"Sure, the contest will be unrated."

It's kind of 2-SAT.

Let's iterate over all pairs of necklaces. For each pair $$$p$$$ and $$$q$$$, look at the similarities of $$$p, q$$$ and $$$p, rev(q)$$$. If this pair can only exist when both are in the same direction, draw a white edge between them; if this pair can only exist when one of them is reversed, draw a black edge between them; if this pair can't exist either way, report a failure. After this reduction it is fairly similar to e.g. 553C - Love Triangles.

We want to split the resulting graph into two parts $$$S$$$ and $$$T$$$, such that there are only black edges between $$$S$$$ and $$$T$$$, and only white edges within $$$S$$$ and $$$T$$$; also, we want to minimize the size of $$$T$$$ (i.e. the number of reversed necklaces). First check if there is a cycle with odd number of black edges, if there is, report a failure. Otherwise, we have a bunch of "bipartite" connected components. Put the smaller half of each component into $$$T$$$ and we have our answer.

E had very odd bounds as it can be solved in quadratic time.

For every pair of necklaces, we either get that

1. Neither or both necklaces must be flipped (xor must be 0)
2. Exactly one of the necklaces must be flipped (xor must be 1)
3. The pair of necklaces imposes no conditions
4. No amount of flipping makes them similar

If 4 happens even once, answer impossible. Now knowing the orientation of one necklace in a connected component (with conditions as edges) determines the orientations of the rest. Just take the orientation that minimises flipped necklaces.

oh I'll check it. Thank you so much -is-this-fft- mango_lassi

you need at least k/2 buses and at most k buses since k<=8000 iterate on each k to find the optimal one

there were a lot of test cases

You delete a bridge.

Iterate over components, run dfs.

Then you can always delete the last vertex in dfs (Try to think why it is true). So connect any vertex from next component with parent of last vertex.

You don't need to find bridges or something like this.

For C, What I did was, start by sorting queries by their r value. then i created a segment tree which tells the maximum possible answer with some l value for fixed r. segment tree was range update and point maximum query segment tree which also tells the index of maximum. now the main problem was how to store answer in segment tree so for fixed right index =x the index x would have -k + sum of segments which lie in (x-1,x) and index x-1 would have -2*k + sum of segments which lie in(x-2,x) and so on. we can do this possible by iterating over r and updating the range (1,r) with -k everytime.

Heres my code- https://www.ideone.com/g2j16q

https://pastebin.com/UNyjNnuu

Here hope this helps

Key observation is that to reach row_index rb from ra all the elements lying between will be of same type(if element at ra is even then all elements will be even and vice versa)..same thing applies for column too... I am limited by my formatting skills otherwise I could have explained better

If the number of connected components in the graph is $$$c$$$, then the answer is $$$c-1$$$.
Just connect all the other components to the first component. To join the component:
1) If it is a tree, change any leaf-vertex's edge
2) otherwise, change any edge that belongs to a cycle

I: my solution was the weirdest solution I've ever came up with

First to make it simpler we'll take the complement of the set, so we'll order by lesser size and greater sum and the restriction goes from SET <= K to SUM — SET >= K.

Now, let's try and get TLE with a bruteforce solution like bf(i, remaining that need to be taken, current sum) + ordering the values before bruteforce.

Here comes the trick, if we compress a tree to not have any internal nodes with degree 2 then it has O(N) leafs and O(N) nodes, using this fact we can compress the segments that we don't have a choice but to choose the values by using a binary search to remove that segment.

This isn't enough since it might have values mixed in that are smaller than what should be the last value, so we do a binary search to find the largest value that doesn't use all M things and then do a special iteration to get some value that's exactly what the last should be.

Complexity is something like O(N * logN * logV) or something like that lol.

M is nice.

For each bit $$$i$$$, let L = set of indices $$$x$$$ with bit $$$i = 0$$$, R = set of indices $$$y$$$ with bit $$$i = 1$$$ and $$$y + 1$$$ also has bit $$$i = 1$$$.

Similarly, for each bit $$$i$$$, let L = set of indices $$$x$$$ with bit $$$i = 1$$$, R = set of indices $$$y$$$ with bit $$$i = 1$$$ and $$$y + 1$$$ also has bit $$$i = 0$$$.

Now, we used 26 sets.

Observe that only pairs $$$(x, y)$$$, with y = 01111...1 (the string of 1s is nonempty) and x = 1 such that there is at least one bit 1 in the suffix, are uncovered.

Thus, it is sufficient to make $$$12$$$ more queries that cover all y of the form 0111...1 for a fixed 1-suffix length and then put all x that can be matched on the left hand side.

In total, my solution uses $$$38$$$ queries.

Simpler solution for I:

Sort all values. We loop through the number of elements from N to 1. Let's say now we are considering the ways to choose k elements. Start with the smallest k elements. We build our solutions progressively using 2 types of steps: Consider a pointer starting at the largest element. Each step consists of either increasing the current element to the next possible element, or move the pointer lefft and increase the current element to the next possible element. We can push the current sum, current position of pointer, position of previous element, position of current element into a priority queue to simulate this expansion. To restore the solution, just create an ID for each state and keep track of which move is used to reach the current state, and we can backtrack at the end.

Complexity: $$$O(M\log M)$$$

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define fbo find_by_order
#define ook order_of_key
 
typedef long long ll;
typedef pair<ll,int> ii;
typedef vector<int> vi;
typedef long double ld; 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
 
pair<int,int> par[3001101];
 
void solve()
{
	int n; ll k; int m; scanf("%d %I64d %d",&n,&k,&m);
	vector<pair<ll,int> > vec;
	for(int i=0;i<n;i++)
	{
		ll x; scanf("%I64d",&x);
		vec.pb({x,i+1});
	}
	sort(vec.begin(),vec.end());
	vector<pair<ll,ll> > concerts;
	ll cursum=0;
	for(int i=0;i<n;i++) cursum+=vec[i].fi;
	int lasid=0;
	for(int siz=n;siz>=1;siz--)
	{
		if(concerts.size()<m)
		{
			if(cursum>k) 
			{
				cursum-=vec[siz-1].fi; continue;
			}
			int curid=0;
			priority_queue<tuple<ll,int,int,int,int>,vector<tuple<ll,int,int,int,int> >,greater<tuple<ll,int,int,int,int> > > pq;
			pq.push({cursum,0,n-1,siz-1,siz-1});
			while(!pq.empty()&&concerts.size()<m)
			{
				auto tmp=pq.top(); pq.pop();
				ll S=get<0>(tmp);
				int id=get<1>(tmp);
				int maxnw=get<2>(tmp);
				int cur=get<3>(tmp);
				int curele=get<4>(tmp);
				lasid=id;
				concerts.pb({siz,S});
				//proceed with current (type=0)
				if(cur<maxnw&&S+vec[cur+1].fi-vec[cur].fi<=k)
				{
					pq.push({S+vec[cur+1].fi-vec[cur].fi,++curid,maxnw,cur+1,curele});
					par[curid].fi=id;
					par[curid].se=0;
				}
				//go next
				if(curele>0&&cur>curele&&S+vec[curele].fi-vec[curele-1].fi<=k)
				{
					pq.push({S+vec[curele].fi-vec[curele-1].fi,++curid,cur-1,curele,curele-1});
					par[curid].fi=id;
					par[curid].se=1;
				}
			}
			cursum-=vec[siz-1].fi;			
		}
	}
	printf("%d\n",int(concerts.size()));
	for(int i=0;i<concerts.size();i++)
	{
		printf("%I64d %I64d\n",concerts[i].fi,concerts[i].se);
	}
	//print ans
	if(!concerts.empty())
	{
		int siz=concerts.back().fi;
		vi seq;
		while(lasid>0)
		{
			seq.pb(par[lasid].se);
			lasid=par[lasid].fi;
		}
		vi ans;
		for(int i=0;i<siz;i++) ans.pb(i);
		int curid=siz-1;
		for(int i=int(seq.size())-1;i>=0;i--)
		{
			if(seq[i]) {curid--; ans[curid]++;}
			else ans[curid]++;
		}
		for(int x:ans)
		{
			printf("%d ",vec[x].se);
		}
		printf("\n");
	}
}
 
int main()
{
	int t; scanf("%d",&t);
	while(t--) solve();
}

In fact, codes of a problem in this contest will be visible after you solve it.

of course not

in J you should binary search the amount of soldiers in one row. in B you should sort teams by size and consider you put 1st and last together, then 2nd and last-1 ans so on for each capacity you must check amount of rides and minimal capacity*number of rides among them will be the answer

Enumerate on the position of last Reset operation, then we can use binary search to determine if this is possible. Also we need to know at least how many Reset operations we need to do before this one, which can be preprocessed using dynamic programming. The transition can be done using either binary search or two pointers. Time complexity should be $$$O(n\log{n})$$$.

Just find the answer with a binary search.

We can assume that $$$a \geq c$$$ because they are symmetric, then for an answer $$$k$$$ wee need to check whether we can satisfy the condition.

Suppose we have all $$$a$$$ in set $$$A$$$, $$$b$$$ in set $$$B$$$ and $$$c$$$ in set $$$C$$$, then we need to adjust it.

If $$$a \leq k$$$, obviously $$$c \leq a \leq k$$$, we only need to check if we can put some person b to $$$A$$$ or $$$C$$$.

If $$$a \geq c > k$$$, we failed, otherwise we put $$$a - k$$$ person a in $$$B$$$, and we can take at most $$$b$$$ person b from $$$B$$$ to $$$C$$$, just to find out whether it is possible.

You solution prints lines like 999998 999998000000000000 in the bottom of the output.

Here it is.

104527338

https://blog.csdn.net/hrd535523596/article/details/109006239 Maybe this will help, use google translate