Idea: adedalic
Tutorial
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Solution (adedalic)
#include<bits/stdc++.h>
using namespace std;
int main() {
int t; cin >> t;
while (t) {
int k; cin >> k;
cout << 100 / gcd(100, k) << endl;
}
return 0;
}
Idea: BledDest
Tutorial
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Solution (Neon)
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
while (t) {
int n;
scanf("%d", &n);
vector<int> a(n);
for (int &x : a) scanf("%d", &x);
int ans = 2;
if (is_sorted(a.begin(), a.end()))
ans = 0;
else if (a[0] == 1  a[n  1] == n)
ans = 1;
else if (a[0] == n && a[n  1] == 1)
ans = 3;
printf("%d\n", ans);
}
}
Idea: BledDest
Tutorial
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Solution (awoo)
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
struct bot{
int x, d;
};
int main() {
int t;
cin >> t;
forn(_, t){
int n, m;
scanf("%d%d", &n, &m);
vector<bot> a(n);
forn(i, n) scanf("%d", &a[i].x);
forn(i, n){
char c;
scanf(" %c", &c);
a[i].d = c == 'L' ? 1 : 1;
}
vector<int> ord(n);
iota(ord.begin(), ord.end(), 0);
sort(ord.begin(), ord.end(), [&a](int x, int y){
return a[x].x < a[y].x;
});
vector<int> ans(n, 1);
vector<vector<int>> par(2);
for (int i : ord){
int p = a[i].x % 2;
if (a[i].d == 1){
if (par[p].empty())
par[p].push_back(i);
else{
int j = par[p].back();
par[p].pop_back();
ans[i] = ans[j] = (a[i].x  (a[j].d == 1 ? a[j].x : a[j].x)) / 2;
}
}
else{
par[p].push_back(i);
}
}
forn(p, 2){
while (int(par[p].size()) > 1){
int i = par[p].back();
par[p].pop_back();
int j = par[p].back();
par[p].pop_back();
ans[i] = ans[j] = (2 * m  a[i].x  (a[j].d == 1 ? a[j].x : a[j].x)) / 2;
}
}
forn(i, n){
printf("%d ", ans[i]);
}
puts("");
}
return 0;
}
Idea: BledDest
Tutorial
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Solution (BledDest)
#include <bits/stdc++.h>
using namespace std;
const int INF = int(1e9);
int main()
{
int n;
cin >> n;
vector<int> a(n);
for(int i = 0; i < n; i++)
cin >> a[i];
vector<int> pos;
for(int i = 0; i < n; i++)
if(a[i] == 1)
pos.push_back(i);
int k = pos.size();
vector<vector<int>> dp(n + 1, vector<int>(k + 1, INF));
dp[0][0] = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j <= k; j++)
{
if(dp[i][j] == INF) continue;
dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]);
if(j < k && a[i] == 0)
dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + abs(pos[j]  i));
}
cout << dp[n][k] << endl;
}
Idea: BledDest
Tutorial
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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 pair<int, int> pt;
const int MOD = 998244353;
int norm(int a) {
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 binPow(int a, int k) {
int ans = 1;
while (k > 0) {
if (k & 1)
ans = mul(ans, a);
a = mul(a, a);
k >>= 1;
}
return ans;
}
int inv(int a) {
return binPow(a, MOD  2);
}
vector< vector<int> > d;
int n, m;
inline bool read() {
if(!(cin >> n >> m))
return false;
d.resize(n, vector<int>(m));
fore (i, 0, n) fore (j, 0, m)
cin >> d[i][j];
return true;
}
inline void solve() {
int invFact = 1;
fore (i, 1, n + 1)
invFact = mul(invFact, i);
invFact = inv(invFact);
int E = 0;
fore (j, 0, m) {
vector<int> cnt(n + 1, 0);
fore (i, 0, n)
cnt[n + 1  d[i][j]]++;
vector<int> d(n + 1, 0);
d[0] = 1;
int rem = 0;
fore (i, 0, n) {
rem += cnt[i];
d[i + 1] = norm(d[i + 1] + mul(d[i], rem));
rem = max(0, rem  1);
}
// cerr << d[n] << "  " << norm(1  mul(d[n], invFact)) << endl;
E = norm(E + 1  mul(d[n], invFact));
}
cout << E << 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: BledDest
Tutorial
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Solution (BledDest)
#include <bits/stdc++.h>
using namespace std;
const int N = 543;
const long long INF = (long long)(1e18);
struct Matching
{
int n1, n2;
vector<set<int>> g;
vector<int> mt, used;
void init()
{
mt = vector<int>(n2, 1);
}
int kuhn(int x)
{
if(used[x] == 1) return 0;
used[x] = 1;
for(auto y : g[x])
if(mt[y] == 1  kuhn(mt[y]) == 1)
{
mt[y] = x;
return 1;
}
return 0;
}
int calc()
{
init();
int sum = 0;
for(int i = 0; i < n1; i++)
{
used = vector<int>(n1, 0);
sum += kuhn(i);
}
return sum;
}
void remove_vertex(int v, bool right)
{
if(right)
{
for(int i = 0; i < n1; i++)
g[i].erase(v);
}
else
g[v].clear();
}
void add_edge(int x, int y)
{
g[x].insert(y);
}
Matching() {};
Matching(int n1, int n2) : n1(n1), n2(n2)
{
g.resize(n1);
};
};
int n, m, k;
long long dp[N][N];
int p[N][N];
vector<int> g[N];
long long x[N], y[N];
int main()
{
cin >> n >> m >> k;
for(int i = 0; i < m; i++)
{
int u, v;
cin >> u >> v;
u;
v;
g[u].push_back(v);
}
for(int i = 0; i < k; i++)
cin >> x[i] >> y[i];
Matching mt(n, n);
for(int i = 0; i < n; i++)
for(auto j : g[i])
mt.add_edge(i, j);
int cnt = mt.calc();
int cur = cnt;
vector<int> seq;
while(cur > 0)
{
int idx = 0;
for(int i = 0; i < n; i++)
{
Matching mt2 = mt;
mt2.remove_vertex(i, false);
if(mt2.calc() < cur)
idx = i + 1;
mt2 = mt;
mt2.remove_vertex(i, true);
if(mt2.calc() < cur)
idx = (i + 1);
}
assert(idx != 0);
seq.push_back(idx);
mt.remove_vertex(abs(idx)  1, idx < 0);
cur;
}
reverse(seq.begin(), seq.end());
for(int i = 0; i <= k; i++)
for(int j = 0; j <= cnt; j++)
dp[i][j] = INF;
dp[0][cnt] = 0;
for(int i = 0; i < k; i++)
for(int j = 0; j <= cnt; j++)
{
if(dp[i][j] == INF) continue;
for(int z = 0; z <= j; z++)
{
if(i + 1 + z >= n) continue;
int t = j  z;
long long add = max(0ll, x[i]  t * y[i]);
if(dp[i + 1][z] < dp[i][j] + add)
{
dp[i + 1][z] = dp[i][j] + add;
p[i + 1][z] = j;
}
}
}
cur = max_element(dp[k], dp[k] + cnt + 1)  dp[k];
vector<int> res;
for(int i = k; i > 0; i)
{
res.push_back(0);
for(int j = p[i][cur]  1; j >= cur; j)
res.push_back(seq[j]);
cur = p[i][cur];
}
reverse(res.begin(), res.end());
cout << res.size() << endl;
for(auto x : res) cout << x << " ";
cout << endl;
}
Thanks for this nice and clear tutorial
The author's implementation of task C is much better than mine.
Thanks, that's why we didn't question it being C.
Finally editorial is out.
Is this contest rated?
In Problem D I find this sentence very confusing:
$$$ $$$ "let $$$dp_{i, j}$$$ be the minimum time if we considered $$$i$$$ first positions and picked $$$j$$$ of them as the ending ones."
What does "pick $$$j$$$ of them (them='$$$i$$$ positions'?) as the ending ones" mean? Shouldn't it be:
$$$ $$$ " let $$$dp_{i, j}$$$ be the minimum time if we considered $$$i$$$ first starting positions and $$$j$$$ first ending positions."
This state represents a situation when we've considered the armchairs $$$0, 1, 2, \dots, i  1$$$ and exactly $$$j$$$ of them are chosen as ending positions (so, after the whole process is done, exactly $$$j$$$ people will sit on the segment $$$[0, i  1]$$$ of the armchairs).
I didn't get the thing. What exactly dp[i][j] is representing? Can someone help? Thanks!
The representation of dp[i][j] is that we have considered first i seats and we have made the consideration of the first j seats on which the people were sitting initially and have given them some new position.
Not first j, in total j.
I've managed to solve this task now. My understanding/interpretation is not exactly the same as in the tutorial, but maybe this image (which I also drew for myself while solving) will help you.
I counted the ones and the zeros in two seperate arrays. the $$$x_i$$$ and $$$y_j$$$ save the index in the original array (so $$$x_1=0$$$ and $$$y_2=2$$$ in my picture). Then $$$i$$$ says us, how many People we chose, $$$j$$$ tells us, how many empty chairs we chose. And $$$dp_{i,j}$$$ tells us the cheapest solution to place those $$$i$$$ people on those $$$j$$$ chairs.
Some examples:
$$$dp_{1,1}$$$ choses only one person $$$x_1$$$ and one chair $$$y_1$$$. What is the best solution in this case? It's $$$abs(x_1y_1)$$$ since we have only one possibility. Same with e.g. $$$dp_{5,5}$$$, we place the first 5 people on the first 5 chairs. There is only one possibility.
$$$dp_{1,max(j)}$$$ choses only one person $$$x_1$$$ and all empty chairs. What is the best solution in this case? For each pair of $$$x_1$$$ and some chair $$$y_j$$$ we measure the distance. The smallest distance is the answer.
$$$dp_{4,3}$$$ choses 4 people and tries to sit them on 3 chairs. This is impossible, so we assign the value $$$\infty$$$ here.
Now the interesting part is the transition. how would we calculate $$$dp_{3,5}$$$? We want to find the cheapest solution to place 3 people on 5 chairs. We compare two steps. We could try $$$dp_{3,4}$$$ because it is also a solution for $$$dp_{3,5}$$$ (not neccessarily the optimal one) but it is a possible way to place 3 people on 4 places, so its also a valid distribution to place those 3 people on 5 places. Or we could try $$$dp_{2,4}$$$ and add person $$$x_3$$$ on place $$$y_5$$$. This has the cost $$$dp_{2,4} + abs(x_3y_5)$$$. This way we obtain: $$$dp_{3,5}=min(dp_{3,4} \,; dp_{2,4} + abs(x_3y_5))$$$. This relation is enough to obtain the solution, which will be $$$dp_{max(i),max(j)}$$$
I really hope this helps and doesn't just confuse even more.
Great help, thx.
That is really a good explanation, I have written AC code following this exact explanation if anyone is interested.
D
Yes it helped. Thanks
greatly explained !!
perfect explaination doesnt exi.......
wouldnt there be any case where suppose we shift index i+1 to i+2 and then move i to i+1. here i think we are only considering the space that are initially blank?? @OleschY
That's why you have to update the $$$dp_{i,j}$$$ in a specific order (as is customary for DPSolutions). More specifically, $$$dp_{i,j}$$$ depends on $$$dp_{i,j1}$$$ and $$$dp_{i1,j1}$$$. So the two latter values have to be updated before $$$dp_{i,j}$$$. You can achieve this e.g. by iterating $$$i$$$ from $$$1$$$ to $$$max(i)$$$ in the outer loop and iterating $$$j$$$ from $$$i$$$ to $$$max(j)$$$ in the inner loop. Then it all works out!
wow, this was a brilliant explanation. Thanks a ton :)
The startingpositionsmapped toendingpositions part in the editorial was a bit confusing.
occupiedchairsmappedtounoccupiedones would have been better.
Ah, I think D is easier than C.
Maybe ABDCEF is a better choice.
I thike E is also easier than C.If you're familiar with expectations and probabilities,you can solve it easily.
Can you list some tutorials or articles?
This is a good one.
I think everything is easier than c in this world
In the B part problem if the question was changed a little bit that instead of rearranging the elements of subarray we have to choose some subarray and reverse it . Then what will be the approach for that problem.
Check code jam qualification round 2021, you will find the exact problem with analysis.
It took over 24 hours for the Editorial to appear :)
The problems were cool, tho... I really enjoyed that round.
Thanks for the clean explanation and solution. But I still don't know why did I get a WA test 8 at D tho...
haha, everyone who wrote greedy soln. faced that
Same
6
0 0 1 1 1 0
this case is failing for greedy solution of my ans. I think you must have done same mistake
Thanks! I couldn't figure out much from the 8th test case tbh but yours helped
Fast Indeed...!
NGL soln of C is pretty neat. I hated this problem before reading the editorial.
For Problem E, why do we subtract 1 in the second step, 2 in the third step and so on?
In 2nd step,the nth city is fixed.And in 3rd step the nth point and the n1th city is fixed.
For problem A, you can figure out by the observation that the answer will always be the denominator of the fraction k/100, as that's the least sum possible. I don't know the actual working of the function, but I used the fractions library in python to Get the denominator. Solution: 116323887
in problem D , why dp is initialised by infinity ??
because you want to minimize the final ans
The C is so cool.
does anyone know why some problems use 10^9 + 7 as mod while other problems use 998244353 as a mod?
Taking MOD to reduce a number makes sense only if done with prime number and 10^9 + 7 and 998244353 are both prime numbers of order 10^9 so typical int range. I hope it makes sense now. They are just arbitrary prime numbers.
It has to do with FFT, as I understand. See here.
thanks!
In problem D if time take to move from $$$i$$$ to $$$j$$$ is some function $$$f(i,j)$$$ which is quadratic or of higher degree then this $$$dp$$$ solution will not work right?(because we can't say "the leftmost starting position is matched with the leftmost ending" for optimal solution). How we can solve in this case?
I'd like to mention that it is well known in China how to solve problem D in O(n) time, and other generalisations of the problem are also studied and known. See This PDF (Problem 2 is problem D). The pdf is in Chinese but I don't have a better resource for it, maybe people interested can try google translate. Maybe you already know it, but I'm posting for those who don't.
Is there anything that is nonstandard for chinese guys ?
I hope that they ask the same question in some other contest with bigger constraints: Expected solution O(n). That way, I will learn the method by reading blogs and editorials.
Reading, translating, and understanding a Chinese document seems too difficult for me.
Could you give me some other access to the pdf without vpn in China? I can hardly download it.
Could you give me some other access to the pdf without using vpn in China? I can hardly download it.
Oh, I borrow someone's VPN acount and download it, thank you for sharing this algorithm!
at nocriz if you can elaborate a bit on this (if you know and wont take much of your time), would be helpful.
Even though I'm Chinese I still can't understand it :(
Nice.
Hey, I'm trying to understand but not getting the intuition behind A, can someone help me ?
Every percent can be written as a fraction.
25% > 25/100
40% > 40/100
Remember u can always simplify the fraction. So 25/100 becomes 1/4, 40/100 becomes 2/5 etc. Amount of moves you need to do is in denominator (w + e).
So all you need to do is simplify fraction to the simplest form and you use GCD for it.
25/100 > GCD is equal to 25. Because all you need is the denominator u divide 100 by GCD and that's your answer.
Thank You So Much, I have now understood the idea behind the gcd.
Can someone please explain in detail how to solve D using flows? Also, flows tag has been assigned to the problem in the Problemset.
Make a graph with n+2 nodes, one for each armchair, a source and a sink. Add an edge between adjacent armchairs with cost 1 and infinite capacity. For occupied chairs, add an edge from the source to that chair with cost 0 and capacity 1, same for unoccupied chairs to the sink. Now you can run mincost maxflow on this.
Can you elaborate on the complexity of this solution ?
and how it can pass under those constraints !
So each chair has one edge either from the source, or to the sink, that's $$$N$$$ edges. Additionally, each pair of adjacent chairs has two edges, one in each direction. So that's $$$N + 2(N  1) = 3N  2$$$ edges. Kactl's MCMF runs in $$$O(E^2) = O(2 * 10^8)$$$ but I guess the implementation ended up being too slow. AtCoder Library has one that runs in $$$O(F(V+E)\log(V + E))$$$, and was fast enough to pass. Here, $$$F$$$ is the max flow which we know is at most $$$\frac{N}{2}$$$.
In this problem even MCMF which uses SPFA will be acceptable and it isn't hard to prove. Here can be n/2 edges from source at maximum (occupied chairs) and in worst case the flow will visit n1 edges between chairs + n1 edges from unoccupied chairs to sink. So the time complexity will be O( n/2 * (n1+n1) ) = O(n^2) = O(2*10^7).
Heyy,
I don't know much about this topic. Can Someone please share some resource or questions from where I can get to know more about it
Solution for C was nice :D . But can someone tell me why I am getting this runtime error on test 4 without any diagnostics for my solution? My Submission NVM. I got it. IGNORE :)
For Problem D is there any greedy solution?
Can anyone tell me how to prepare for problems like C?
In editorial of problem E, Can anyone explain the meaning of this sentence "Let's for each turn k∈[0,n) calculate the number of cities that you can build Monument in starting this turn as cnt[k]"? What exactly are we storing in cnt array?
I have the same question
I have added few comments in the code snippet which is necessary to understand the explanation further.
For each point $$$j, 0 <= j < m$$$ (0 based indexing), $$$cnt[i]=$$$ number of cities in which we can build a monument on the $$$i^{th}$$$ day, such that the $$$j^{th}$$$ point won't be captured by any of the cities $$$\epsilon$$$ $$$cnt[i]$$$.
Can anyone help me out to find my mistake for solution of Problem D. What I am doing is that for every position already occupied, I at first find a position of the leftmost available place to that and the rightmost available place. If any of these two positions is already occupied, then I simply choose the one available, and in case if both are available, then I simply pick the position which takes the minimum time. This is my code : 116381349
6
0 0 1 1 1 0
This case is failing for your answer and also for almost every greedy solution.
Thanks for the reply.I have found my mistake
nocriz already mentioned $$$O(n)$$$ solution for D, but sadly not everyone can read chinese. I will explain another $$$O(n)$$$ solution from frodakcin
Let's look at the resulting solution and analyze it. I claim that resulting array can be divided into segments such that each segment is either
1) All cells in a segment are not people and are not new places for them  useless cells. All segments of this type will have length one, and if there are $$$m$$$ people, there will be exactly $$$n2m$$$ such segments.
2) Segment of cells such that it has even length at it has same amount of people as new seats for them inside and people match places in this segment (for example, segment has length 4, and there are 2 people who moved to 2 places)
To prove, let's find first not useless cell and go to the right, maintaining balance. Each person gives $$$+1$$$, and each taken place gives $$$1$$$. If we arrive at balance 0, we got segment of type 2 and can start new one. If we didn't arrive at balance 0, there are two cases
1) we reached the end of array. That is just impossible, because number of taken places is equal to the number of people
2) we reached useless cell with nonzero current balance. Again, that is impossible because that would mean that we have nonzero balance on a segment, so someone (person or place) from this segment is matched to something to the right of empty cell and this is just not an optimal answer (we can match with empty cell instead)
Now you can notice that there are only $$$O(n)$$$ interesting segments. All segments of type 1 are just single cells, and all segments of type 2 are $$$[l, r]$$$ such that for each $$$l$$$, $$$r$$$ is minimal such that
balance([l, r]) = 0
. And we have to "construct" our array from these segments. In other words, we can replace segments $$$[l, r]$$$ with cost $$$c$$$ with edge from $$$l$$$ to $$$r+1$$$ with cost $$$c$$$ and find shortest path from $$$1$$$ to $$$n+1$$$. Obviously you can find shortest path in $$$O(n)$$$ if you know costs of all edges.Cost of type1 edge is $$$0$$$. To calculate cost of type2 edge, let's write it: it is sum of $$$abs (person\_position  place\_position)$$$. Since this is minimal segment with balance $$$0$$$, you can see that all values $$$(person\_position  place\_position)$$$ will have the same sign, so you just need $$$abs(sum(person\_positions)  sum(empty\_places\_positions))$$$, and sum of such positions you can easily calculate with prefix sums.
You can look at my submission with some comments here
.
I found problem E very interesting. This was the first time I came across a probabilty related problem in competitive programming. [ still a noob:) ]
Problem C is pretty amazing actually. Kudos to the author!!
can anybody please give a more detailed explanation for D as I am not good with DP , after reading this tutorial for many times still I am stuck , Thanks in advance.
This is my Logic
And this is my Submission:116702763
Thanks, it really helped.
why problem D follows dp and not greedy ? like how we identify the property that tells us that this problem exhibits dp property ?
How on earth is E rated just 2100?
is there anywhere I can find an explanation for C
Problem 1525C  Robot Collisions is very interesting and it's tutorial is more interesting. Thanks a lot BledDest

I am wondering what would the solution be for problem D if the problem is extended to 2d array.
[moving cost being L1]
can anyone help me find the problem with my implementation of problem E?119103848
If problem D is modeled as a linear sum assignment problem (a bipartite graph with 2n vertices and connection between vertices i and j+n if a[i] = 1 and a[j] = 0), there is some general linear sum assignment algorithm that doesn't give TLE? I tryed successive shortest path (with O(E*logV) Dijkstra as intermediate step). I think it solves the general linear sum assignment problem in O(n²*logn) but it gives TLE on testcase 31 for this problem (code). So I wondered if it's some problem in my implementation or just the fact that the time limit for this problem is tight enough to avoid solutions using a general linear sum assignment algorithm.
In C, how did this formula come?