Идея: BledDest
Разбор
Tutorial is loading...
Решение (awoo)
for _ in range(int(input())):
n, k = map(int, input().split())
for x in range(2):
if n - x * k >= 0 and (n - x * k) % 2 == 0:
print("YES")
break
else:
print("NO")
Идея: BledDest
Разбор
Tutorial is loading...
Решение (awoo)
for _ in range(int(input())):
a, b = map(int, input().split())
ans = a + b
for m in range(1, 100000):
ans = min(ans, (a + m - 1) // m + (b + m - 1) // m + (m - 1))
print(ans)
Идея: BledDest
Разбор
Tutorial is loading...
Решение (BledDest)
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
scanf("%d", &t);
for(int i = 0; i < t; i++)
{
int n;
scanf("%d", &n);
vector<int> s(2);
for(int j = 0; j < 2; j++)
scanf("%d", &s[j]);
vector<pair<int, int>> a(n);
for(int j = 0; j < n; j++)
{
scanf("%d", &a[j].first);
a[j].second = j + 1;
}
sort(a.begin(), a.end());
reverse(a.begin(), a.end());
vector<vector<int>> lists(2);
for(int j = 0; j < n; j++)
{
int cost1 = s[0] * (lists[0].size() + 1);
int cost2 = s[1] * (lists[1].size() + 1);
if(cost1 < cost2)
lists[0].push_back(a[j].second);
else
lists[1].push_back(a[j].second);
}
for(int j = 0; j < 2; j++)
{
cout << lists[j].size();
for(auto x : lists[j]) cout << " " << x;
cout << endl;
}
}
}
Идея: adedalic
Разбор
Tutorial is loading...
Решение (adedalic)
fun main() {
repeat(readln().toInt()) {
val (n, k) = readln().split(' ').map { it.toInt() }
val f = readln().split(' ').map { it.toLong() }
val d = readln().split(' ').map { it.toLong() }
fun checkAround(pos : Long) : Int {
val qs = Array(2 * k + 1) { MutableList(0) { 0 } }
fun inside(x : Long, pos: Long) = (pos - k <= x) && (x <= pos + k)
for (i in f.indices) {
var newD = maxOf(1L, pos / f[i])
if (newD != d[i] && newD + 1 != d[i] && inside(d[i] * f[i], pos)) {
val id = (i + 1)
qs[(d[i] * f[i] - pos + k).toInt()].add(id)
}
repeat(2) {
if (inside(newD * f[i], pos)) {
val id = if (newD == d[i]) (i + 1) else -(i + 1)
qs[(newD * f[i] - pos + k).toInt()].add(id)
}
newD++
}
}
val cntPerId = IntArray(n) { 0 }
var cntDistinct = 0
var cntGood = 0
fun addToSeg(cid : Int) {
val id = -1 + if (cid > 0) cid else -cid
val c = if (cid > 0) 1 else 0
if (cntPerId[id] == 0)
cntDistinct++
cntPerId[id]++
cntGood += c
}
fun eraseFromSeg(cid : Int) {
val id = -1 + if (cid > 0) cid else -cid
val c = if (cid > 0) 1 else 0
cntPerId[id]--
if (cntPerId[id] == 0)
cntDistinct--
cntGood -= c
}
var ans = 0
for (p in 0 until k)
qs[p].forEach { addToSeg(it) }
for (p in 0..k) {
qs[p + k].forEach { addToSeg(it) }
if (cntDistinct == n)
ans = maxOf(ans, cntGood)
qs[p].forEach { eraseFromSeg(it) }
}
return n - ans
}
var ans = n
for (i in f.indices) {
ans = minOf(ans, checkAround(d[i] * f[i]))
}
println(ans)
}
}
Идея: BledDest
Разбор
Tutorial is loading...
Решение (BledDest)
#include<bits/stdc++.h>
using namespace std;
typedef long long li;
const li INF = (li)(1e18);
const int N = 200043;
struct data
{
array<array<long long, 2>, 2> d;
data() {
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++)
d[i][j] = INF;
};
data(li x)
{
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++)
d[i][j] = INF;
d[0][0] = 0;
d[1][1] = x;
};
data(data l, data r)
{
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++)
d[i][j] = INF;
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++)
for(int k = 0; k < 2; k++)
for(int x = 0; x < 2; x++)
{
if(j == 0 && k == 0) continue;
d[i][x] = min(d[i][x], l.d[i][j] + r.d[k][x]);
}
};
};
data T[4 * N];
li a[N];
void recalc(int v)
{
T[v] = data(T[v * 2 + 1], T[v * 2 + 2]);
}
void build(int v, int l, int r)
{
if(l == r - 1) T[v] = data(a[l]);
else
{
int m = (l + r) / 2;
build(v * 2 + 1, l, m);
build(v * 2 + 2, m, r);
recalc(v);
}
}
void upd(int v, int l, int r, int pos, li val)
{
if(l == r - 1)
{
a[l] = val;
T[v] = data(a[l]);
}
else
{
int m = (l + r) / 2;
if(pos < m)
upd(v * 2 + 1, l, m, pos, val);
else
upd(v * 2 + 2, m, r, pos, val);
recalc(v);
}
}
int main()
{
int n;
scanf("%d", &n);
for(int i = 0; i < n - 1; i++)
{
scanf("%lld", &a[i]);
}
build(0, 0, n - 1);
int q;
scanf("%d", &q);
for(int i = 0; i < q; i++)
{
int x;
li k;
scanf("%d %lld", &x, &k);
upd(0, 0, n - 1, x - 1, k);
printf("%lld\n", T[0].d[1][1] * 2);
}
}
Идея: Neon
Разбор
Tutorial is loading...
Решение (Neon)
#include <bits/stdc++.h>
using namespace std;
#define sz(a) int((a).size())
const int N = 200002;
const int V = 50 * N;
using pt = pair<int, int>;
int n, m;
pt a[N];
vector<pt> t[4 * N];
int p[N], rk[N], vs[N];
int cntV;
pt g[V];
bool ans[V];
void upd(int v, int l, int r, int L, int R, pt e) {
if (L >= R) return;
if (l == L && r == R) {
t[v].push_back(e);
return;
}
int m = (l + r) / 2;
upd(v * 2 + 1, l, m, L, min(m, R), e);
upd(v * 2 + 2, m, r, max(m, L), R, e);
}
int k;
int* ptr[V];
int val[V];
void upd(int &a, int b) {
ptr[k] = &a;
val[k] = a;
k += 1;
a = b;
}
int getp(int v) {
return p[v] == v ? v : getp(p[v]);
}
void unite(int v, int u) {
v = getp(v), u = getp(u);
if (v == u) return;
if (rk[v] > rk[u]) swap(v, u);
upd(p[v], u);
g[cntV] = {vs[v], vs[u]};
upd(vs[u], cntV++);
if (rk[v] == rk[u])
upd(rk[u], rk[u] + 1);
}
void solve(int v, int l, int r) {
int cur = k;
for (auto& [v, u] : t[v])
unite(v, u);
if (l + 1 == r) {
ans[vs[getp(0)]] = 1;
} else {
int m = (l + r) / 2;
solve(v * 2 + 1, l, m);
solve(v * 2 + 2, m, r);
}
while (k > cur) {
k -= 1;
(*ptr[k]) = val[k];
}
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> n >> m;
for (int i = 0; i < n; ++i)
cin >> a[i].first >> a[i].second;
for (int i = 0; i < m; ++i) {
int x, y;
cin >> x >> y;
--x; --y;
int L = max(a[x].first, a[y].first);
int R = min(a[x].second, a[y].second);
upd(0, 0, N, L, R + 1, {x, y});
}
for (int i = 0; i < n; ++i) {
p[i] = i;
rk[i] = 1;
vs[i] = i;
g[i] = {-1, -1};
}
cntV = n;
solve(0, 0, N);
queue<int> q;
for (int i = 0; i < cntV; ++i) if (ans[i])
q.push(i);
while (!q.empty()) {
int v = q.front(); q.pop();
for (int u : {g[v].first, g[v].second}) {
if (u != -1 && !ans[u]) {
ans[u] = true;
q.push(u);
}
}
}
for (int i = 0; i < n; ++i) if (ans[i])
cout << i + 1 << ' ';
}
I know it isn't an "issue", but in problem B the variable names are a,b while here they are n,m
In problem B, My thought was ternary search since the values tend to decrease at first and then increase but the problem I faced is that around the middle of the curve, the values don't necessarily keep changing in the same way (X-axis for max K length and Y-axis for cost)
As you notice some irregularities, can this be handled using only ternary search or by iterating over a much smaller range of K's?
I tried a weaker 1-D case (go from 0 to n)
there $$$\sqrt{n}$$$ seems to be the best leg size. (floor or ceil I don't remember)
(somewhat related to the fact that $$$x + a/x$$$ minimizes at $$$x = \sqrt{a}$$$ and
similar is the case with $$$x-1 + \lceil n/x \rceil$$$)
But ofc the problem is going up a notch to the 2-D case.
(new to code-forces and I don't know how to write in laTex etc, pardon me for that).
problem EF video solution for Chinese：
BiliBili
why the rating result are not visible now?
because it's an unrated round
Can anyone please elaborate on the solution of Problem D, please?
A can be solved by just checking parity of n and k
Can you please give an explanation of this? Also, what is meant by parity check?
parity means it's odd or even.
if n is even, it can be made with 2 coins.
if n is odd then k must be odd as well otherwise it is not possible.
Let's check for specific N and all values of k. let N=7
k=2 not possible
k=3
7= 3*1+2*2
k=4 not possible
k=5
7=5*1+2*1
k=6 not possible
k=7
7=7*1
For a more formal proof
let n be written as coins of 2 denominations and k denominations as - n=2*a +k*b. Let k be even then n can be written as n=2*a+2*b*c n=2(a+b*c) Hence, if k is even then n must be even as well otherwise it's not possible.
if k is odd then you can get all numbers from k to n by using one k coin and the other 2 denomination coins. k,k+2, k+4,k+6,k+8....n
since it's given k<=n we don't need to check cases where k>n
can anyone please explain me why this code will not work for problem B?
The cost is different depending on where you put it. Say that the current number of elemnts in list_a is k1, and in list_b is k2, then the cost of adding an element S is S * (k1 + 1) * S1 in the first case, and S * (k2 + 1) * S2 in the second case.
Can anyone tell me why we need to fix the value of K in B.
Let's say we are calculating answer for $$$b$$$ only (you can extend this to $$$a$$$ also).
Now, let's say you are using 2 values of $$$k$$$ i.e. $$$k_1 < k_2$$$. So your answer would be kind of $$$b/k_1 + b/k_2$$$, here if you just replace $$$b/k_1$$$ with $$$1$$$ you would get better answer and we can always do this since we will always increase value of $$$k$$$ from 1 to $$$k_2$$$ and $$$k_1$$$ is working on just the remainder of $$$b/k_2$$$.
E Has a nice observation, we would never take 3 consecutive edges. Any elegant solution using this observation.
(Take , NotTake, Take) is better than (Take, Take, Take) and both achieve our goal. f(i) = min(2*e(i,i+1) + f(i+2) , 2*[e(i,i+1)+e(i+1,i+2)] + f(i+3))
Sorry for necroposting, but could you expand on how we can utilize this to quickly answer update queries? Does it still require the "classical" segment tree sol mentioned in edi?
Solution to problem B seems pretty complicated. Can someone please explain a simpler approach?
It's not difficult. You should just know the answer is $$$\lceil\frac{a}{k}\rceil + \lceil\frac{b}{k}\rceil + (k - 1)$$$ when you add your feet to k long. Then if you remove the ceil, you can get $$$\frac{a + b}{k} + k - 1$$$, it's just an inequality $$$a + b \ge 2 \times \sqrt{a \times b}$$$ when $$$a = b$$$ the equality is achieved. Therefore, for this function, it's $$$\frac{a + b}{k} = k$$$, .i.e $$$k^2 = a + b$$$. So $$$k = \sqrt{a + b}$$$, you can get the smallest value. However, you remove the ceil, so there maybe 2 off the correct answer. You should check around. Anyway, 1e5 is enough.
How did you come to the conclusion that
(A+b)/k)=k ?
Because the inequality I wrote abode. $$$\frac{a + b}{k} + k \ge 2 \times \sqrt{a + b}$$$, and when $$$\frac{a + b}{k} = k$$$ the equality takes.
can someone recommend more problems similar to E
If the Editorial has atleast transistion states you could even try to reduce to matrix form The editorial is as good as problem statement. It's one of the kind we call it Skind S means not "shit".It means Super
Sorry for necroposting, but I want to share my solution of F without using KRT.
Consider doing lazy tag on rollback DSU, when we visit to a certain time, add a tag with time stamp on the representative of vertex $$$1$$$, and when we rollback(assume this edge point to $$$boss$$$), if this edge is added before the tag on $$$boss$$$ is added, then push down the tag. At the end just check whether each vertex have tag on it.
I found it is quite amazing that we can do lazy tag on DSU, maybe there also have some potential application of it?
Can anyone please tell me why my submission TLEs on test 5 for Problem D? It uses the same solution as editorial. Submission: 239734235
What is the recurrence relation used in Problem E ? what Dp transistions should one speed up ? It would be better to add equations and how they were transformed into matrix form. IMO this is the only step which is key in the matrix expo problems and Editorial is missing them
Sorry for necroposting, but here are some equations I found. Notice that, of the edges you choose, you never choose 3 in a row. Therefore, there will never be more than 2 chosen edges between two unchosen edges. Also notice that you can never not choose two adjacent edges, because that will leave 1 chip in its original place. You can see that this means we must distribute unchosen edges so that they are either 1 edge or 2 edges apart from each other.
Our answer can be expressed as two times the sum of the edges minus two times the maximum sum of edges we don't choose.
Let's say x is one of the edges you do not choose, and _ is one of the edges you do choose. In each node of the tree, store 4 values for the segment $$$[l, r]$$$
a. the maximum sum of x's where one x is on $$$l + 1$$$ and one x is on $$$r - 1$$$: _ x ? ? ? ? ? x _
b. the maximum sum of x's where one x is on $$$l$$$ and one x is on $$$r - 1$$$.: x ? ? ? ? ? ? x _
c. the maximum sum of x's where one x is on $$$l + 1$$$ and one x is on $$$r$$$: _ x ? ? ? ? ? ? x
d. the maximum sum of x's where one x is on $$$l$$$ and one x is on $$$r$$$: x ? ? ? ? ? ? ? x
To combine two children nodes $$$left$$$ and $$$right$$$ into a parent node, we can do this (you can draw it out to see for yourself if you need to):
$$$parent_a = \max(left_a + right_a, left_a + right_b, left_c + right_a)$$$
$$$parent_b = \max(left_b + right_a, left_b + right_b, left_d + right_a)$$$
$$$parent_c = \max(left_c + right_c, left_a + right_c, left_a + right_d)$$$
$$$parent_d = \max(left_d + right_c, left_b + right_d, left_b + right_c)$$$
The new maximum will be the $$$\max(root_a, root_b, root_c, root_d)$$$.
Can anyone explain for me problem C because its kinda vague, I dont really understand the given testcases and the meaning of array R.
In problem B, can the robot take smaller steps than it's current leg length of m?