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1971A - My First Sorting Problem
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200'007;
const int MOD = 1'000'000'007;
void solve() {
int x, y;
cin >> x >> y;
cout << min(x, y) << ' ' << max(x, y) << '\n';
}
int main() {
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200'007;
const int MOD = 1'000'000'007;
void solve() {
string s;
cin >> s;
bool ok = false;
for (int i = 1; i < (int)(s.length()); i++) {
if (s[i] != s[0]) {swap(s[i], s[0]); ok = true; break;}
}
if (!ok) {
cout << "NO\n"; return;
}
cout << "YES\n";
cout << s << '\n';
}
int main() {
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200'007;
const int MOD = 1'000'000'007;
void solve() {
int a, b, c, d;
cin >> a >> b >> c >> d;
string s;
for (int i = 1; i <= 12; i++) {
if (i == a || i == b) {s += "a";}
if (i == c || i == d) {s += "b";}
}
cout << (s == "abab" || s == "baba" ? "YES\n" : "NO\n");
}
int main() {
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200'007;
const int MOD = 1'000'000'007;
void solve() {
string s;
cin >> s;
int res = 1;
bool ex = false;
for (int i = 0; i + 1 < (int)(s.size()); i++) {
res += (s[i] != s[i + 1]);
ex |= (s[i] == '0' && s[i + 1] == '1');
}
cout << res - ex << '\n';
}
int main() {
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
Idea: mesanu
Tutorial
Tutorial is loading...
Solution
#include <iostream>
#include <algorithm>
#include <vector>
#include <array>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <chrono>
#include <random>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <iomanip>
#include <bitset>
#include <cassert>
typedef long long ll;
using namespace std;
void solve()
{
int n, k, q;
cin >> n >> k >> q;
vector<long long> a(k+1), b(k+1);
a[0] = 0;
b[0] = 0;
for(int i = 1; i <= k; i++)
{
cin >> a[i];
}
for(int i = 1; i <= k; i++)
{
cin >> b[i];
}
for(int i = 0; i < q; i++)
{
long long c;
cin >> c;
int l = 0, r = k;
while(l <= r)
{
int mid = l+r>>1;
if(a[mid] > c)
{
r = mid-1;
}
else
{
l = mid+1;
}
}
if(a[r] == c)
{
cout << b[r] << " ";
continue;
}
long long ans = b[r] + (c-a[r])*(b[r+1]-b[r])/(a[r+1]-a[r]);
cout << ans << " ";
}
cout << endl;
}
int32_t main(){
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
Idea: mesanu
Tutorial
Tutorial is loading...
Solution
#include <iostream>
using namespace std;
void solve()
{
long long r;
cin >> r;
long long height = r;
long long ans = 0;
for(long long i = 0; i <= r; i++)
{
while(i*i+height*height >= (r+1)*(r+1))
{
height--;
}
long long cop = height;
while(i*i+cop*cop >= r*r && cop > 0)
{
cop--;
ans++;
}
}
cout << ans*4 << endl;
}
int32_t main(){
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
Idea: mesanu
Tutorial
Tutorial is loading...
Solution
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
using namespace std;
void solve()
{
int n;
cin >> n;
vector<int> a(n);
map<int, priority_queue<int>> mp;
for(int i = 0; i < n; i++)
{
cin >> a[i];
mp[a[i]>>2].push(-a[i]);
}
for(int i = 0; i < n; i++)
{
cout << -mp[a[i]>>2].top() << " ";
mp[a[i]>>2].pop();
}
cout << endl;
}
int32_t main(){
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
#include <algorithm>
#include <utility>
#include <vector>
namespace atcoder {
namespace internal {
template <class E> struct csr {
std::vector<int> start;
std::vector<E> elist;
csr(int n, const std::vector<std::pair<int, E>>& edges)
: start(n + 1), elist(edges.size()) {
for (auto e : edges) {
start[e.first + 1]++;
}
for (int i = 1; i <= n; i++) {
start[i] += start[i - 1];
}
auto counter = start;
for (auto e : edges) {
elist[counter[e.first]++] = e.second;
}
}
};
// Reference:
// R. Tarjan,
// Depth-First Search and Linear Graph Algorithms
struct scc_graph {
public:
scc_graph(int n) : _n(n) {}
int num_vertices() { return _n; }
void add_edge(int from, int to) { edges.push_back({from, {to}}); }
// @return pair of (# of scc, scc id)
std::pair<int, std::vector<int>> scc_ids() {
auto g = csr<edge>(_n, edges);
int now_ord = 0, group_num = 0;
std::vector<int> visited, low(_n), ord(_n, -1), ids(_n);
visited.reserve(_n);
auto dfs = [&](auto self, int v) -> void {
low[v] = ord[v] = now_ord++;
visited.push_back(v);
for (int i = g.start[v]; i < g.start[v + 1]; i++) {
auto to = g.elist[i].to;
if (ord[to] == -1) {
self(self, to);
low[v] = std::min(low[v], low[to]);
} else {
low[v] = std::min(low[v], ord[to]);
}
}
if (low[v] == ord[v]) {
while (true) {
int u = visited.back();
visited.pop_back();
ord[u] = _n;
ids[u] = group_num;
if (u == v) break;
}
group_num++;
}
};
for (int i = 0; i < _n; i++) {
if (ord[i] == -1) dfs(dfs, i);
}
for (auto& x : ids) {
x = group_num - 1 - x;
}
return {group_num, ids};
}
std::vector<std::vector<int>> scc() {
auto ids = scc_ids();
int group_num = ids.first;
std::vector<int> counts(group_num);
for (auto x : ids.second) counts[x]++;
std::vector<std::vector<int>> groups(ids.first);
for (int i = 0; i < group_num; i++) {
groups[i].reserve(counts[i]);
}
for (int i = 0; i < _n; i++) {
groups[ids.second[i]].push_back(i);
}
return groups;
}
private:
int _n;
struct edge {
int to;
};
std::vector<std::pair<int, edge>> edges;
};
} // namespace internal
} // namespace atcoder
#include <cassert>
#include <vector>
namespace atcoder {
// Reference:
// B. Aspvall, M. Plass, and R. Tarjan,
// A Linear-Time Algorithm for Testing the Truth of Certain Quantified Boolean
// Formulas
struct two_sat {
public:
two_sat() : _n(0), scc(0) {}
two_sat(int n) : _n(n), _answer(n), scc(2 * n) {}
void add_clause(int i, bool f, int j, bool g) {
assert(0 <= i && i < _n);
assert(0 <= j && j < _n);
scc.add_edge(2 * i + (f ? 0 : 1), 2 * j + (g ? 1 : 0));
scc.add_edge(2 * j + (g ? 0 : 1), 2 * i + (f ? 1 : 0));
}
bool satisfiable() {
auto id = scc.scc_ids().second;
for (int i = 0; i < _n; i++) {
if (id[2 * i] == id[2 * i + 1]) return false;
_answer[i] = id[2 * i] < id[2 * i + 1];
}
return true;
}
std::vector<bool> answer() { return _answer; }
private:
int _n;
std::vector<bool> _answer;
internal::scc_graph scc;
};
} // namespace atcoder
using namespace std;
using namespace atcoder;
void solve() {
int n;
cin >> n;
int b[3][n];
for (int i = 0; i < 3; i++) {
for (int j = 0; j < n; j++) {
cin >> b[i][j];
}
}
two_sat ts(n);
for (int j = 0; j < n; j++) {
for (int i = 0; i < 3; i++) {
int nxt = (i + 1) % 3;
ts.add_clause(abs(b[i][j]) - 1, b[i][j] > 0, abs(b[nxt][j]) - 1, b[nxt][j] > 0);
}
}
cout << (ts.satisfiable() ? "YES\n" : "NO\n");
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
🌹
My post contest discussion stream
What if you changed thte black backlight to a light white one
My joke solution to problem C:
https://codeforces.com/contest/1971/submission/260436859
Bro can you explain your solution?
He converts given integers to points on plane to find the answer with geometry
I was gonna ask you to atleast give me a keyword to google. But i asked chat gpt. Only thing i understood is it's not worth it at my level. lol
Anyway, Thank you for your time.
e^(i*omega) produces a point in the unit circle in the complex plane. Here I use it to produce the position of 1-12 o’clock. Then, there is a triangle area formula that produce positive value if 3 points are given in counterclockwise order, negative value clockwise. So I use it to calculate the area of Δabc and Δabd. It’s obvious that c and d are on opposite of line ab iff one is positive and the other is negative.
he just , gave the hands of click different coordinates on a plane , then plotted the time(points) on the given graph , then check if there is any point of intersection between the given two lines , an interesting way XD.
Dude, you included the wrong tutorial for problem B.
problem b editorial for now is an editorial of an old div4b problem, fix pls
upd1: fixed now
Too Fast!
https://codeforces.com/contest/1971/submission/260448426
-> this is giving out of bound
https://codeforces.com/contest/1971/submission/260447515
-> this is not giving out of bond
Both code are same,, just the writing style is different..
Why is this happening?? Thanks in advance..
flamestorm(could you plz help sir)
The difference lies in how division is performed: the first snippet directly divides by a variable (speed) that could be close to zero, risking a division by zero or resulting in large values leading to out-of-bounds errors when accessing the 't' vector, while the second snippet avoids this issue by rearranging the calculation to use multiplication and division together.
the second code that passed should get out of bound
he is using upper bound take for example this case 10 1 1 10 10 10
it should give RE but it's not because I tried to hack so why ??
The second code is avoiding the expected out-of-bounds error because it decrements the index i after upper_bound, making it within bounds. However, this behavior is unreliable and could lead to issues with different inputs or compiler optimizations.
he decrements the index i but he is using (i+1)
In the second code, even though the index 'i' is decremented after upper_bound, it's still used to access t[i], which could lead to accessing an out-of-bounds element since i is decremented but used as 'i+1'. This behavior is indeed a vulnerability in the code, which could lead to unpredictable results or runtime errors, depending on the input.
https://codeforces.com/contest/1971/submission/260423806 This friend of mine did the same thing using another vector but his code didn't show any kind of this behaviour
This code ensures that mid always points to the correct segment in the array 'a', preventing out-of-bounds issues when accessing elements from 'a' and 'b'. Therefore, the code doesn't exhibit the same behavior as the previous codes.
The first code also decrement i to prevent out of bound.. And, I said the division by smaller number logic is invalid since he did the same thing...
You're correct. I apologize for that. Both the first and second snippets decrement the index i to mitigate the risk of accessing out-of-bounds elements. Additionally, you rightly pointed out that the division by a smaller number logic doesn't apply since both codes essentially use the same approach.
In that case, it seems that the difference in behavior between the two codes might be due to other factors, such as input data or compiler optimizations
Input data is same brother
weird
took 30 min to solve F+G, still have no idea how I got WA from E 260451462
too bad actually
Solutions that use floating points will eventually get hacked.
please tell me why this is not getting accepted but (PROBLEM E)
long long ans = brr[it] + ((drr[i]-arr[it]) / ((arr[it+1]-arr[it])/(brr[it+1]-brr[it])));this is accepted? how?
long long ans = brr[it] + (drr[i]-arr[it]) * (brr[it+1]-brr[it])/(arr[it+1]-arr[it]);260452416 260452100
please, put your code in spoiler next time, or at least on different lines.
from what i can see, in the first one you should have used floating point arithmetic, but it would be a bad idea, cuz solutions with floating point are getting hacked now.
the second one just makes your original formula compatible with integer arithmetics in c++, so it works just right
I am sorry, for the bad structure of the comment, will do the needful next time.. "the second one just makes your original formula compatible with integer arithmetics in c++, so it works just right" how you know that, like how can I think like that during the contest.
try to just simplify formula as mush as you can (and also try to use less division), that's what i usually do
thankyou very much :)) wishing you loads of +ve delta
thank you!
hey, I am not getting why i am getting TLE for the same problem 261422309. I even implemented the same binary search as the official solution
in the binary search function, every time that you use it, you create a new copy of a vector. what you should do instead is pass it by reference, just changing the vector v to vector &v should work
here is your solution when vector is passed by reference 261565356
Never knew making copies would take up so much time. Thank a lot mate.
Same problem and really learned a lot! Thanks!
yeah similar problem same testcase failed.
void main() { long long s = 999999937; long long a = (double)1.0 / s * s; long long b = s * 1 / s; printf("%lld %lld", a, b); }
Detailed Video Editorials (BCDEFG)
https://www.youtube.com/playlist?list=PLbAEGNVuy98kfnqOz26QPV4cSpj9j-3tY
Language — Hindi
nice editorial but why does always with c++ why dont in python or other languages
+1
its better for the purpose ig
Can anyone explain me please(with reference to problem E) — as the answer needs to be rounded off to nearest integer value...so let say total comes out to be like 1.6 then the rounded off answer wouldn't be 2 instead of 1. And without using round() 1 will be the ans, then how it's correct without using round function? Am i misunderstood the word rounded *down(instead off) which means to always round off to floor value? But i sumbitted with using round() also..and why ac for that?
i think that's just a typo, because russian statement says it should be rounded down
Then why it got also ac with using round() In case like 1.6 it gives 2 but the ans would be 1 yeah?
Now i got the clarity. I used int and before rounding off it already converted into int .so there is no effect of round() in this situation..lol blunder by me
XD
I wrote this code for problem F During the contest.. But now I really have no idea way it works !!!!!!
There is a case when r*r-x*x-1 becomes negative and then sqrt(negative) !!
I didn't notice that during the contest but I am really surprised to see it works!
the second thing is why the answer is 2 ? when I wrote this weird solution I noticed that it get the write answer — 2 so I added two to the answer
Very weired
1 . Yes, $$$r^2 - x^2 - 1$$$ becomes $$$-1$$$ when $$$x = r$$$, then it will become -2147483648 (saw from code), but then you multiply it by 2, so it becomes 0 (because of the overflow).
You can add this in your loop to check it out for yourself.
cout << x << " " << (int)sqrt(r*r-x*x-1) << " " << (int)sqrt(r*r-x*x-1)*2 << endl;2 . It comes from first (why you need to add 2). When $$$x = ±r$$$, you have $$$\sqrt{2r} - 0$$$ instead of $$$\sqrt{2r} - (-1)$$$, so you loose 1 point for both $$$x = r$$$, and $$$x = -r$$$.
Very helpful. Thanks alot
In the Editorial Implementation for the problem F, when we say the overall answer is the answer per quadrant multiplied by 4, aren't we over-counting at the edges of the semi-disc (like x=0 or x=r) ?. Could anyone help figure out what I might be missing. Thanks.
We have counted at x = 0, but not at h = 0 (in second while statement there is h > 0 which ensures that we are not crossing x-axis).
Didn't see that. That's slick! Thanks!
Shouldn't this editorial be published after the end of hacking phase? [In Educational Rounds and Div3s, I usually don't see Editorials to be published before the end of Hacking Phase]
This is my solution for G if using maps, if you don't know priority_queue .
I learnt programming a month ago. I gave my first contest(944 div.4) yesterday n I solved only 3 problems. I read that div.4 is rated for unrated players. But my account is still unrated. can anyone explain why? Why I didnt get any rating?
raiting change will came after open hacks whic is 12 hours, and then system testing, which is also 4-5 hours, so wait
260473966 why is this failing? I cant seem to find the issue.
res = b[prevdist] + (d - a[prevdist]) * ((b[prevdist + 1] - b[prevdist]) / (a[prevdist + 1] - a[prevdist]));is wrong since you do the division first, which would lose precision.correct version:
res = b[prevdist] + (d - a[prevdist]) * (b[prevdist + 1] - b[prevdist]) / (a[prevdist + 1] - a[prevdist]);Learned this three things: 1. simplifying the equation so that division operations are less resulting less floating point errors. (Though not much experienced in this, I'll be glad to accept tips!) 2. precedence (* comes before /) 3. associativity (/ comes before *)
Thanks a lot, your comment helped!
it run well and output correct answer in my c++ ide ,but wrong on test 31, 260475302
In C, you say "Below is the shortest solution we know of". What about my solution? https://codeforces.com/contest/1971/submission/260294806
For Problem C 1971C - Clock and Strings, We can:
1.Perform a,b,c,d = min(a,b), max(a,b), min(c,d), max(c,d)
2.Check if a<c<b and !(c<b<d) then YES else NO
interesting , can u show ur submission ?
If anyone is interested in solving problem H without fully implementing 2-SAT algo, here is a solution.
Those who used float/double in E ------------------> :(
Then what to use there & why not use double/float?
Thanks
There has been a lot of discussion on this. Check out the discussions on this blog, and you'll understand. It's actually about precision errors that occur when you convert to double/float.
Got Hacked in "E — Find the Car" due to unnecessary cast to double :(
my submission got hacked it gave wrong answer on test case 21 but when i run that test case on my ide other than that of cf it is giving the expected output . Can somebody help https://codeforces.com/contest/1971/submission/260487849
You are given the whole input of TC21 and it's easy to calculate the real answer by hand, so you should be able to debug what went wrong in your code. The expected answer for the fourth query is: $$$\displaystyle \frac{45 \cdot 2}{6} = 15$$$. Your code produces inaccurate results due to the use of
double.Basically, you shouldn't use
doublein this problem. It's very hard to make calculation usingdoubleright (there are more tricky edge cases; see hacks). It's possible to solve this problem with just integer arithmetic, which gives precise answers.For someone who need a small hack test case for E
answer is 29, you might get 28 due to precision errors.
in problem F if we can implement a function g that returns the number of points that have a distance d<r then the answer is g(r+1)-g(r) this function can easily be made with binary search this is my submission 260491421
In Problem E ,I submitted the following code : My Submission
but it gives wrong answer on test case : 1 6 1 1 6 45 2
the expected output is 15 but my code's output is 14
When i debug my code, i found that whole error occurs in the line long long ans=floorl(time); bcoz the value of time before was 15 but after taking floorl it became 14.I don't understand why this is happening on my system and also on online judge.Is this a bug of C++ 20?? or i have done something wrong in my code.I have also used fixed<<setprecision(0) to avoid printing of answer in scientific notation like 1e9 instead of 1000000000 . But it does not affect my answer but floorl does. Kindly help me out and explain this unexpected behaviour of floorl which can be used for long double numbers as per documentation.
The value of
timebeforefloorlis 14.999999999999999999, sofloorl(time)is 14. Floating point arithmetic is inaccurate by design, so this is expected behavior.what best practice should we follow when dealing with floating point arithmetic.
Ok,got it.Thanks.But don't understand while debugging why it shows time 15 instead of 14.99999999 .
When
cout/cerrprints a floating point numbers, the default precision is only 6 digits (IIRC), which is insufficient fordouble/long doubles. To get sufficiently accurate representation, use:or
(the latter requires C++20).
Thanks,very informative.
I don't know why but changing
((dis-a[ind-1])/speed)to((dis-a[ind-1])*(b[ind]-b[ind-1]))/(a[ind]-a[ind-1]);worksCan someone prove why it's working ?
double x=1/3; double y=3; int ans=(floor)(x*y); cout<<ans<<endl;
see this example .33333 * 3 == .9999999 that will round off to 0.
you can't trust doubles, there can be errors. better way would be this
Can someone explain why this is not working ?
I thought long double's precision is enough .
submission : 260494210
flamestorm can you give solution without using atcoder library of problem H??
i wrote my own 2-sat in H after contest, you can look at my solution: 260463835
Here i got a shorter solve of problem C 260325595
flamestorm
Nice
Problem F reminded me of my Computer Graphics course, where we discussed how to draw a circle efficiently on the screen. (It involves avoiding FP computation and replacing multiplication with addition.)
So my solution works as follows. Let point P start at $$$(0, r)$$$. For every step, checks if $$$P$$$ falls in the range. Then, P moves to the right 1 unit, and if $$$|OP| \ge r+1$$$, revert the previous move and move P down 1 unit instead. Repeat until you reach $$$(r, 0)$$$, and multiply the answer by 4.
This algorithm works because the width is 1. The reason why CG introduced this algorithm is that it never requires calculating the $$$|OP|^2$$$ by performing 2 multiplications, but you can instead calculate the increment of $$$|OP|^2$$$ in O(1) additions for each iteration.
this is my first contest on codeforces...can anybody tell me why i don't get the score yet? i did particitate in the contest during the time it was held.
the contest is now retested using tests included in hacks. after this retesting phase you will get your rating
Could someone explain for me this statement from editorial of H:
I mean, without using truth table
0 true variables is obvious
if only one variable is true (let it be $$$x$$$) is true, then 2 out of 3 statements will evalutate to true becuase $$$x$$$ is present in two equations.
but if you turn two variables true (let them be $$$x$$$ and $$$y$$$), then all 3 equations will evaluate to true (2 of which because of variable $$$x$$$ and another one because of variable $$$y$$$ which we couldn't use before)
It's tedious but you can obtain that by using the distributive law of boolean logic: $$$a \land (b \lor c) = (a \land b) \lor (a \land c)$$$ and $$$a \lor (b \land c) = (a \lor b) \land (a \lor c)$$$, where $$$\land$$$ means logical "and" and $$$\lor$$$ means logical "or".
First, the condition "at least two of x, y, z are true" can be translated to: $$$(x \land y) \lor (y \land z) \lor (z \land x)$$$. What we want to prove is $$$(x \land y) \lor (y \land z) \lor (z \land x) = (x \lor y) \land (y \lor z) \land (z \lor x)$$$.
Let's consider just the first two terms. By the distributive law, we get $$$(x \land y) \lor (y \land z) = (x \lor (y \land z)) \land (y \lor (y \land z)) = (x \lor y) \land (x \lor z) \land (y \lor y) \land (y \lor z)$$$. Since $$$y \lor y = y$$$, and $$$y$$$ must be true in this conjunction, $$$x \lor y$$$ and $$$y \lor z$$$ are always true, thus they are redundant and can be removed. Therefore we get: $$$(x \land y) \lor (y \land z) = (x \lor z) \land y$$$.
You can derive $$$((x \lor z) \land y) \lor (z \land x) = (x \lor y) \land (y \lor z) \land (z \lor x)$$$ by expanding the expression in a similar manner. I leave this as your homework because I've written too much $$$\TeX$$$ s already :)
Forward implication is trivial since atmost 1 variable can be false.
Backward implication is also trivial. Assume 2 are false, wlog let them x, y. Then clearly there are 2 things wrong in (x, y) pair. Contradiction
Why am i gettng wrong answer on test case 6 in F, where the value of r is 1e5. Submission link and when I am printing it for the 1e5 case, it gives correct answer. Submission link for accepted
Int the problem F wouldn't be the area of the region be 2*pie*r + pie
In problem E
include<bits/stdc++.h>
using namespace std; // #include <ext/pb_ds/assoc_container.hpp> // #include <ext/pb_ds/tree_policy.hpp> // using namespace __gnu_pbds;
// #define ordered_set tree<int, null_type,less, rb_tree_tag,tree_order_statistics_node_update>
define f0r(a, b) for (long long a = 0; a < (b); ++a)
define f1r(a, b, c) for (long long a = (b); a < (c); ++a)
define f0rd(a, b) for (long long a = (b); a >= 0; --a)
define f1rd(a, b, c) for (long long a = (b); a >= (c); --a)
define ms(arr, v) memset(arr, v, sizeof(arr))
define send {ios_base::sync_with_stdio(false);}
define help {cin.tie(NULL); cout.tie(NULL);}
define fix(prec) {cout << setprecision(prec) << fixed;}
define mp make_pair
define all(v) v.begin(), v.end()
define getunique(v) {sort(all(v)); v.erase(unique(all(v)), v.end());}
define readgraph(list, edges) for (int i = 0; i < edges; i++) {int a, b; cin >> a >> b; a--; b--; list[a].pb(b); list[b].pb(a);}
define vsz(x) ((long long) x.size())
typedef long long ll; typedef long double lld; typedef unsigned long long ull; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef vector vi; typedef vector vl; typedef vector vpi; typedef vector vpl; typedef long long ll; typedef unsigned long long ull; typedef long double lld;
void __print(int x) {cerr << x;} void __print(long x) {cerr << x;} void __print(long long x) {cerr << x;} void __print(unsigned x) {cerr << x;} void __print(unsigned long x) {cerr << x;} void __print(unsigned long long x) {cerr << x;} void __print(float x) {cerr << x;} void __print(double x) {cerr << x;} void __print(long double x) {cerr << x;} void __print(char x) {cerr << ''' << x << ''';} void __print(const char *x) {cerr << '"' << x << '"';} void __print(const string &x) {cerr << '"' << x << '"';} void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V> void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';} template void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << '}';} void _print() {cerr << "]\n";} template <typename T, typename... V> void print(T t, V... v) {_print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
ifndef ONLINE_JUDGE
define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
else
define debug(x...)
endif
void solve() { // TODO: Add your code here int n,k,q; cin>>n>>k>>q; vector a(k+1); vector b(k+1);
a[0]= 0; b[0]= 0; for(int i=1;i<=k;i++){ cin >> a[i]; } for(int i=1;i<=k;i++) { cin>>b[i]; } while(q--){ int d; cin>>d; int it = lower_bound(a.begin(), a.end(), d) - a.begin(); double t2,x2; if(d==a[it]){ cout << b[it] << ' '; continue; } else it--; ll ans = b[it]; ll t = ((d-a[it])*(b[it+1]-b[it])) / (a[it+1]-a[it]); // if(ans - int(ans) >= 0.5) ans = int(ans + 1); // else ans = int(ans); cout << ans + t << ' '; } cout << '\n';}
int main() { #ifndef ONLINE_JUDGE freopen("Error.txt", "w", stderr); #endif send help; int _t;cin>>_t;while(_t--) solve(); return 0; }
ABOVE SOL got accepted
include<bits/stdc++.h>
using namespace std; // #include <ext/pb_ds/assoc_container.hpp> // #include <ext/pb_ds/tree_policy.hpp> // using namespace __gnu_pbds;
// #define ordered_set tree<int, null_type,less, rb_tree_tag,tree_order_statistics_node_update>
define f0r(a, b) for (long long a = 0; a < (b); ++a)
define f1r(a, b, c) for (long long a = (b); a < (c); ++a)
define f0rd(a, b) for (long long a = (b); a >= 0; --a)
define f1rd(a, b, c) for (long long a = (b); a >= (c); --a)
define ms(arr, v) memset(arr, v, sizeof(arr))
define send {ios_base::sync_with_stdio(false);}
define help {cin.tie(NULL); cout.tie(NULL);}
define fix(prec) {cout << setprecision(prec) << fixed;}
define mp make_pair
define all(v) v.begin(), v.end()
define getunique(v) {sort(all(v)); v.erase(unique(all(v)), v.end());}
define readgraph(list, edges) for (int i = 0; i < edges; i++) {int a, b; cin >> a >> b; a--; b--; list[a].pb(b); list[b].pb(a);}
define vsz(x) ((long long) x.size())
typedef long long ll; typedef long double lld; typedef unsigned long long ull; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef vector vi; typedef vector vl; typedef vector vpi; typedef vector vpl; typedef long long ll; typedef unsigned long long ull; typedef long double lld;
void __print(int x) {cerr << x;} void __print(long x) {cerr << x;} void __print(long long x) {cerr << x;} void __print(unsigned x) {cerr << x;} void __print(unsigned long x) {cerr << x;} void __print(unsigned long long x) {cerr << x;} void __print(float x) {cerr << x;} void __print(double x) {cerr << x;} void __print(long double x) {cerr << x;} void __print(char x) {cerr << ''' << x << ''';} void __print(const char *x) {cerr << '"' << x << '"';} void __print(const string &x) {cerr << '"' << x << '"';} void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V> void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';} template void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << '}';} void _print() {cerr << "]\n";} template <typename T, typename... V> void print(T t, V... v) {_print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
ifndef ONLINE_JUDGE
define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
else
define debug(x...)
endif
void solve() { // TODO: Add your code here int n,k,q; cin>>n>>k>>q; vector a(k+1); vector b(k+1); vector dp(k+1, 0); a[0]= 0; b[0]= 0; for(int i=1;i<=k;i++){ cin >> a[i]; }
for(int i=1;i<=k;i++) { cin>>b[i]; } while(q--){ double d; cin>>d; int it = lower_bound(a.begin(), a.end(), d) - a.begin(); double t2,x2; if(d==a[it]){ cout << b[it] << " "; continue; } else it--; ll ans = b[it]; ll t = ((d-a[it])*(b[it+1]-b[it])) / (a[it+1]-a[it]); // if(ans - int(ans) >= 0.5) ans = int(ans + 1); // else ans = int(ans); cout << ans + t << " "; } cout << '\n';}
int main() { #ifndef ONLINE_JUDGE freopen("Error.txt", "w", stderr); #endif send help; int _t;cin>>_t;while(_t--) solve(); return 0; }
ABOVE got rejected
only difference is the double d and int d in the while loop query can anyone tell me whats wrong
I got hacked my initial cases were passed correctly
Pro tip :always share your code as a submission link or no one is reading it.
like this 260430183
What is the best way to deal with precision issues when using floating points like in problem E? MY SOLUTION
Thanks in advance
My very intuitive solution of problem c : 260585157
Can anyone help me indentifying why my code is giving runtime error for this following code in Problem G: "https://codeforces.com/contest/1971/submission/260600724"
I modified your submission in 2 places. Now it can AC. See submission
Hey thanks, it got an AC. I removed the equality sign from comparator and got correct answer, but still can't figure out why removing equality sign from comparator got an AC.
(Disclaimer: my claim has a little bit of rule of thumbs and informal terminologies here, since I was half unsure, half wanting to simplify the topic)
Normally for programming languages, defined comparator that takes two entities $$$a$$$ and $$$b$$$ should return
trueif and only if $$$a$$$ should be placed before $$$b$$$, and it will look upon those comparisons returningtrueto dictate the final order of the list/array.Thus if you consider $$$a \le b$$$ as
true, and in fact $$$a = b$$$, comparator will take bothcompare(a,b)andcompare(b,a)astrue, thus sending those two entities into an endless loop of placing one before another. This part is an undefined behavior, that different compiler implementations will have different responses to this (and in this exact case, most implementations will break your code, either through TLE or RTE).Removing equal sign will break that loop, ensuring that the array/list can be sorted in a deterministic way in the comparator's viewpoint.
Yes, your point makes lot of sense but if both the "compare" functions should return true, then the compiler should ideally make both the elements stay at their own place. Why it should end in an infinite loop ?
Like it can also occur in cases when sorting an array with all elements equal
This part, I apologize, I can't explain further, coz' I also don't know in-depth. It's not really supposed to be compiler's doing at this point, but how the sorting algorithm in the library works internally, and it is over my head (usually in a programming language, sorting algorithm is a hybrid of things, so without digging in I can't tell for sure).
A comparator (in c++ atleast) is supposed to return true IFF a should be placed before b. This is the standard of the language.
By returning true when a = b, you are basically telling the compiler to put a before b, and also to put b before a.
https://codeforces.com/problemset/submission/1971/260624825
can anyone hep me how to deal with this error (wrong output format Expected integer, but "1e+009" found)
thank you
cout << int(b[k]) << " ";
also change
(b[i + 1] - b[i]) / (a[i + 1] - a[i]) * (c - a[i]);to(c - a[i])*(b[i + 1] - b[i]) / (a[i + 1] - a[i]);why are we using -a[i] in editorial soln of problem G ` ~~~~~ mp[a[i]>>2].push(-a[i]); ~~~~~
`
So that priority queue gives min value instead of max value.
~~~~~
include <bits/stdc++.h>
using namespace std; const int M = 1000000007;
void hello() { int n; cin >> n; vector a(n); map<int, deque> f; for(int i = 0; i < n; i++) { cin >> a[i]; f[a[i] >> 2].push_back(a[i]); } for(int i = 0; i < n; i++) { if(f[a[i] >> 2].size() > 1) { sort(f[a[i] >> 2].begin(), f[a[i] >> 2].end()); } } for(int i = 0; i < n; i++) { cout << f[a[i] >> 2].front() << " "; f[a[i] >> 2].pop_front(); } cout << "\n"; }
int32_t main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int t = 1; cin >> t; while(t--) { hello(); } return 0; }~~~~~
can anyone plz help me with this, getting tle on test 13
can anybody tell me what is wrong with my code, its giving wa on test case 3
for clarification i got my mistake for every clause A v B
not A -> B not B -> A
both these edges should be added but i just added not A -> B
How to correct this precision error? 260750048 ```
...
void solve() {
ll n,k,q; cin>>n>>k>>q; vectora(k),b(k);
for(int i=0;i<k;i++){ cin>>a[i]; } for(int i=0;i<k;i++){ cin>>b[i]; } a.insert(a.begin(),0); b.insert(b.begin(),0); vector<float>speed; k++; for(int i=1;i<k;i++){ ll dis=a[i]-a[i-1]; ll time = b[i]-b[i-1]; float spd = dis/(float)time; speed.push_back(spd); } speed.insert(speed.begin(),0); while(q--){ ll num; cin>>num; ll idx = lower_bound(a.begin(),a.end(),num)-a.begin(); ll total= a[idx]-a[idx-1]; if(num==a[idx]){ cout<<b[idx]<<" "; continue; } else{ ll rem = total - (a[idx]-num); double tm = static_cast<double>(rem) / speed[idx]; cout << fixed << setprecision(10); cout << static_cast<ll>(b[idx-1] + tm) << " "; } } cout<<nl;}
first of all never use float always use double or long double .But in this problem that is not enought.
just write the whole formula
tm = rem * (b[idx+1]-b[idx])/(a[idx+1]-a[idx])Thanks, it all worked!
Hey, a question regarding problem G (in C++). My code gives TLE when using unordered_map, but is accepted when using map. Why is it so? Most of the time I am using unordered_map and in this case, although the solution is correct, it is a bit of a bummer that it gives TLE (because how would I know that the problem is the use of unordered_map).
I'd assume that there are some tests (either from main or hacking phase) that exploits the hashing algorithm of
unordered_map, causing enough collisions to make its operations $$$\mathcal{O}(n)$$$ instead of the average $$$\mathcal{O}(1)$$$.mapis $$$\mathcal{O}(\log n)$$$ constantly so such tests are harmless to it.For C can anyone find error in my code https://codeforces.com/contest/1971/submission/260905825
Mostly precision errors to me.
A few rules of thumbs:
float, usedoubleandlong doublewhen needed.float's fp16 nature has very low precision.int/long long, just make sure you do the correct calculating order (there is only one dividing operation per query, so do it last).1971H - ±1
Why in the four testcase of problem H, the case 6 1 3 -6 2 5 2 1 3 -2 -3 -6 -5 -2 -1 -3 2 3 1
Say no?, the array [1, 1, 1, 1, 1, -1] doesn't works?
col3 is -6 -2 -3, with array [1, 1, 1, 1, 1, -1], it would be 1 -1 -1. Sort them, we find the middle row is -1.
'each column', yeah i dont realized that until now, thanks a lot for answering and your help :D
6 1 3 -6 2 5 2 1 3 -2 -3 -6 -5 -2 -1 -3 2 3 1 explain this test help me
I think it cout YES with [1,1,1,1,1,-1]
The Sort Is in each columna, no the all of the matriz, so su need each columna only have a -1, ( the case 4, in third column dont work with [1, 1,1,1,1,-1], because you have un that column 1, -1, -1, AND sorted, you have -1, -1, 1, where you have a -1 in the middle and can't win)
Amazing tutorials and solutions. Thanks.
For G, why I use map is correct, but when I use unordered_map is timeout?
261159622
Could please someone explain me what is wrong with my 2-SAT implementation for the problem H? My attempt fails on the test 3 with wrong answer: https://codeforces.com/contest/1971/submission/261300539. I guess there should be a problem somewhere in a Tarjan's algorithm implementation, because earlier I used less optimized way to find SCC, and it worked up to test 7, where it timed out.
Okay, I just tried to rewrite the code, and now my solution passes. If someone is interested in another correct code for problem H: https://codeforces.com/contest/1971/submission/261434608
My second solution to clock & string
Here's a neat little solution for F, I observed that all points on the circle perimeter are touching one another, so we can just "trace" the circle and bfs every point within the correct distance. Very easy solution when you do it this way.
Hi All,
Can anyone help why this solution is incorrect for problem F
this was nice contest.
Solution for G is damn Good!
https://codeforces.com/contest/1971/submission/262898473
Here is my solution for problem C. I think it is much shorter and easier to understand!
is the clock string soln correct bro?
Why can't D be solved by counting number of "01"s?
for example, grader says 101 must need 3 pieces, but you can cut like this 1|01 then arrange into 01|1, which shows you only need 2 pieces