1569A - Сбалансированная подстрока
Идея: BledDest
Разбор
Tutorial is loading...
Решение (awoo)
for _ in range(int(input())):
n = int(input())
s = input()
for i in range(n - 1):
if s[i] != s[i + 1]:
print(i + 1, i + 2)
break
else:
print(-1, -1)
Идея: BledDest
Разбор
Tutorial is loading...
Решение (Neon)
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n;
string s;
cin >> n >> s;
vector<int> id;
for (int i = 0; i < n; ++i) if (s[i] == '2')
id.push_back(i);
int k = id.size();
if (k == 1 || k == 2) {
cout << "NO\n";
continue;
}
vector<string> t(n, string(n, '='));
for (int i = 0; i < n; ++i) t[i][i] = 'X';
for (int i = 0; i < k; ++i) {
int x = id[i], y = id[(i + 1) % k];
t[x][y] = '+';
t[y][x] = '-';
}
cout << "YES\n";
for (int i = 0; i < n; ++i) cout << t[i] << '\n';
}
}
Идея: BledDest
Разбор
Tutorial is loading...
Решение (Neon)
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n);
for (int &x : a) cin >> x;
int mx = *max_element(a.begin(), a.end());
int cmx = count(a.begin(), a.end(), mx);
int k = count(a.begin(), a.end(), mx - 1);
int ans = 1, sub = 1;
for (long long i = 1; i <= n; ++i) {
ans = ans * i % MOD;
if (i != k + 1) sub = sub * i % MOD;
}
if (cmx == 1) ans = (ans - sub + MOD) % MOD;
cout << ans << '\n';
}
}
Идея: BledDest
Разбор
Tutorial is loading...
Решение (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 all(a) (a).begin(), (a).end()
#define x first
#define y second
typedef long long li;
typedef pair<int, int> pt;
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << ", " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
fore(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
const int INF = int(1e9);
const li INF64 = li(1e18);
int n, m, k;
vector<int> x, y;
vector<pt> ps;
inline bool read() {
if(!(cin >> n >> m >> k))
return false;
x.resize(n);
fore (i, 0, n)
cin >> x[i];
y.resize(m);
fore (i, 0, m)
cin >> y[i];
ps.resize(k);
fore (i, 0, k)
cin >> ps[i].x >> ps[i].y;
return true;
}
inline void solve() {
li ans = 0;
fore (_i, 0, 2) {
vector<int> cntY(m, 0);
sort(all(ps));
vector<pt> bord(k);
int u = 0;
while (u < k) {
int v = u;
while (v < k && ps[v].x == ps[u].x)
v++;
fore (i, u, v) {
int r = int(lower_bound(all(y), ps[i].y) - y.begin());
int l = r;
if (y[l] > ps[i].y)
l--;
assert(y[l] <= ps[i].y && ps[i].y <= y[r]);
bord[i] = {l, r};
}
fore (i, u, v) if (bord[i].x < bord[i].y)
ans += cntY[bord[i].x];
fore (i, u, v) if (bord[i].x < bord[i].y)
cntY[bord[i].x]++;
u = v;
}
fore (i, 0, k)
swap(ps[i].x, ps[i].y);
swap(x, y);
swap(n, m);
}
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);
int t; cin >> t;
while (t--) {
read();
solve();
#ifdef _DEBUG
cerr << "TIME = " << clock() - tt << endl;
tt = clock();
#endif
}
return 0;
}
1569E - Восстановление турнирной таблицы
Идея: BledDest
Разбор
Tutorial is loading...
Решение (BledDest)
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
int add(int x, int y)
{
x += y;
while(x >= MOD) x -= MOD;
while(x < 0) x += MOD;
return x;
}
int mul(int x, int y)
{
return (x * 1ll * y) % MOD;
}
int binpow(int x, int y)
{
int z = 1;
while(y)
{
if(y & 1) z = mul(z, x);
x = mul(x, x);
y >>= 1;
}
return z;
}
vector<int> evaluate(int n, int choice_mask)
{
int cur_place = n / 2 + 1;
int cur_bit = n - 2;
vector<int> p(n);
vector<int> c(n);
for(int i = 0; i < n; i++)
c[i] = i;
while(c.size() != 1)
{
vector<int> nc;
for(int i = 0; i < c.size(); i += 2)
{
if(choice_mask & (1 << cur_bit))
{
p[c[i]] = cur_place;
nc.push_back(c[i + 1]);
}
else
{
p[c[i + 1]] = cur_place;
nc.push_back(c[i]);
}
cur_bit--;
}
c = nc;
cur_place /= 2;
cur_place++;
}
p[c[0]] = 1;
return p;
}
vector<int> adjust(int n, vector<int> p, bool winning)
{
for(int i = 0; i < n; i++)
{
if(p[i] == 1)
{
if(!winning) p[i]++;
}
else
p[i] = p[i] * 2 - 1;
}
return p;
}
int get_hash(int n, vector<int> p, int A, bool partial = false, bool winning = false, int shift = 0)
{
if(partial)
p = adjust(n, p, winning);
int res = 0;
for(int i = 0; i < n; i++)
res = add(res, mul(add(i + 1, shift), binpow(A, p[i])));
return res;
}
int main()
{
int k, A, h;
cin >> k >> A >> h;
if(k <= 4)
{
for(int i = 0; i < (1 << ((1 << k) - 1)); i++)
{
vector<int> p = evaluate(1 << k, i);
if(get_hash(1 << k, p, A) == h)
{
for(auto x : p) cout << x << " ";
cout << endl;
return 0;
}
}
}
else
{
int mask_left = -1;
int mask_right = -1;
bool left_win = false;
for(int c = 0; c < 2; c++)
{
map<int, int> left_map;
for(int i = 0; i < (1 << 16); i++)
{
vector<int> p = evaluate(16, i);
int left_hash = get_hash(16, p, A, true, c == 0, 0);
left_map[left_hash] = i;
}
for(int i = 0; i < (1 << 16); i++)
{
vector<int> p = evaluate(16, i);
int right_hash = get_hash(16, p, A, true, c == 1, 16);
int left_hash = add(h, MOD - right_hash);
if(left_map.count(left_hash))
{
mask_left = left_map[left_hash];
mask_right = i;
left_win = (c == 0);
}
}
}
if(mask_left != -1)
{
vector<int> ans_left = evaluate(16, mask_left);
vector<int> ans_right = evaluate(16, mask_right);
ans_left = adjust(16, ans_left, left_win);
ans_right = adjust(16, ans_right, !left_win);
for(auto x : ans_left)
cout << x << " ";
for(auto x : ans_right)
cout << x << " ";
return 0;
}
}
cout << -1 << endl;
return 0;
}
1569F - Палиндромный гамильтонов путь
Идея: BledDest
Разбор
Tutorial is loading...
Решение (awoo)
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
vector<vector<char>> g;
map<long long, bool> dp;
void brute(int n, vector<int> &p){
int x = find(p.begin(), p.end(), -1) - p.begin();
if (x == int(p.size())){
vector<vector<char>> dp2(1 << n, vector<char>(n));
vector<int> pos1(n), pos2(n);
forn(i, p.size()){
pos1[p[i]] = pos2[p[i]];
pos2[p[i]] = i;
}
forn(i, n) if (g[pos1[i]][pos2[i]]) dp2[1 << i][i] = true;
forn(mask, 1 << n) forn(i, n) if (dp2[mask][i]){
forn(j, n) if (!((mask >> j) & 1)){
dp2[mask | (1 << j)][j] |= (g[pos1[i]][pos1[j]] && g[pos2[i]][pos2[j]]);
dp2[mask | (1 << j)][j] |= (g[pos1[i]][pos2[j]] && g[pos2[i]][pos1[j]]);
}
}
forn(i, n) if (dp2[(1 << n) - 1][i]){
long long num = 0;
for (int x : p) num = num * 6 + x;
dp[num] = true;
break;
}
return;
}
for (int y = x + 1; y < int(p.size()); ++y) if (p[y] == -1){
p[x] = p[y] = n;
brute(n + 1, p);
p[x] = p[y] = -1;
}
}
bool dfs(vector<int> p){
vector<int> used(int(p.size()), -1);
int cnt = 0;
forn(i, p.size()) if (used[p[i]] == -1)
used[p[i]] = cnt++;
long long num = 0;
for (int& x : p){
x = used[x];
num = num * 6 + x;
}
if (dp.count(num)) return dp[num];
bool res = false;
vector<int> cur(cnt);
forn(i, p.size()) ++cur[p[i]];
forn(i, p.size()) if (cur[p[i]] > 2){
int x = p[i];
for (int j = i + 1; j < int(p.size()); ++j) if (p[j] == p[i]){
p[i] = p[j] = cnt;
if (dfs(p)){
res = true;
break;
}
p[i] = p[j] = x;
}
break;
}
return dp[num] = res;
}
void brute2(int n, vector<int> &p){
int x = find(p.begin(), p.end(), -1) - p.begin();
if (x == int(p.size())){
dfs(p);
return;
}
forn(i, n + 1){
for (int y = x + 1; y < int(p.size()); ++y) if (p[y] == -1){
p[x] = p[y] = i;
brute2(max(n, i + 1), p);
p[x] = p[y] = -1;
}
}
}
int main() {
int n, m, k;
scanf("%d%d%d", &n, &m, &k);
g.resize(n, vector<char>(n));
forn(_, m){
int v, u;
scanf("%d%d", &v, &u);
--v, --u;
g[v][u] = g[u][v] = 1;
}
vector<int> cur(n, -1);
brute(0, cur);
brute2(0, cur);
vector<long long> fact(k + 1);
fact[0] = 1;
for (int i = 1; i <= k; ++i) fact[i] = fact[i - 1] * i;
long long ans = 0;
for (auto it : dp) if (it.second){
long long num = it.first;
long long mx = 1;
while (num){
mx = max(mx, num % 6 + 1);
num /= 6;
}
if (mx <= k){
ans += fact[k] / fact[k - mx];
}
}
printf("%lld\n", ans);
return 0;
}
Fastest editorial I've seen
Stop disliking him.
Lesson learned: Don't use segment trees unless required.
(My segment tree implementation for D gave TLE on system testing.)
Interesting. I used segment trees and maps as well, and the highest runtime was ~1 second.
My submission 128269298
code and explanation for problem D
Just separate the points into two parts one where the x of the point matches with any of the x of vertical roads and other part where the y of the point matches with the y of any horizontal road. these two parts may or may not be disjoint as it is also possible that point's x as well as y matches with any vertical and horizontal road then this point will be the part of both parts.
Now solve the answers for both the parts separately ->
for 1st part Sort it using a comparater which sorts first according to y and then according to x. Now just iterate one by one on the points since the points are sorted using the above comparater so the points will be divided into groups which lie on the same horizontal line(same y) and this group is also sorted according to x also. Now we can see that the vertical roads divides the x axis into some segments. (x0,x1) (x1,x2) (x2,x3) ........
for calculating the answer we will maintain one array f of size (n-1) where ith element represents the number of points that lie in the ith segment and whose y is also less than the current y(on which we are currently at in the iteration).
f[i]=number of points which lie in the segment (x[i],x[i+1]) end points not included.
for finding the segment-number of a point (a,b) we can easily find it using binary search(using lower_bound in c++)
now let us assume we are at a point (a,b) so to find out the pairs that this point will make with other points that are already visited in the iteration we can make use of the array f.
first we find the segment to which the point (a,b) lies and then the bad pairs which it makes is equal to f[segment-number]. because f[segment-number] stores the number of points that lie in that segment in the previous horizontal line (y<b). This can be easily observed that bad pairs are the points which strictly lie inside same segment(end points not included) and which do not lie on the same line.
Now when we move to the next horizontal line (y>b) then we need to add the point (a,b) to its respective segment(increment value of f[segment-number]) so that it can contribute the points which lie on the horizontal line above it.
what we can do is we can maintain a cnt variable which stores the number of points which lie on the same horizontal line and in the same segment of x axis.When we move to next segment then we can just update the value of f[prev-segment-number]+=cnt.
for 2nd part-
Sort it using a comparater which sorts first according to x and then according to y. It's exactly same the difference is just the change of axis now we will iterate vertical line by line and in increasing order of x.Now the y axis is divided into segments.
Link to my code- https://mirror.codeforces.com/contest/1569/submission/128325617
I am facing some difficulty in understanding the equation in Problem C: $$$A_n^{n-k-1} = \frac{n!}{(k+1)!}$$$ What is $$$A$$$ here and how did we get $$$\frac{n!}{(k+1)!}$$$?
$$$A_x^y$$$ is the number of ordered ways to choose exactly $$$y$$$ different objects from $$$x$$$. So, it's like a binomial coefficient $$${x}\choose{y}$$$, but with also considering the order in which we choose the object. Hence the formula for $$$A_x^y$$$ is $$${{x}\choose{y}} \cdot y! = \frac{x!}{(x-y)!}$$$: there are $$${x}\choose{y}$$$ ways to choose exactly $$$y$$$ objects out of $$$x$$$, and $$$y!$$$ ways to order them.
why to select n-k-1 from n we know which elements are n-k-1( n-k-1 elements are those which are not equal to ax or ax-1,where ax is the maximum element);
It is like considering the positions; instead of the elements. There are in total n positions. We wish to pick n-k-1 positions among them for the n-k-1 elements. There will be k+1 gaps after that. The last gap is fixed for the max element. The rest of the k elements goes k! ways into the rest of k gaps. I was also confused at first. Hope this helps.
nPr formula is n!/(n-r)! where nPr represents permutating r numbers in n places. It is basically n p (n-k-1).
is it true? nPr= rAn ? bcoz n should be greater than r.
Problem C : if (cmx == 1) ans = (ans — sub + MOD) % MOD; I don't know why we need plus MOD. I thought we just need : ans = ( ans — sub ) % MOD Thanks for someone explain this (^^)
It's because ans and sub are both %MOD, and if (ans%MOD — sub%MOD) < 0, so ( ans — sub ) % MOD will be < 0 too, that's because the % of negative numbers is also negative.
for example, if MOD = 1e9+7, sub = 5 and ans = 1e9 + 8 (but in your code ans = 1, because you do ans%= MOD): ans = (1 — 5)%MOD = -4%MOD = -4
But if you add MOD this never you be negative, because sub < MOD.
I hope you understood and sorry for my bad english, I'm not fluent.
I get that you are doing this because ans-sub can be negative but how does
MOD+(ans-sub)give the right answer whenans-subis negative. Is this a property of MOD?The "taking value modulo MOD" operation basically tells you the distance between your number and the previous number that divides MOD. So in case of negative numbers (in range (-MOD; 0)) you need to find the distance between -MOD and your number. If you shift both values by adding MOD it becomes the distance between 0 and (your number + MOD) which is of course (your number + MOD).
By the way, answer can be simplified to n! * k / (k+1). That way you don't need to subtract modulo MOD :)
In problem C my approach giving the wrong answer on test 2 but I couldn't understand why it is giving this error.
If the maximum number in the array exists more than once we can simply print N!. Because they will go together to end of the discussion, However, if it exists only once there should be maximum-1 in front of the maximum. So we can find unwanted situations and we can subtract it from N!. We can replace other numbers except maximum and maximum-1. we can do this in:
Permutation(N!,number_of_maximum + number_of_maximum-1) different ways.
After we have replaced others, we can put the numbers with values maximums and maximum-1. If we don't have any maximum-1 in front of maximum it is an unwanted situation. Hence, we can replace them in:
(number_of_maximum-1)! different ways.
Because we can put the maximum in the end and we can permutate others.
I tried to explain it well. I hope you got it and you can help me.
UPD: I added accepted solution so you can check my solution to understand the problem.
It's because of this
int x = (fact(n) / fact(mp[maxi] + mp[maxi-1])) % mod;fact(n) is divisible by fact(mp[maxi] + mp[maxi-1] if you don't use mod, but with mod it doesn't, for example:
16 is divisible by 8 but if i use mod 3 (16%3 = 1 and 8%3 = 2): 1 isn't divisible by 2
To fix this we can use module inverse, you can see it here or here
I hope you understood and sorry for my bad english.
upt: I don't know if the rest of your solution is correct
Thanks for your answer I got it. I didn't consider errors about mods.
For problem A you can use two pointers.
Is it worth ?
There are 4 methods for this question. 1. O(n) — simple traversal as described in editorial. 2. O(n^2) — brute force — generating all subarrays 3. O(n) — using two pointers. 4. O(n) — using traversal + map used to store sum till particular index
What about finding count of 'a' and 'b' for each range using segment/fenwick trees ? Isn't that useful ?
I have a different implementation for E , I am traversing on 1 match per function call by finding the people who haven't lost yet . It was giving TLE until I figured out ( yes figured out , because I didn't Google it ) how to traverse on Only set bits of a number in a given range ( range of bits ) . My submission
I think there exists a method for question D that is a little faster than the tutorial. The method used in the tutorial is to enumerate points, but I think it would be faster to enumerate edges. We divide the points into two groups, on x and on y, and discard the ones that satisfy both, since they cannot be composed. This way a pointer can be used to model the points that are between the two edges. Here the practice of my idea link
I think there exists a method for question D that is a little faster than the tutorial. The method used in the tutorial is to enumerate points, but I think it would be faster to enumerate edges. We divide the points into two groups, on x and on y, and discard the ones that satisfy both, since they cannot be composed. This way a pointer can be used to model the points that are between the two edges. Here the practice of my idea link
Done E with a really funny complexity -> submission
Worst case of $$$O(2^{23}*23)$$$
Your codes for D/E/F are soo looooong
Easy D approach -
Count no of points between each adjacent vertical lines. Let this count be n. Total no of pairs will be n*(n-1)/2. Subtract those pairs which lie on same horizontal line. This can be easily done using map.
Do the same for each adjacent horizontal lines
128428875
Hello, Can anybody please explain what does (1<<(1<<k))-1 represent in problem E??
It means $$$2^{2^k}-1$$$. The << (bitwise shift) operator shifts the bits in the left number by the right number of times.
[deleted]
Both $$$m$$$ and $$$n$$$ can be upto $$$2e5$$$. So yes, $$$O(mn)$$$ will exceed both memory and time limits.
Hi I was upsolving the C problem, can someone tell me why this wont pass the second test case? Thanks in advanced! my solution
If you want to modulo division, you should use the inverse.
128528815
problem C is interesting !
Can someone help me with my solution for problem C. So I came up with the same idea as the solution that a nice permutation must have at least 1 pair of i, j that i < j, a[i] is the largest number of the array and a[j] = a[i] — 1. I will call the largest number M. So using that, I will choose 2 position for M and M-1 , so there will be n*(n-1)/2 ways to place M and M-1. Suppose there are k number equal to M-1, there will be k*n*(n-1)/2 such ways. The remaining element can be placed anywhere so there will be (n-2)! ways. So in total there will be k*n*(n-1)/2*(n-2)! pair of nice permutation. But this formula gives the wrong answer for the 4th example test. I haven't known what is wrong with my solution yet so I hope u guys will help me with it. Thanks in advance!
I'm struggling with the same sitaution have you figured out what's wrong with the approach @Dyln
So when we use $$$C_n^2$$$ and multiply it with $$$(n-2)!$$$ it will count some cases multiple times because the positions of the second max elements are not fixed (ie. when the array is 1 1 2). Therefore, we have to fixed $$$k+1$$$ elements first and the max element will be placed anywhere but the last position, so our formula will be $$$A_n^{n-k-1}k!k$$$, which will simplify to $$$\frac{n!}{k+1}k$$$.
Hey its still doubtful as by doing nC2 we ensure that the second (atleast) one is behind the max one. So how can it count multiple cases
So let take a look at the array i mentioned: 1 1 2. First we take the first 1 (1st index) and the max value, then pick two random positions for them, let assume they are 1 and 2, so our permutation will be 3 1 2. But when we consider the second 1 (2nd index) and the max value, if we take position 1 and 3 then our permutation will also be 3 1 2. This permutation is counted twice because we didnt fix the position of the 1s. Therefore, the original formula is wrong although intuitively, it seems right XD.
There is a typo in the last line of paragraph 5, 1569D. strret->street :D @awoo
In C, this line of code
if (cmx == 1) ans = (ans - sub + MOD) % MOD;Why do we use mod like this: (.. + MOD % MOD)? We already calculated 'ans' and 'sub' under MOD, why do we use it again in the if statement?you need to +MOD because ans-sub may be a negative number , you need to %MOD because if ans-sub is non-negative , the former +MOD will give a value greater than ( or equal to ) MOD so you have to modulo again