Idea: Roms

**Tutorial**

Tutorial is loading...

**Solution (Roms)**

```
#include<bits/stdc++.h>
using namespace std;
int n;
string s;
int main(){
int tc;
cin >> tc;
for(int t = 0; t < tc; ++t){
cin >> n >> s;
int pos = n;
for(int i = 0; i < n; ++i)
if(s[i] == '8'){
pos = i;
break;
}
if(n - pos >= 11)
cout << "YES\n";
else
cout << "NO\n";
}
return 0;
}
```

Idea: vovuh

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
vector<int> p = {4, 8, 15, 16, 23, 42};
int ans[4];
int main()
{
for(int i = 0; i < 4; i++)
{
cout << "? " << i + 1 << " " << i + 2 << endl;
cout.flush();
cin >> ans[i];
}
do
{
bool good = true;
for(int i = 0; i < 4; i++)
good &= (p[i] * p[i + 1] == ans[i]);
if(good) break;
}
while(next_permutation(p.begin(), p.end()));
cout << "!";
for(int i = 0; i < 6; i++)
cout << " " << p[i];
cout << endl;
cout.flush();
return 0;
}
```

Idea: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
const int N = 1000043;
vector<int> g[N];
int color[N];
int siz[N];
int n, m;
int cc = 0;
int dfs(int x)
{
if(color[x])
return 0;
color[x] = cc;
int ans = (x < n ? 1 : 0);
for(auto y : g[x])
ans += dfs(y);
return ans;
}
int main()
{
scanf("%d %d", &n, &m);
for(int i = 0; i < m; i++)
{
int k;
scanf("%d", &k);
for(int j = 0; j < k; j++)
{
int x;
scanf("%d", &x);
--x;
g[x].push_back(i + n);
g[i + n].push_back(x);
}
}
for(int i = 0; i < n; i++)
{
if(!color[i])
{
cc++;
siz[cc] = dfs(i);
}
printf("%d ", siz[color[i]]);
}
}
```

**Solution (PikMike)**

```
#include <bits/stdc++.h>
using namespace std;
#define forn(i, n) for(int i = 0; i < int(n); i++)
const int N = 500 * 1000 + 13;
int n, m;
int rk[N], p[N];
int getP(int a){
return (a == p[a] ? a : p[a] = getP(p[a]));
}
void unite(int a, int b){
a = getP(a), b = getP(b);
if (a == b) return;
if (rk[a] < rk[b]) swap(a, b);
p[b] = a;
rk[a] += rk[b];
}
int main(){
scanf("%d%d", &n, &m);
forn(i, n) p[i] = i, rk[i] = 1;
forn(i, m){
int k;
scanf("%d", &k);
int lst = -1;
forn(j, k){
int x;
scanf("%d", &x);
--x;
if (lst != -1)
unite(x, lst);
lst = x;
}
}
forn(i, n)
printf("%d ", rk[getP(i)]);
puts("");
}
```

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())
int n;
string s;
inline bool read() {
if(!(cin >> n))
return false;
cin >> s;
return true;
}
inline void solve() {
string t(n, '0');
int bal = 0;
fore(i, 0, n) {
if(s[i] == ')')
bal--;
t[i] = char('0' + (bal & 1));
if(s[i] == '(')
bal++;
}
cout << t << endl;
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
int tt = clock();
#endif
if(read()) {
solve();
#ifdef _DEBUG
cerr << "TIME = " << clock() - tt << endl;
tt = clock();
#endif
}
return 0;
}
```

Idea: Roms

**Tutorial**

Tutorial is loading...

**Solution (Roms)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = int(1e6) + 99;
int n, x;
int a[N];
vector <int> pos[N];
int prefMax[N];
int main() {
scanf("%d %d", &n, &x);
for(int i = 0; i < n; ++i){
scanf("%d", a + i);
pos[a[i]].push_back(i);
prefMax[i] = max(a[i], (i > 0 ? prefMax[i - 1] : a[i]));
}
int p = 1;
int lst = n + 5;
for(int i = x; i >= 1; --i){
if(pos[i].empty()){
p = i;
continue;
}
if(pos[i].back() > lst) break;
p = i;
lst = pos[i][0];
}
long long res = 0;
lst = -1;
for(int l = 1; l <= x; ++l){
int r = max(l, p - 1);
if(lst != -1) r = max(r, prefMax[lst]);
res += x - r + 1;
if(!pos[l].empty()){
if(pos[l][0] < lst) break;
lst = pos[l].back();
}
}
cout << res << 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())
typedef long long li;
const int MOD = int(1e9) + 7;
int add(int a, int b) {
a += b;
while(a >= MOD)
a -= MOD;
while(a < 0)
a += MOD;
return a;
}
int mul(int a, int b) {
return int(a * 1ll * b % MOD);
}
int n;
vector<int> a;
inline bool read() {
if(!(cin >> n))
return false;
a.resize(n);
fore(i, 0, n)
cin >> a[i];
return true;
}
vector<li> f;
void inc(int pos, int val) {
for(; pos < sz(f); pos |= pos + 1)
f[pos] += val;
}
li sum(int pos) {
li ans = 0;
for(; pos >= 0; pos = (pos & (pos + 1)) - 1)
ans += f[pos];
return ans;
}
vector<int> S[2];
inline void solve() {
fore(k, 0, 2) {
S[k].assign(n, 0);
f.assign(n, 0);
vector<int> dx(a.begin(), a.end());
sort(dx.begin(), dx.end());
fore(i, 0, n) {
int pos = int(lower_bound(dx.begin(), dx.end(), a[i]) - dx.begin());
S[k][i] = int(sum(pos) % MOD);
inc(pos, i + 1);
}
reverse(a.begin(), a.end());
}
reverse(S[1].begin(), S[1].end());
int ans = 0;
fore(i, 0, n)
ans = add(ans, mul(a[i], add(mul(i + 1, n - i), add(mul(S[0][i], n - i), mul(S[1][i], i + 1)))));
cout << ans << 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);
if(read()) {
solve();
#ifdef _DEBUG
cerr << "TIME = " << clock() - tt << endl;
tt = clock();
#endif
}
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;
const int INF = int(1e9);
const li INF64 = li(1e18);
const int N = 200 * 1000 + 555;
const int D = 7077;
int n, d, m;
int a[N], qs[N];
inline bool read() {
if(scanf("%d%d", &n, &d) != 2)
return false;
fore(i, 0, n)
scanf("%d", &a[i]);
assert(scanf("%d", &m) == 1);
fore(i, 0, m)
scanf("%d", &qs[i]);
return true;
}
bitset<D> L, R, cur;
int dist[N];
set<int> has;
void shiftL(int s, int t, int &uk) {
L <<= t - s;
for(; uk < n && a[uk] < t; uk++) {
if(t - a[uk] - 1 < D)
L[t - a[uk] - 1] = 1;
if(t - a[uk] < D)
L[t - a[uk]] = 1;
}
}
void shiftR(int s, int t, int &uk) {
R >>= t - s;
if(!has.count(t))
R[0] = 0;
for(; uk < n && a[uk] < t + D; uk++) {
if(a[uk] - t >= 0) {
R[a[uk] - t] = 1;
if(a[uk] - t + 1 >= 0)
R[a[uk] - t + 1] = 1;
}
}
}
inline void solve() {
fore(i, 0, m) {
dist[i] = INF;
int pos = int(lower_bound(a, a + n, qs[i]) - a);
if(pos > 0)
dist[i] = qs[i] - a[pos - 1] - 1;
if(pos < n)
dist[i] = min(dist[i], a[pos] - qs[i]);
}
has = set<int>(a, a + n);
int uL = 0, uR = 0;
L.reset();
R.reset();
fore(i, 0, m) {
shiftL(i > 0 ? qs[i - 1] : -D, qs[i], uL);
shiftR(i > 0 ? qs[i - 1] : -D, qs[i], uR);
double ans = atan2(1, dist[i]);
cur = L & R;
int pos = (int)cur._Find_first();
if(pos < D) {
int ds = pos;
ans = max(ans, 2 * atan2(1, ds));
}
printf("%.15f\n", ans);
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
int tt = clock();
#endif
cout << fixed << setprecision(15);
if(read()) {
solve();
#ifdef _DEBUG
cerr << "TIME = " << clock() - tt << endl;
tt = clock();
#endif
}
return 0;
}
```

Editorial of F is soooo goooooddd. thx man

F is a little overcomplicated I think. Easier way to think about it is to see that every contribution of the lower-valued items on the left is that their contribution is just the number of spaces on the left of each of them plus one (if example $$$a_j < a_i$$$ and $$$j < i$$$, contribution is $$$j + 1$$$, since this is the number of intervals that the lower-valued item gets included in). Multiply this by the number of intervals on the right (

`n - index`

), since it contributes that many more times. Then easy done. Do this with the segment tree by updating the position index in fenwick tree with itself to get contribution over all of them fast (contribution from left side is`query(index) * (n - index)`

). Then do same thing on the right side. (All values are zero indexed in this explanation)UPD: Ok, someone has asked for a clearer explanation, so I have taken a little more time to write it out. It is very large, so I spoilered it.

better explanationEach number contributes to the final sum a certain number of times (multiplier, or coefficient). So the end result is sum of all $$$a_i * mult_i$$$ where a is our array. We will explain this with zero indexing.

First, note that the total number of items to the left of index $$$i$$$, including $$$i$$$, is just $$$i + 1$$$, and the total number of items on the right, including $$$i$$$, is $$$n - i$$$.

Now we consider that the multiplier is contributed to by $$$(i + 1) * (n - i)$$$, because this is the total number of segments the number is included in.

Now consider the cases where contribution is increased by an element less than our current element at position $$$i$$$ in the array. Let's call this element as having position $$$j$$$.

Add to the multiplier the number of contributions from the left hand side: each number on the left hand side less than us contributes to the multiplier by the number of times it is seen. This is equal to the number of spaces on the left of the number, including that number. Then multiply this count by the number of items on the right, including the one at position $$$i$$$, because this is the number of segments in which $$$a_j$$$ is included. This contribution will be $$$(j + 1) * (n - i)$$$.

A similar thing happens with the right hand side, except on the right hand side the contribution to the coefficient is $$$(i + 1) * (n - j)$$$.

Now we have to compute these values for all $$$j$$$ in fast time. We can do this with a fenwick tree (also called BIT, binary indexed tree) or segment tree. Process the values from small to large and keep two fenwick trees. When you have finished processing index $$$i$$$, just update the fenwick tree representing the left nodes with $$$i + 1$$$ at index i, and update the right fenwick tree with $$$n - i$$$. Then, we can query really fast, because we only need the

sumof the values given by each individual $$$a_j$$$ to find the final multiplier.The final multiplier will be

`(i + 1) * (n - i) + leftTree.queryRange(0, i - 1) * (n - i) + rightTree.queryRange(i + 1, n - 1) * (i + 1)`

, where queryRange is inclusive.wouldn't this solution overcount?

ohhhh nevermind this is genius thank you!

just a minor correction in last line: (i + 1) * (n — i) + leftTree.queryRange(0, i) * (n — i) + rightTree.queryRange(i + 1, n) *

(i + 1). Thnx for the explanation.Oh shoot, thanks!!!

(also there were some other typos in that last line i corrected)

Please explain problem C .The problem as well as input ,output is not clear to me. How become the output 4 4 1 4 4 2 2 for sample test case ?

Use some pen & paper, you'll get eventually.

Read the problem attentively, you'll understand input & output.

Firstly you read the number of people and the number of groups, then for each group you read the which people belong to this group. In then end for each people $$$i$$$ you wish to find the number of people that are share a group with $$$i$$$.

plz someone xplain hep ..i also stuck that how is this output for sample test case ?plz help

Can problem C be solved using DSU data structure ?

Yes. Initially, you want to add everyone in a group of friends do the same component. Then, when someone is into another group, you connect the components.

This is exactly what the editorial says in the last line.

I think one person cannot be in more than one component as if he's in more than one component those components cannot be disjoint as he'll be friends with people from both components. I'm still not very clear about problem C. Can somebody explain the solution once.

Let's say there are two main groups on the input: group

Aconsisting ofp,qandrand groupBwith two other guys:uandv.If the next line of the input says that there exists another group,

C, consisting ofrandu, that means there's an edge connecting them and, therefore, their groups. Now, there is a connection betweenpandqwithr, fromrtou(just became friends), and fromutov.That means we need a fast way to unite those disjoint sets of friends.

In the end, we are required to say how many people will know about the news if person

istarts spreading it. From the statement, we know that one guy tells the news to everyone in his groups of friends, so we only need to print the size of the component containingi-th person.You can check my submission (54194939). It's pretty easy to read and there are lots of comments.

Thanks for the explanation! I get it now

I'm really sad that this is the first time I did an interactive problem and the tutorial in the contest show

`fflush(stdout)`

to flush output in C++. I kept getting idleness error in the contest and couldn't figure out why (WA is much easier to deal with). After this editorial released, I changed`fflush(stdout)`

to`cout.flush()`

and I could debug my old code in a blink. You guys should update the tutorial for the interactive problem. It was frustrating.No. It's your fault for using IO optimizations without understanding what they do minimally. https://codeforces.com/contest/1167/submission/54261003 fflush + commenting the optimizations.

My tip is always using endl instead of '\n' since code stays the same and endl flushes the output. Example: https://codeforces.com/contest/1167/submission/54205213

With this being said this is worth mentioning in that blog.

Thanks. Now I know. Good lesson tho.

Use endl only in interactive problems! It's slow so you risk TLE.

I meant always using endl in interactive problems, my bad I was careless in the comment.

If you use

`cout`

and`endl`

then you don't need an extra`cout.flush()`

in C++.The link to this blog has not yet been mentioned in the contest page. Should you not add it to the contest page as soon as the editorial is published? Just asking.

For Lost Numbers I used almost the same trick.But struck with one problem.I was using GCD to get to know unique number but it turns out that it is not.let say the array is [a,b,c,d,e,f] and x=a*b and y=b*c .So GCD(x,y) should give me "b" but i was getting multiple of b.So is there a way to resolve this issue.Thanks is advance.

gcd(x,y) gives you b only if gcd(a,c)=1. Otherwise, it will give you gcd(a,c)*b, which is a multiple of b.

The explanation to the solution of problem C is not very clear. Can someone explain the approach in simpler words ?

Use a DSU to connect all of the groups. This can be done in O(group size) for each group by connecting the first person in each group to everyone else. Then find the size of the group each person is in to get the answer.

I managed to pass G with 2 pointers (breaking early helped a lot). I do prefer the editorial solution though. Thanks!

Why is it enough to maintain 7000 positions to the left and right?

Oh, I totally forgot to mention this in editorial. Basically, positions further than that won't ever be needed. Check the angle you get for the closest collision between a building and a ray. It's less or equal to $$$arctan(\frac 1 {3500})$$$. Find the maximum $$$x$$$ such that $$$2 arctan (\frac 1 x) > arctan (\frac 1 {3500})$$$ so that two buildings collinding produce greater angle. That $$$x$$$ is 7000, thus further collisions will never affect the answer.

Bitsets can be updated in (d/32). Can you give any reference for this? I didn't get this. Thanks for the editorial.

Can someone explain the editorial of problem E (Range Deleting)?

Having a difficult time to understand it or is there any other approach which is simpler and easy to explain.

Hello, I solved Problem C using DFS in Java. I saw one "Accepted" submission in GNU C++ with the same logic. I am getting time limit exceeded on test case 6. I have tried to speed up the code as much as possible. Is it possible that the same solution in Java gives a time limit but not in GNU C++?

Whenever your program might go in deep recursion or dfs you might get RunTime Error. This is due to StackOverFlow Error. As stack size of Java is very low, hence, either you should avoid deep recursion and use iteration or increase stack size dynamically.

Source :

May you attach this tutorial link to the contest material section?

It might really be a mess (a search will work :3) for future participants to look for it (for practice).

Problem F

Many people got WA on test 35 (me too) as we didn't mod while adding inside BIT

awoo adedalic

Can you please explain the editorial of problem E (Range Deleting)? Not able to get it

can anyone explain problem D properly?? here is my solution.. and it is getting wrong on test case 17Your text to link here...

Please help! I have a question about question C solution. Why my code get TLE on 6th test case and is slow in general because I do not understand:(. The code -> https://codeforces.com/contest/1167/submission/54572709 The solution inspired by BledDest solution, just written in Java, not line by line so there might be difference but it should do roughly the same. It creates graph where person nodes and groups nodes are mixed with each other and to recognize what is what the groups nodes are enumerated from (n+1) to (n + m + 1) and person nodes from 1 to n. verts array holds colors of nodes after DFS. So if there is only one connected component in graph, then DFS should run only once. DFS returns the number of person nodes that the code encounter during search.

Can anyone explain

problem E54435380 In C, Why am I getting memory limit over and over again? Can anyone please help?

You are declaring a vector of size

zinside youforloop — and you are doing itmtimes. Two global arrays of size 5*1e5 andmvectors with variable sizes (but sum equal to 1e5) are using too much memory.You can declare the vector "vec" outside of the

forloop and just reuse it — or don't use it at all.OK thanks :D

here is my submission of problem C News Distribution http://codeforces.com/problemset/submission/1167/55075296 please, can someone tell what's wrong in it its showing tle and I don't know why?

can anyone explain problem c sample test case 4 4 1 4 4 2 2 in some details? how to get this output? please.

Can someone please check why 57224389 for 1167E - Удаление отрезка gets WA on test #51?

I used a greedy approach similar to tutorial and some two pointers. The overall time complexity is $$$O(n)$$$.

I am trying to understand the issue of my submissions for problem C, that hit Memory Limit. It would be nice if someone can help with my understanding.

Submission #1 , Submission #2

The only difference between them, is the allocation of:

In #1,

`N = 1e6 + 5`

. In #2,`N = 2e5 + 5`

. So I thought #2 would require less memory than #1, but it turned out #1 Accepted, while #2 is over limit. Can someone help me understand the issue?Help me in my code of problem c, i am getting TLE

Can anyone explain why my I'm getting RTE on test case 4 for problem D.

https://codeforces.com/contest/1167/submission/89328218