Idea: fcspartakm

**Tutorial**

Tutorial is loading...

**Solution (fcspartakm)**

```
#include <iostream>
#include <sstream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <string>
#include <cstring>
#include <cassert>
#include <iomanip>
#include <algorithm>
#include <set>
#include <map>
#include <ctime>
#include <cmath>
#define forn(i, n) for(int i=0;i<n;++i)
#define fore(i, l, r) for(int i = int(l); i <= int(r); ++i)
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()
#define pb push_back
#define mp make_pair
#define x first
#define y1 ________y1
#define y second
#define ft first
#define sc second
#define pt pair<int, int>
template<typename X> inline X abs(const X& a) { return a < 0? -a: a; }
template<typename X> inline X sqr(const X& a) { return a * a; }
typedef long long li;
typedef long double ld;
using namespace std;
const int INF = 1000 * 1000 * 1000;
const ld EPS = 1e-9;
const ld PI = acos(-1.0);
li n;
inline void read() {
cin >> n;
}
inline void solve() {
if (n == 1 || n == 2 || n == 4) {
cout << -1 << endl;
return;
}
if (n % 3 == 0) {
cout << n / 3 << ' ' << 0 << ' ' << 0 << endl;
} else if (n % 3 == 1) {
cout << (n - 7) / 3 << ' ' << 0 << ' ' << 1 << endl;
} else {
cout << (n - 5) / 3 << ' ' << 1 << ' ' << 0 << endl;
}
}
int main () {
#ifdef fcspartakm
freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
cerr << setprecision(10) << fixed;
int t; cin >> t;
while(t--) {
read();
solve();
}
//cerr << "TIME: " << clock() << endl;
}
```

Idea: fcspartakm

**Tutorial**

Tutorial is loading...

**Solution (fcspartakm)**

```
#include <iostream>
#include <sstream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <string>
#include <cstring>
#include <cassert>
#include <iomanip>
#include <algorithm>
#include <set>
#include <map>
#include <ctime>
#include <cmath>
#define forn(i, n) for(int i=0;i<n;++i)
#define fore(i, l, r) for(int i = int(l); i <= int(r); ++i)
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()
#define pb push_back
#define mp make_pair
#define x first
#define y1 ________y1
#define y second
#define ft first
#define sc second
#define pt pair<int, int>
template<typename X> inline X abs(const X& a) { return a < 0? -a: a; }
template<typename X> inline X sqr(const X& a) { return a * a; }
typedef long long li;
typedef long double ld;
using namespace std;
const int INF = 1000 * 1000 * 1000;
const ld EPS = 1e-9;
const ld PI = acos(-1.0);
const int N = 200 * 1000 + 13;
int n, k;
int a[N];
inline void read() {
cin >> n >> k;
for (int i = 0; i < n; i++) cin >> a[i];
}
inline void solve() {
sort(a, a + n);
reverse(a, a + n);
li ans = 0;
for (int i = 0; i <= k; i++) {
ans += a[i];
}
cout << ans << endl;
}
int main () {
#ifdef fcspartakm
freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
cerr << setprecision(10) << fixed;
int t; cin >> t;
while(t--) {
read();
solve();
}
//cerr << "TIME: " << clock() << endl;
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (fcspartakm)**

```
#include <iostream>
#include <sstream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <string>
#include <cstring>
#include <cassert>
#include <iomanip>
#include <algorithm>
#include <set>
#include <map>
#include <ctime>
#include <cmath>
#define forn(i, n) for(int i=0;i<n;++i)
#define fore(i, l, r) for(int i = int(l); i <= int(r); ++i)
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()
#define pb push_back
#define mp make_pair
#define x first
#define y1 ________y1
#define y second
#define ft first
#define sc second
#define pt pair<int, int>
template<typename X> inline X abs(const X& a) { return a < 0? -a: a; }
template<typename X> inline X sqr(const X& a) { return a * a; }
typedef long long li;
typedef long double ld;
using namespace std;
const int INF = 1000 * 1000 * 1000;
const ld EPS = 1e-9;
const ld PI = acos(-1.0);
int n;
inline void read() {
cin >> n;
}
inline void solve() {
multiset<int> was;
for (int i = 1; i <= n; i++) {
was.insert(i);
}
vector<pair<int, int> > ans;
for (int i = 0; i < n - 1; i++) {
auto it = was.end();
it--;
int a = *it;
was.erase(it);
it = was.end();
it--;
int b = *it;
was.erase(it);
was.insert((a + b + 1) / 2);
ans.pb(mp(a, b));
}
cout << *was.begin() << endl;
for (int i = 0; i < sz(ans); i++) {
cout << ans[i].ft << ' ' << ans[i].sc << endl;
}
}
int main () {
#ifdef fcspartakm
freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
cerr << setprecision(10) << fixed;
int t; cin >> t;
while(t--) {
read();
solve();
}
//cerr << "TIME: " << clock() << endl;
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
char buf[200043];
int main()
{
int t;
scanf("%d", &t);
for(int tc = 1; tc <= t; tc++) {
int n;
scanf("%d", &n);
scanf("%s", buf);
string s = buf;
queue<int> q;
int cur = 0;
for(int i = 0; i < n; i++)
{
if(i > 0 && s[i] != s[i - 1])
cur++;
if(i > 0 && s[i] == s[i - 1])
q.push(cur);
}
int deleted = 0;
int score = 0;
for(int i = 0; i < n; i++)
{
if(q.empty())
break;
q.pop();
deleted++;
score++;
while(!q.empty() && q.front() == i)
{
q.pop();
deleted++;
}
deleted++;
//cerr << deleted << endl;
}
score += (n - deleted + 1) / 2;
printf("%d\n", score);
}
}
```

Idea: fcspartakm

**Tutorial**

Tutorial is loading...

**Solution (fcspartakm)**

```
#include <iostream>
#include <sstream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <string>
#include <cstring>
#include <cassert>
#include <iomanip>
#include <algorithm>
#include <set>
#include <map>
#include <ctime>
#include <cmath>
#define forn(i, n) for(int i=0;i<n;++i)
#define fore(i, l, r) for(int i = int(l); i <= int(r); ++i)
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()
#define pb push_back
#define mp make_pair
#define x first
#define y1 ________y1
#define y second
#define ft first
#define sc second
#define pt pair<int, int>
template<typename X> inline X abs(const X& a) { return a < 0? -a: a; }
template<typename X> inline X sqr(const X& a) { return a * a; }
typedef long long li;
typedef long double ld;
using namespace std;
const int INF = 1000 * 1000 * 1000;
const ld EPS = 1e-9;
const ld PI = acos(-1.0);
const int N = 200 * 1000 + 13;
int n;
string s;
string revS;
vector<int> posS[30];
vector<int> posT[30];
int cnt[30];
int t[N];
inline int sum (int r) {
int result = 0;
for (; r >= 0; r = (r & (r+1)) - 1)
result += t[r];
return result;
}
inline void inc (int i, int d) {
for (; i < n; i = (i | (i+1)))
t[i] += d;
}
int sum (int l, int r) {
return sum (r) - sum (l-1);
}
inline void read() {
cin >> n >> s;
}
inline void solve() {
revS = s;
reverse(all(revS));
for (int i = 0; i < sz(s); i++) {
posS[s[i] - 'a'].pb(i);
posT[revS[i] - 'a'].pb(i);
}
li ans = 0;
for (int i = 0; i < sz(revS); i++) {
int let = revS[i] - 'a';
int cur = posS[let][cnt[let]];
int oldC = cur;
cur += sum(cur, n - 1);
int need = i;
ans += cur - need;
inc(oldC, 1);
cnt[let]++;
}
cout << ans << endl;
}
int main () {
#ifdef fcspartakm
freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
srand(time(NULL));
cerr << setprecision(10) << fixed;
read();
solve();
//cerr << "TIME: " << clock() << endl;
}
```

Idea: BledDest

**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;
const int INF = int(1e9);
const li INF64 = li(1e18);
const ld EPS = 1e-9;
int n, k;
vector<int> l, r, a;
inline bool read() {
if(!(cin >> n >> k))
return false;
l.resize(n);
r.resize(n);
a.resize(n);
fore(i, 0, n)
cin >> l[i] >> r[i] >> a[i];
return true;
}
inline void solve() {
vector<li> d(n + 1, INF64);
d[0] = 0;
li ans = INF64;
fore(i, 0, n) {
li rem = k, total = d[i];
for (int j = i; j < n; j++) {
li cntReloads = (max(0LL, a[j] - rem) + k - 1) / k;
if (cntReloads > r[j] - l[j])
break;
li newRem = (rem + cntReloads * k) - a[j];
total += a[j];
if (j + 1 < n) {
if (l[j] + cntReloads < l[j + 1])
d[j + 1] = min(d[j + 1], total + newRem);
} else
ans = min(ans, total);
rem = newRem;
}
}
if (ans > INF64 / 2)
ans = -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);
cout << fixed << setprecision(15);
if(read()) {
solve();
#ifdef _DEBUG
cerr << "TIME = " << clock() - tt << endl;
tt = clock();
#endif
}
return 0;
}
```

1430G - Yet Another DAG Problem

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
typedef long long li;
const int N = 18;
const int M = (1 << N);
const li INF64 = li(1e18);
int n, m;
vector<int> g[N];
li sum[N];
int need_mask[N];
li dp[N + 1][N + 1][M];
bool p[N + 1][N + 1][M];
vector<int> order;
vector<int> used;
void dfs(int x, bool build_topo)
{
if(used[x])
return;
used[x] = 1;
for(auto y : g[x])
dfs(y, build_topo);
if(build_topo)
order.push_back(x);
}
int main()
{
cin >> n >> m;
for(int i = 0; i < m; i++)
{
int x, y, w;
cin >> x >> y >> w;
--x;
--y;
sum[x] += w;
sum[y] -= w;
g[x].push_back(y);
}
used.resize(n);
for(int i = 0; i < n; i++)
dfs(i, true);
for(int i = 0; i < n; i++)
{
used = vector<int>(n, 0);
dfs(i, false);
for(int j = 0; j < n; j++)
if(j != i && used[j] == 1)
need_mask[i] |= (1 << j);
}
for(int i = 0; i <= n; i++)
for(int j = 0; j <= n; j++)
for(int k = 0; k < (1 << n); k++)
dp[i][j][k] = INF64;
dp[0][0][0] = 0;
reverse(order.begin(), order.end());
for(int i = 0; i < n; i++)
for(int j = 0; j <= n; j++)
for(int k = 0; k < (1 << n); k++)
{
if(dp[i][j][k] > INF64 / 2)
continue;
if(j == n)
{
if(dp[i + 1][0][k] > dp[i][j][k])
{
dp[i + 1][0][k] = dp[i][j][k];
p[i + 1][0][k] = false;
}
}
else
{
int v = order[j];
li add = sum[v] * i;
if(dp[i][j + 1][k] > dp[i][j][k])
{
dp[i][j + 1][k] = dp[i][j][k];
p[i][j + 1][k] = false;
}
if(((k & (1 << v)) == 0) && ((need_mask[v] & k) == need_mask[v]))
{
int nk = k | (1 << v);
if(dp[i][j + 1][nk] > dp[i][j][k] + add)
{
dp[i][j + 1][nk] = dp[i][j][k] + add;
p[i][j + 1][nk] = true;
}
}
}
}
vector<int> ans(n);
int i = n;
int j = 0;
int k = (1 << n) - 1;
while(i > 0 || j > 0 || k > 0)
{
if(j == 0)
{
j = n;
i--;
}
else
{
if(p[i][j][k])
{
int v = order[j - 1];
ans[v] = i;
k ^= (1 << v);
}
j--;
}
}
for(int i = 0; i < n; i++)
cout << ans[i] << " \n"[i == n - 1];
}
```

Editorial in Russian will be posted soon. Please be patient!

My solution of A got hacked! Lesson learnt.

I'm saying it every contest)

A,B,C were easy to do but couldn't solve out D in time :(

same pinch bruh !!!

Does anyone give a f?

did I ask for anyone to give f?

Can A be solved using a linear diophantine equation in 3 variables?

Yes. Nugget number tells us that all number more than or equal to $$$3*5-3-5 = 8$$$ could be written in the form $$$3a+5b$$$ where $$$a,b \geq 0.$$$ So, now you get the general formula

1,2,4 -> no answer

7 -> 0,0,1

The rest could be either brute force or notice that the smallest $$$a$$$ is either $$$0,1,2,3$$$ or $$$4.$$$ So, you could solve this in $$$O(1)$$$

Thanks for helping :)

An easier O(1) solution to A..

Thanks it was helpful.

I converted it to diophantine by doing this: 3x+5y=n-7z for z = 0 to n/7 Then using the extended euclids algorithm I found x0 and y0 st 3x0+5y0=gcd(3, 5)=1 and multiplied x0 and y0 by n-7z. Then I tried to find t such that x0+5t, y0-3t are positive. We can use some inequalities to determine range of t and loop over these t and check if the current is correct if yes return. Actually all values in the range are correct but i looped over some modification of the range because I could not think whether to +1 at each iteration or -1(i am new to CP and CS in general) so i checked the cases. My algorithm was correct(you can check my profile for code) but it was very messy, hard to implement for someone not good in math, a lot of code so I don't suggest this.

Can F be solved using BinarySearch? I tried but because of poor implementation and slow speed, got wrong answer on test case 6? Did anyone else try using binary search?

If I am not Wrong this is your Fake Account to hack your own solutions and try some alternate solutions if passed you can just make some changes submit with your original ID so that your rating will not be disturbed.

Lol! Bro you can search this username. It's everywhere. Facebook, Instagram, LinkedIn, Codeforces, Codechef, AtCoder, XDA-Dev, Hackerrank, HackerEarth, LeetCode, and any possible site where one can register and choose a username irrespective of whether it's a coding platform or not. Cant think of any reason why you thought so, but that's not the case. Peace :)

moli2398 I see your solution for problem B got hacked. Is it because of

Arrays.sort()beingO(N^2)? Just curious.Yeah I guess!

Can someone please tell their solution for G using Flows?

I solved G with Simplex. I don't have any proof that the answer will always be an integer though. I guess I just got lucky that it was in this case.

Same here.

I think it can be converted into an equivalent mincost maxflow

Yes it can.

The linear dual of this problem is actually maximizing the sum of flows along all edges — not just the source. And from each node with negative $$$c_v$$$ you need to send $$$-c_v$$$ units of flow more than goes into this node and each node with positive $$$c_v$$$ needs to receive $$$c_v$$$ flow more than it sends.

I see some people solving E without merge sort tree or fenwick tree . Can someone explain their approach . I know Merge sort tree is the most standard approach for counting inversions but in this case string will have max 26 distinct characters . Can this constraint be used as an advantage to solve in more simpler way ?

These are the 2 approaches I know for calculating inversions in an array other than merge sort and Fenwick tree(If it's relevant to the current discussion).

DIVIDE_AND_CONQUER_SOLUTIONPBDS_SOLUTIONYou can also do coordinate compression + segment tree / BIT. Note that if it's a permutation (or something you can transform to a permutation easily) you do not need coordinate compression.

I have an alternative solution to G:

First assign to each node the number given by the following greedy algorithm: Process all nodes in order of topological sort starting from the leaves and assign 0 to a leaf and maximum of values of all the children plus 1 if it's not a leaf.

But this greedy will be not optimal in cases like this: There are four nodes, and there is an edge from node 1 to node 2, another from node 2 to node 3, and another to node 1 to node 4. Here the optimal solution is 2 1 0 1 while the greedy gives 2 1 0 0. But this can be improved with the following greedy brute force-like algorithm:

Select some subset of nodes and try to sum 1 to all nodes in this subset, check if it not ruins anything and if the profit(change to the total answer if we sum 1 to all nodes in this subset) is positive, then if the profit is positive and we can apply the operation we will apply them.

We can do this step to all 2^n subsets, and when it's finished repeat them n — 3 times(because it may be optimal add more than 1 to some nodes). The overall complexity is 2^n*n^3 but in practice is much lower, i have passed the time limits without many optimizations.

However, i have no idea of why it works, so, if anyone can prove it or hack me, i would be really grateful.

I had the same idea. It's correct because you're effectively emulating the simplex algorithm for linear programming.

Wow, I didn't knew about simplex algorithm, i will find it in google, thanks.

Can someone point whats wrong in my submission for D? https://codeforces.net/contest/1430/submission/95385972

your solution will fail in these kind cases: 01110 expected output: 3 your output: 2

Here is another BIT solution for problem E, which I found more intuitive as a speedup on the naive O($$$n^2$$$) solution.

Explanation:

In the naive solution, you would like to iterate in the reverse direction on the reversed string $$$t$$$, at each position check which character should be there, and then try to move the closest character to its position. That would take $$$i-j$$$ swaps to move the character form position $$$i$$$ to $$$j$$$. But when we do this swap, the positions of the elements after it get messy so if we did it in a dumb way, we would need O(n) to maintain the new positions.

To get around this, we can use a BIT. Store 1 in all positions [0,n-1]. Now, when we query on any position, we would get its one based index in the string. The next observation is that we are always fixing the suffix of the string, so when we swap, we are effectively

deletingit from out original string. To simulate this, we can make its value in the BIT 0, and now we can once again find every character's position in O(logn). This way, we could speed it up to O(nlogn)I did this as well. My solution is here 170995085 Note that it uses a segment tree instead of a BIT

my solution to D here

which uses a simple visited array to keep track of deleted numbers by the first operation

can you explain it, i didn't understood D

You see the second operation always strips away the first continuous segment of the string, so to maximize operations we need to delete using first operations in a way in which the number of continuous segments doesn't decrease a lot

For example in 1011 if you remove 0 you'll be left with 111 which is one segment but if you remove the last one you'll get 101 which has three continuous segments So what we wanna do here is in first operation if possible choose a position which has the same value at its adjacent, because deleting it won't decrease the number of segments such as in 1101 deleting first one will result in 101 which still has 3 segments So in my code I use a visited array to keep track of what positions I deleted with my first operation I try to find what is the first position which is the same as the number after it, and I visit it, after that I use the j variable and traverse it until the end of the first continuous segment, now in next iteration i need to choose another position so it needs to start from i+1 but also after j so i updated it accordingly

In the end i might run out but we might still have j left this will happen in cases like 1010101 In these cases we can simple cut one by one

Thank you so much sir, now i got it

I request all the people contributing to the editorial to not use their templates while giving solutions. It would be very helpful to see whole for loop written instead of just seeing inc(l,r) because we have to search through your templates where that is, and then we have to put that code along with your other code and then see it in continuation with that code. It would be easier for better coders to see through it, but for noobs like me, it is just confusing.

Can anybody provide me with proof as to why choosing the two biggest numbers in problem C always leads to the smallest number remaining at last? It was intuitive but why does that greedy actually works? Thanks in advance!

If you look at the sum of the entire array, you'll see that after n-1 operations this sum is the number itself, which we have to minimize At each operation we are removing (a + b) /2

From the sum, so to minimize the sum you'll choose the largest 2 numbers

I think I have a better solution for F in implement,if the n is 5000 , we can use HashMap or other "almost" O(1) datastruct to replace map.Let's get down to our business,we just naive denote dp[i][j] is we finish i task and we left j bullets , we cost the minmum bullets , and normal transtation which should consider if we finish the i in advance or R[i]!=L[i+1] we can refresh the dp[i+1][k] the Code: https://codeforces.net/contest/1430/submission/95434956

Does anyone have flow solution for problem 'G'?

discussed

D was a very good greedy, thanks, but why shouldn't we delete the first character if no 2 adjacent characters are equal ?

Extremely simple O(1) soln for A 95195199

It may sound stupid but anyone can explain or proof why problem E

"so the first character a in the original string is the first character a in the resulting string, the second character a in the original string is the second character a in the resulting string, and so on."

will give the optimal solution?

I was trying to debug my Solution for E for the past hour. I realized that I declared the vector as char, instead of int -_- (95816212)

I love how bugs in Competitive Programming makes you want to kill yourself!

Stop writing "it's easy to see that" in tutorials. Makes me feel dumb.

Yep, apparently if you are good at something, you have to tell others that it is easy, in order to increase your ego.

Can someone explain E?

I don't know if you already solved, but i'll try to explain in a different way from the editorial.

Let's call

tthe inverted strings, and let's define an arraypof n positions where:p[i]=position of the character of s that is in positionioft.It can be seen that if

p= [n, n-1, n-2, ..., 3,2,1] we would have the stringsinverted.Example:

It can also be seen that the number of inversions in this array is equal to the number of operations required to place all letters in their initial positions.

But here we are faced with a problem, we can see that the number of inversions of this array is maximum (read the article in the link to understand), so we need to somehow try to minimize it to the maximum, how to do this?

Since some characters in the string

scan be repeated, let's take advantage of this.Example:

Note that in the example above the two arrays

pcreate the same stringt, butp'needs a smaller number of operations to be ordered (it has less inversions). This is because the 'a' appears twice ins. After that, it is possible to notice that if there are repeated letters, their positions inpmust be in ascending order in the order in which they appear ins, thus the number ofpinversions will always be minimized.The final answer will be the number of inversions in

p.My solution (don't forget to use long long): 97206684

Thank you for your detailed explanation! Please allow me to add my understanding for Problem E:

When sorting an array by bubble-sort, the number of swap() operations is equal to the number of inversions in this array. Proof: 1. Assume this array is non-empty 2. This array has no inversions if and only if this array is already sorted. 3. Each legit swap() operation will reduce the number of inversions by 1.

I have been stuck by this proof for hours and finally understood how straightforward it is. Hope my posting could help someone else.

Ref: https://stackoverflow.com/a/32119927/1166518

nice contest

what is dp array in the solution code, seems very different from the tutorial

In the editorial of D, para 3, it is written, "It's easy to see that we should choose a character from the leftmost such block since that block is the earliest to be deleted (and if we want to make the same action later, we might be unable to do it).".

My doubt is why is it optimal to delete from left only? Why does it matter which block we delete?

A slightly different solution for F, it runs in $$$O(N^2logN)$$$, but could be sped up. We can simply store the dp parameters in a set — the parameters are $$${x, y}$$$, which means that in order to have x left-over bullets, we need to waste y bullets. The transitions are simple to do in $$$O(NlogN)$$$, and this problem can probably be solved in $$$O(NlogN)$$$ with slope trick, but I will try to do so later.

Working code: https://codeforces.net/contest/1430/submission/209195512

Can anyone please explain the C problem?

The test case they have given is not correct I think. Please correct me