Thanks for taking part in the round. I hope you enjoyed the round. It happens that the statements were really easy to understand (thanks to testers). We've got only 38 questions during a round!
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Code
int n, m;
cin >> n >> m;
int result = -1;
if (m % n == 0) {
result = 0;
int d = m / n;
while (d % 2 == 0)
d /= 2, result++;
while (d % 3 == 0)
d /= 3, result++;
if (d != 1)
result = -1;
}
cout << result << endl;
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Code
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++)
cin >> a[i];
int result = 0;
int len = 0;
for (int i = 0; i < 2 * n; i++) {
if (a[i % n] == 1) {
len++;
result = max(result, len);
} else {
len = 0;
}
}
cout << result << endl;
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Code
int n;
cin >> n;
vector<int> q(n - 1);
long long sum = 0;
long long min_val = 0;
for (int i = 0; i + 1 < n; i++) {
cin >> q[i];
sum += q[i];
if (sum < min_val)
min_val = sum;
}
vector<long long> p(n);
p[0] = 1 - min_val;
forn(i, n - 1)
p[i + 1] = p[i] + q[i];
bool ok = true;
for (int i = 0; i < n; i++)
if (p[i] < 1 || p[i] > n)
ok = false;
if (ok)
ok = set<long long>(p.begin(), p.end()).size() == n;
if (ok) {
forn(i, n)
cout << p[i] << " ";
} else
cout << -1 << endl;
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Code
#define forn(i, n) for (int i = 0; i < int(n); i++)
const int A = 26;
...
int n;
cin >> n;
string l;
cin >> l;
vector<vector<int>> left(A);
vector<int> wl;
forn(i, n)
if (l[i] != '?')
left[l[i] - 'a'].push_back(i);
else
wl.push_back(i);
string r;
cin >> r;
vector<vector<int>> right(A);
vector<int> wr;
forn(i, n)
if (r[i] != '?')
right[r[i] - 'a'].push_back(i);
else
wr.push_back(i);
vector<pair<int,int>> p;
vector<int> cl(A), cr(A);
forn(i, A) {
forn(j, min(left[i].size(), right[i].size()))
p.push_back(make_pair(left[i][j] + 1, right[i][j] + 1));
cl[i] = cr[i] = min(left[i].size(), right[i].size());
}
forn(i, A)
while (cl[i] < left[i].size() && !wr.empty()) {
p.push_back(make_pair(left[i][cl[i]] + 1, wr[wr.size() - 1] + 1));
cl[i]++;
wr.pop_back();
}
forn(i, A)
while (cr[i] < right[i].size() && !wl.empty()) {
p.push_back(make_pair(wl[wl.size() - 1] + 1, right[i][cr[i]] + 1));
wl.pop_back();
cr[i]++;
}
forn(j, min(wl.size(), wr.size()))
p.push_back(make_pair(wl[j] + 1, wr[j] + 1));
cout << p.size() << endl;
for (auto pp: p)
cout << pp.first << " " << pp.second << endl;
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Code
long long H;
int n;
cin >> H >> n;
vector<long long> a(n);
long long sum = 0;
long long gap = 0;
long long h = H;
for (int i = 0; i < n; i++) {
cin >> a[i];
sum -= a[i];
h += a[i];
if (h <= 0) {
cout << i + 1 << endl;
return 0;
}
gap = max(gap, sum);
}
if (sum <= 0) {
cout << -1 << endl;
return 0;
}
long long whole = (H - gap) / sum;
H -= whole * sum;
long long result = whole * n;
for (int i = 0;; i++) {
H += a[i % n];
result++;
if (H <= 0) {
cout << result << endl;
break;
}
}
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Tutorial is loading...
Code
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++)
cin >> a[i];
map<int, vector<pair<int,int>>> segs;
for (int r = 0; r < n; r++) {
int sum = 0;
for (int l = r; l >= 0; l--) {
sum += a[l];
segs[sum].push_back({l, r});
}
}
int result = 0;
vector<pair<int,int>> best;
for (const auto& p: segs) {
const vector<pair<int,int>>& pp = p.second;
int cur = 0;
int r = -1;
vector<pair<int,int>> now;
for (auto seg: pp)
if (seg.first > r) {
cur++;
now.push_back(seg);
r = seg.second;
}
if (cur > result) {
result = cur;
best = now;
}
}
cout << result << endl;
for (auto seg: best)
cout << seg.first + 1 << " " << seg.second + 1 << endl;
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Code
int n, k, r;
vector<vector<pair<int,int>>> g;
int D;
vector<int> col;
void dfs(int u, int p, int f) {
int color = 0;
for (auto e: g[u])
if (p != e.first) {
if (color == f) {
color = (color + 1) % D;
f = -1;
}
col[e.second] = color;
dfs(e.first, u, color);
color = (color + 1) % D;
}
}
int main() {
cin >> n >> k;
g.resize(n);
vector<int> d(n);
for (int i = 0; i + 1 < n; i++) {
int x, y;
cin >> x >> y;
x--, y--;
g[x].push_back({y, i});
g[y].push_back({x, i});
d[x]++, d[y]++;
}
map<int,int> cnt;
for (int dd: d)
cnt[dd]++;
int kk = n;
D = 0;
for (auto p: cnt)
if (kk > k)
D = p.first,
kk -= p.second;
else
break;
col = vector<int>(n - 1);
dfs(0, -1, -1);
cout << D << endl;
for (int i = 0; i + 1 < n; i++)
cout << col[i] + 1 << " ";
}
MikeMirzayanov, in code of 1141B - Maximal Continuous Rest should be
if (a[i % n] == 1)but there isif (a[i % 2] == 1).Thanks, fixed.
why is this on my 'recent actions' page
Thank you so much MikeMirzayanov for this round... I enjoy the problems a lot...
E can be solved using binary search also .
Yes, see this
can you please explain , how are you applying binary search, I couldn't solve it in contest because I couldn't find sum to be monotonically increasing or decreasing
The answer exists only when H becomes zero in the first round, or $$$\sum_{i=1}^{\ n} d_i <0 $$$ .
If $$$\sum_{i=1}^{\ n} d_i >= 0 $$$ , print "-1".
So, after each round H decreases by some amount, now binary search for the first round in which H becomes negative.
I also thought to use binary search but could not come up with any idea so i did greedy. Also your code uses the similar idea so binary search can be avoided. Anyway nice to see a binary search solution.
+1 for username.
:)
Yes but its complexity would be high
no
You are doing a binary search and in each search, you are iterating the whole array to find the answer.
nlogn
That's what I am saying. It can be done in O(n)
Thank you for this good editorial.
Can someone please this, why I'm having Memory Limit Exceeded, on the implementation I did, while editorial says the same.
51582194
Thanks, in advance.
Please see the memory taken by your code, it is more than 250 MB.
:D that is ~250mb not gb.
Yeah thanks, I just wrongly read that.
Try using
intNothing happened, did give a try.
Do not use deque
Any specific reason??
I have heard that deque uses more memory. I just found this also https://stackoverflow.com/questions/16252183/why-is-deque-using-so-much-more-ram-than-vector-in-c Give it a try I think it should work.
BTW, it got accepted, but it'd be too good if you can explain the reason.
Thanks.
Though this implementation is in Java but I think you can look at this also https://codeforces.net/blog/entry/10444
How should I approach Problem D in Java? I m not able to get the structure of code for this problem in java.
Not sure if you're still looking for a response, but my approach was as follows: Use ArrayDeques for each letter or '?' to store the indices. Then, when matching letters I just polled left and right until either went empty and then matched rest with '?' if I had any '?' remaining. In the meantime, you can track how many pairs you have created and use StringBuilder's insert method to put that at the front of the output. Code: https://codeforces.net/contest/1141/submission/51599198
For problem "Same Sum Blocks (Hard)" for finding all the possible sum if I do the loop like this
then I'm trying to find the maximum number of disjoint set but it doesn't give me the optimal answer as you can see here 51596165 can anyone explain why it isn't working?
It's a famous CS problem- Interval Scheduling.
You sort the pairs by the end points and then greedily choose the disjoint intervals. My solution- https://codeforces.net/contest/1141/submission/51669831
Is there an intended slower solution that was meant to pass F1 but not F2 ?
Sure. I did the following. The idea is much more standard but might be trickier to implement if you lack experience at dp. Find all different sums of the segments. Now for each one you can do $$$dp[i]$$$ — the maximal number of disjoint segments of this sum up to position $$$i$$$. Updates can be done straightforwardly by iterating over all $$$j < i$$$ and checking if the sum between $$$j$$$ and $$$i$$$ is the current one or not. That's $$$O(n^4)$$$ at best ($$$O(n^5)$$$ if no prefix sums). 51606155
Btw you can also AC f2 if you do updates in $$$O(1)$$$ instead of $$$O(n)$$$ lol. That would be $$$O(n^3)$$$. 51606140
Why your $$$O(n^3)$$$ solution passes. :xD. Isn't $$$n=1500$$$ too much for $$$n^3$$$.
Welp, I just believed that there are not that many distinct sums such that each of them appear at least twice) Apparently, the number is less than expected $$$O(n^2)$$$. I'd be glad to hear the countertest)
Hi Mike,
Thanks for your reply. I appreciate you taking the time to explain your solutions. :)
Isn't the dp part just like finding the LIS?
1141A - Game 23 can be also solved using dfs.
For problem G I thought of first painting the k vertices which have maximum number of neighbors i.e. to paint the k maximum neighbor vertices with color 1 and then paint rest of the vertices. The number of colors required would be then the maximum number of neighbors of remaining n-k vertices. I don't know how to implement this also whether this approach is correct or not.Has someone else followed this strategy.
hello .......
you can find the (k+1)th largest degree let's say 'd' and this is the maximum number of color needed to color the graph. Now, you can use DFS to color the edges by initiating some variable let's say 'c' with some integer in between 0 to d-1(assuming 0 indexed color) and each time when you move to unvisited vertex make c = (c + 1)%d.
Anyone else having problem with problem G test 24 ?
In code of 1141G — Privatization of Roads in Treeland why was it necessary to set f=-1 after color and f became equal?
Can someone explain me what went wrong in this (my solution for problem E)
In problem G for (auto p: cnt) if (kk > k) D = p.first, kk -= p.second; else break; how we get that D we get from the above loop is the minimum
Interesting problems, thanks.
Binary Search Solution for problem G Simple solution
"vector<pair<int,int»"