1){
int cur = 0;
forn(j, m){
cur += max(0, min(i + k, a[j].r) - max(i, a[j].l));
ans = max(ans, cur + su[j + 1]);
}
}
cout << ans << endl;
return 0;
}
```

Idea: adedalic

**Tutorial**

Tutorial is loading...

**Solution (adedalic)**

```
#include<bits/stdc++.h>
using namespace std;
#define fore(i, l, r) for(int i = int(l); i < int(r); i++)
#define sz(a) int((a).size())
#define x first
#define y second
typedef long long li;
typedef long double ld;
typedef pair<int, int> pt;
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << ", " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
out << "[";
fore(i, 0, sz(v)) {
if(i) out << ", ";
out << v[i];
}
return out << "]";
}
const int INF = int(1e9);
const li INF64 = li(1e18);
const ld EPS = 1e-9;
int n, q;
vector<li> cnt;
inline bool read() {
if(!(cin >> n >> q))
return false;
cnt.assign(n, 0);
fore (i, 0, n)
cin >> cnt[i];
return true;
}
inline void solve() {
fore (qs, 0, q) {
int tp, pos;
li val;
cin >> tp >> pos >> val;
if (tp == 1) {
cnt[pos] = val;
} else {
li small = 0, cur = 0;
fore (i, 0, pos + 1) {
small += cnt[i] * ((1ll << i) - 1);
val -= cnt[i];
}
if (val <= 0) {
cout << 0 << '\n';
continue;
}
int id = pos + 1;
while (id < n) {
li add = 1ll << (id - pos);
li need = min(val / add, cnt[id]);
cur += need * (add - 1);
val -= need * add;
small += need * add * ((1ll << pos) - 1);
if (need == cnt[id])
id++;
else
break;
}
if (val <= 0) {
cout << cur << '\n';
continue;
}
if (id >= n) {
cout << (val > small ? -1 : cur + val) << '\n';
continue;
}
li ans = INF64;
while (id > pos) {
if (small >= val)
ans = min(ans, cur + val);
cur++;
id--;
li add = 1ll << (id - pos);
if (val >= add) {
cur += add - 1;
val -= add;
small += add * ((1ll << pos) - 1);
}
}
assert(val <= 0);
cout << min(ans, cur) << endl;
}
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
int tt = clock();
#endif
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
cout << fixed << setprecision(15);
if(read()) {
solve();
#ifdef _DEBUG
cerr << "TIME = " << clock() - tt << endl;
tt = clock();
#endif
}
return 0;
}
```

Idea: adedalic

**Tutorial**

Tutorial is loading...

**Solution 1 (pikmike)**

```
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
int n;
vector<vector<int>> g;
vector<int> h, pcd, d;
vector<vector<int>> st, vals;
vector<int> a;
int dfs(int v, int s, int &cd, int p = -1){
int sum = 1;
for (int u : g[v]) if (h[u] == -1 && u != p)
sum += dfs(u, s, cd, v);
if (cd == -1 && (2 * sum >= s || p == -1))
cd = v;
return sum;
}
void build(int v, int s, int d, int p = -1){
int cd = -1;
dfs(v, s, cd);
h[cd] = d;
pcd[cd] = p;
for (int u : g[cd]) if (h[u] == -1)
build(u, s / 2, d + 1, cd);
}
vector<char> cur;
void calcd(int v, int p = -1){
for (int u : g[v]) if (u != p && cur[u]){
d[u] = d[v] + 1;
calcd(u, v);
}
}
vector<vector<int>> dist;
void init(){
a.resize(n, -1);
int k;
scanf("%d", &k);
queue<int> q;
forn(i, k){
int v;
scanf("%d", &v);
--v;
q.push(v);
a[v] = 0;
}
while (!q.empty()){
int v = q.front();
q.pop();
for (int u : g[v]) if (a[u] == -1){
a[u] = a[v] + 1;
q.push(u);
}
}
h.resize(n, -1);
pcd.resize(n);
build(0, n, 0);
st.resize(n);
vector<int> nd(n);
forn(v, n){
int u = v;
while (u != -1){
st[u].push_back(v);
if (pcd[u] != -1)
nd[pcd[u]] = max(nd[pcd[u]], nd[u] + 1);
u = pcd[u];
}
}
cur.resize(n);
vals.resize(n);
dist.resize(n);
vector<int> ord(n);
iota(ord.begin(), ord.end(), 0);
sort(ord.begin(), ord.end(), [&nd](int x, int y){
return nd[x] < nd[y];
});
d.resize(n);
for (int v : ord){
for (int u : st[v]) cur[u] = true;
d[v] = 0;
calcd(v);
int mx = 0;
for (int u : st[v]) mx = max(mx, d[u]);
vals[v].resize(mx + 1, 0);
for (int u : st[v]) vals[v][d[u]] = max(vals[v][d[u]], a[u]);
forn(j, mx) vals[v][j + 1] = max(vals[v][j + 1], vals[v][j]);
for (int u : st[v]) cur[u] = false;
for (int u : st[v]) dist[u].push_back(d[u]);
}
}
bool check(int v, int x){
for (int i = 0, u = v; u != -1; u = pcd[u], ++i)
if (x - dist[v][i] >= 0 && vals[u][min(int(vals[u].size()) - 1, x - dist[v][i])] > x)
return true;
return false;
}
int main() {
scanf("%d", &n);
g.resize(n);
forn(i, n - 1){
int v, u;
scanf("%d%d", &v, &u);
--v, --u;
g[v].push_back(u);
g[u].push_back(v);
}
init();
forn(v, n){
int res = 0;
int l = 0, r = n;
while (l <= r){
int m = (l + r) / 2;
if (check(v, m)){
res = m + 1;
l = m + 1;
}
else{
r = m - 1;
}
}
printf("%d ", res);
}
puts("");
}
```

**Solution 2 (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
const int N = 200043;
const int LN = 18;
vector<int> g[N];
vector<int> dist[N];
int sz[N];
int par[N];
bool used[N];
int max_dist[N];
vector<int> val[N];
int calc_size(int x, int p = -1)
{
sz[x] = 1;
for(auto y : g[x])
if(y != p && !used[y])
sz[x] += calc_size(y, x);
return sz[x];
}
int find_centroid(int x, int p, int s)
{
int ans = -1;
bool good = true;
for(auto y : g[x])
if(y != p && !used[y])
good &= sz[y] * 2 <= s;
else if(y == p && !used[y])
good &= (s - sz[x]) * 2 <= s;
if(good)
ans = x;
for(auto y : g[x])
if(y != p && !used[y])
ans = max(ans, find_centroid(y, x, s));
return ans;
}
void calc_dist(int x, int p, int d, int s)
{
dist[x].push_back(d);
for(auto y : g[x])
if(y != p && !used[y])
calc_dist(y, x, d + 1, s);
max_dist[s] = max(max_dist[s], d);
}
int decomposition(int v)
{
calc_size(v);
int c = find_centroid(v, v, sz[v]);
used[c] = true;
for(auto y : g[c])
if(!used[y])
{
par[decomposition(y)] = c;
}
used[c] = false;
calc_dist(c, c, 0, c);
return c;
}
int main()
{
int n;
scanf("%d", &n);
for(int i = 0; i < n - 1; i++)
{
int x, y;
scanf("%d %d", &x, &y);
--x;
--y;
g[x].push_back(y);
g[y].push_back(x);
}
decomposition(0);
for(int i = 0; i < n; i++)
val[i].resize(max_dist[i] + 1);
int k;
scanf("%d", &k);
vector<int> d(n, int(1e9));
queue<int> q;
for(int i = 0; i < k; i++)
{
int x;
scanf("%d", &x);
--x;
q.push(x);
d[x] = 0;
}
while(!q.empty())
{
int x = q.front();
q.pop();
for(auto y : g[x])
if(d[y] > d[x] + 1)
{
q.push(y);
d[y] = d[x] + 1;
}
}
for(int i = 0; i < n; i++)
{
if(d[i] == 0) continue;
int curc = i;
for(int j = 0; j < dist[i].size(); j++)
{
int dd = dist[i][j];
if(dd > d[i] - 1)
{
curc = par[curc];
continue;
}
dd = d[i] - 1 - dd;
if(dd >= val[curc].size())
dd = val[curc].size() - 1;
val[curc][dd] = max(val[curc][dd], d[i]);
curc = par[curc];
}
}
for(int i = 0; i < n; i++)
for(int j = max_dist[i]; j >= 1; j--)
val[i][j - 1] = max(val[i][j], val[i][j - 1]);
for(int i = 0; i < n; i++)
{
int ans = 0;
int curc = i;
for(int j = 0; j < dist[i].size(); j++)
{
int dd = dist[i][j];
ans = max(ans, val[curc][dd]);
curc = par[curc];
}
if(d[i] == 0)
ans = 0;
printf("%d%c", ans, " \n"[i == n - 1]);
}
}
```

What about rating updates?

Who cares about rating, right?

I think, in tutorial G, Alice and Bob had been swapped. It's Alice who has one starting vertex, and it's Bob who chases her, not vice versa.

Oh, indeed. I've been told the problem with them reversed and I haven't read the actual statement haha. Will fix in a sec, thanks.

Can anyone explain problem A's solution a bit more. I couldn't get it!

See let

`min(x,y)=a`

Then,

`we can first go to (a,a) in exactly a*2 steps.`

After that we just have to increment in one direction and the best way to do that is by first waiting then moving because it always takes comparatively less number of steps,i.e, 2.Also note we can always reach (a,a) in two possible ways by first moving right or left, so when we take steps from (a,a) we can always increment one step without waiting.So,

`answer is (a*2)+1+max(y-x-1)*2 when (y-x-1)>=0`

`else answer is simply (a*2) when x==y`

You can look at my submission 98961942. Hope you find it helpful !

can we write second condition when x!=y as 2*max(x,y)-1 ?? if not please give some examples sorry i didn't read the editorial before

why i get tle in test case 20 of problem E

my submission : 99017296

logic

Comparator for sorting should follow strict weak ordering. If $$$a=b$$$ then it should always return false, otherwise it's undefined behavior.

thanks it work now .

If anyone didn't understand B. Here is an easy solution-

`For solution to exist let us consider an element any index i.`

Then, for solution to exist we should be able to increment all values at all indexes to the max value in the array because we can't decrement the max value anyway.So solution boils down to two cases-

1. When`S<a[max]*n-1`

then the answer is just the difference of these two.2. If`S>a[max]*n-1 then (S-a[max]*n-1) should be equally divisible by (n-1)`

, so just increment S until divisible by (n-1) and the answer is difference of the two.You can look at my submission here

I had exactly same thought process, https://codeforces.com/contest/1452/submission/98940264.

Yes, exactly we both used the same logic and so I thought it may help some people if they were thinking this logic but couldn't get it properly. Did I explain it clearly?

yup!

Thanks! really helpful

very good solution!

Thanks a ton! I was really having trouble understanding the editorial! :D

I thought the same but couldn't implement it in the right manner ,I was thinking it as either all the elements have to be made equal to a[max] or equal to a[max]+1 but i couldn't get a clear understanding

I also realized it after the contest and couldn't solve in contest time.

My code gives me wrong answer in test case 3 any one help

Question no Bvoid solve() { ll n; cin>>n;

}

use spoiler please

Try using 0ll, in std:: accumulate, otherwise it gives integer overflow.

Proof by Induction for problem D is great but can someone tell me if there is a mathematical proof for that?

If we have a tower with power p, it will cover 2p-1 spots, thus the length of the subsegment covered by a tower is always odd.Now we need to find a way to divide a length of n into a number of odd length subsegments. This can be done via dynammic programming, which reduces to finding the nth Fibonacci number. In the end , just divide this by 2^n.

I'm sorry but I didn't need the proof for the solution to this problem. I was wondering how did that function get reduced to just finding the Fibonacci number for a particular input.(which was proved by Induction by the Editorialist)

f(n) = f(n-1) + f(n-3) + f(n-5) + .... f(n-2) = f(n-3) + f(n-5) + .... thus f(n) = f(n-1) + f(n-2) Is this what you were looking for?

Yeah exactly, thanks for putting it that way. Made it so easy to understand.

Here's the intuitive proof I came up with while upsolving.

First let's solve it like a normal dp problem (forgetting about the fibonacci numbers). We sweep from left to right. Let $$$dp[i]$$$ mean the number of ways to cover

onlythe first $$$i$$$ positions.How do our transitions work? First, $$$dp[0] = 1$$$ because there is one way to cover the first $$$0$$$ positions (don't have any towers at all). Now loop over all i's (from left to right).

Consider the current $$$dp[i]$$$. We can cover some range with exactly one tower if and only if the length of the segment is odd (so we can put the tower in the midpoint). So $$$dp[i]$$$ is equal to the sum of all dp values where the index has a different parity than i (so $$$dp[i] = dp[i-1]+dp[i-3]+dp[i-5]...$$$). Though this would work, it runs in $$$O(n^2)$$$, so it needs to be optimized.

The optimization is pretty simple. Because you don't care about the actual values of all previous dp's but just the parity. So instead of storing the whole dp array, just have it store 2 values, $$$dp[0]$$$ and $$$dp[1]$$$. $$$dp[0]$$$ stores the numbers of ways to cover an even prefix, and $$$dp[1]$$$ stores the number of ways to cover an odd prefix.

You still loop from 1 to n but the transitions are now $$$dp[i \mod 2] = dp[i \mod 2] + dp[1-(i \mod 2)]$$$. This is exactly the Fibonacci recurrence.

Hope this was helpful! Thanks!

That's a really nice way to think about it. Thanks.

hey thanks for sharing your idea . I have implemented the same idea with bottom up DP. but i am not able to do the same with top down DP. Can you please help me with that ?

@PurpleCrayon how did you come up with the idea that

dp[i] is equal to the sum of all dp values where the index has a different parity than i??Nice explanation!

I had a different approach to this problem which is quite intuitive imo. We are effectively choosing a subset out of $$$n$$$ towers such that the sum of powers of those towers cover the entire range $$$1-n$$$.This is equivalent to number of

oddintegral solutions of the equation $$$ p_1 + p_2 + ... + p_r = n $$$.This can be found out easily by converting it to the form $$$ (2*x_1 + 1) + (2*x_2 + 1) + .. + (2*x_r + 1) = n $$$ and finding non negative integral solution for $$$x_1 , x_2 ,... x_r$$$. Multiply this with the probability of choosing $$$r$$$ towers = $$$1/{2^{n}}$$$ to get the required answer solutionThank you very much! It's a great solution.

i am sorry, this might sound dumb but in you solution , ncr(d+x-1,x-1) here what are you actually choosing and from what

number of non negative integral solutions of the equation $$$x_1 + x_2 + ... +x_r=d$$$ is $$$d+r-1\choose r-1$$$

Why this Solution do not get TLE in problem G? I solved only with bfs/dfs. There is some property?

As in Tutorial, this solution is at most $$$O(nlog^2n)$$$

@BledDest Can you give more explanation how problem D can be solved. I didn't get the point how the probability can be considered as Fibonacci number. Please help me to understand.

In Problem D Editorial

How this statement is true?

Covering all towns can be expressed as splitting n into the sum of several positive odd integersAnyone, Please Explain.

Each town that has power must power itself, and an equal number of towns either side of it.

Suppose it powers K towns to its left, then it must also power K towns to its right. So in total it powers 2*K+1 towns.

All the towers together power N towns, and the set is partitioned such that no town is powered twice. As such,

N = sum (i = 1 to j) (2*Ki + 1), where there are j towns receiving power directly, and the ith town powers Ki either side of it.

The numerator of the answer is the number of unique sets of j and K1,...,Kj such that the equality holds.

Thanks a lot. I got it now.

What is the logic of taking max of the array in calculation of k for B?

We need an end state where each of the n-1 boxes has exactly (TOTAL)/(n-1). Suppose we choose one of the non-max boxes to redistribute: then all the boxes must have at least the max starting value, since we cannot take any away. Therefore max*(n-1) is a lower bound for the total number.

yeah but why would there be atleast max in each box?

Because if there is one box with max, we can't take any out of this box (if redistributing from any of the others). So that box must finish with at least max, and we require all the boxes to finish with the same number, so they must also have at least max.

Understood Thanks!

i successfully solved 1st problem by this=> ```

``` but i am thinking can it be solvable using dp like other grid paths problem? i am new at dp so i am having a hard time to think about it.so can anyone help me with this if it is possible to solve using dp

I have solved using dp. The idea is that the points on the diagonal(assume it is {x,x}) will always take x+x = 2*x. Other points can be reached from diagonal by this sequence. At diagonal -> Move -> Stop -> Move -> Stop

IN PROBLEM B Why (a.sum() + n — 2) / (n — 1) and not (a.sum()) / (n — 1) i didn't get it

the formula to get ceil of(x/y) is (x+(y-1))/y

Okk. But when I write ceil function, ceil(x/y), it throws an error, but writing (x+y-1)/y goes successful.

Maybe you forgot to cast long double/float/double type to this fraction. For example: ceil(5 / 2) = ceil(2) = 2, but ceil(5.0 / 2) = ceil(2.5) = 3.

But actually you should never try to use floating point numbers then use ceil, always opt to integer operations if possible.

Why does this give same results, in modular exponentiation?

than this

The only difference is we do

`y=y%(mod-1)`

in the first function.Edit: figured it out its fermat's theorem, thanks everyone

Why don't people use spoiler while commenting there code?

Isn't your question obvious, let say we want to compute 2^123097521478997542 mod m, then what do you think will be faster computing 2^123097521478997542 mod m or first compute y mod m-1 which will be smaller than equal to y, then compute power?

The answer is obvious, but I remember there was a problem in which the input size it self exceeds mod-1 and everyone who did pow=pow%(mod-1) in advance got WA, be careful.

I'm sorry community, I framed it wrong, I actually meant why both lead to same answer.

A friend of mine later told me its Little Fermat's theorem.

This was my first comment in codeforces, wasn't aware about the comment standards we need to maintain, are there any resources, on how questions be asked and all?

In D problem, if there are 5 cities, the number of ways to break it into sum of odd positive integer is (3)1,3,1 which shows that there are 3 ways,Can anyone help me in understanding how is it the nth fibonacci ie (5)?

Why solutions with complexity O(nmk) can get AC for problem E? The example are below: submission 1 and submission 2

upd: submission 2 has been hacked with the result of WA. But no one could made it TLE during the hacking phase.

Can someone explain problem E's solution a bit more.Thanks in advance! xD!!

in the editorial of problem b the first para [ceil(sum/n-1)] is the condition when the nephew chooses the box containing the maximum block, right?

Yes it is the block with max toys after you have added toys.

Problem E,why the solution is right?I spend a long time understanding the code,but can't prove it's correctness.The tutorial seems to ignore this.

you may think that algorithm is incorrect because of prefix and suffix and that one of them may contain some element which should have been in the other one but it doesn't matter because answer for cases like this will not be the final answer anyway because we are considering all possibilities and final answer will be the case where the conditions are satisfied.

oh!miaoa~zhenbuchuo!thank you for your reply!

Why i get tle for B?

submission : 99050069

whatever error you meet,you can see the data,the data you tle has sum=5e9,and n-1=4,and ma=1e9,your ma should add many times before you can break,so you tle

There is another nice approach to Problem D using combinatorics.

For the problem D

how can we find the number of ways in which N can represented as sum of odd integers as mentioned in the editorial ??: To be specific using the following method

I would like to know which method is this?

can anybody explain me ...in the problem toy blocks , why are you adding (n-2) to the array sum ? i.e. 4th line in fun main() function

ceil function, that is to get the smallest greater integer.

So Can't we use directly ceil() function available in header cmath ?

Since these are integers, ceil function wont work here.

And everyone ignored the pun in editorial of B. Not I but you.

Can anyone help me on the modular division given in solution of the problem d?Specifically why he has used 'mod-2' during the divide process?? Any document is also helpful...

cin >> a >> b; cout << a + b + abs(a — b) — (a ^ b ? 1 : 0) << endl;

Why the divide function in problem D is mul(x, binpow(y, MOD — 2)) ? I understand the idea behind mul() and binpow(), but why they can perform the divide operation?

It uses Fermat's little theorem to calculate Modular multiplicative inverse.