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int rp = 0;
while (contest.end != true) rp++;
cout << rp << endl;

How about D? How to do it?

Problem C: This is some text.

// this is code
void inc(int pos, int d) {
    for (; pos < n; pos |= pos + 1)
        f[pos] += d;
}

The rest of the text.

F is so cooked?

Anybody got some hints? I don't see any observation that makes it simpler?

https://atcoder.jp/contests/abc410/submissions/66780530

what s wrong with this D ?

How to construct the sequence of operations for problem G, second sample input?

I think E is easier than before. I usually cannot solve E in previous ABCs, but this time I got accepted in 52 minutes.

E is very easy.I solve A,B,CandE.I use Dynamic Programming to solve E.

Can't we solve D by having a path xor of 1 to n (let's call this xn) and then using xn = min(xn, xn^b) for all b where b is one of the basis of xor of all the cycles in the graph connected with the path 1 to n?

haven't participated any coding contest for a year, and I just come back here. however i solved E but didn't come up with a solution with D... didn't realize it was just a bfs...

Can anyone give me an idea for problem D? I just can't come up with a solution for the cyclic graph.

very nice and elegant soln for E. Might be easy for some, but I couldn't get ahold of the DP idea till the very end.

does someone how to cheeze the time limit for E with an $$$O(n.H.M)$$$ solution?

Edit: I asked GPT to solve it and this is what it gave. Can someone verify if it is what the editorial meant in their OTHER SOLUTION?

Rapidly solve A-F then stuck at G be like:

Can anyone link their solution for E in Top Down Approach ? I was trying to solve it but failing in memoisation part.

I thought in problem E, you can skip the monster, then, BOOM !!!!

The most correct thing I did is abandon E, then do F in last 25 minutes of contest and finally I get AC in F in the last few minutes. If the ranking is calculated in ICPC rules (which means each problem weight equally), then I will totally be finished even if I solve F at last.

Does anyone have an idea how we could solve E if we were allowed to reorder the monsters before starting the game?

Is problem E solvable assuming you could choose the monsters in any order?

This F is harder than before I think.

Could someone tell me why the Bellman-Ford algorithm cannot be used for Problem D?

Can someone explain why my solution using Dijkstra's for D is incorrect? My submission

any recursive dp solution for problem E (without binary search O(n*m))

What I did for question E reducing state

Initially my dp state was depending on three state , now with the optimization shown in the picture I can convert it to two state ..

How??

each time my dp state is depending upon the previous values of h and m so either we can eliminate h or m

1--> eliminating h will result dp[m] --> what is the maximum health I can restore till I use max possible money in this level . If I cannot make even 0 health left it means I was forced to use my health (assumption) even though it was impossible for me so I can't go further break and return i

2--> similarly I can eliminate my magic or money .

dp[h] what is the maximum magic that I can save in this level .

I will use two dp , dp[h] --> for current magic and nxt[h] --> maximum magic that can saved for next state if the health at the very next state is h . Two transitions are possible I am taking the 2nd case

a--> either I can use my health against monster if(arr[i]<=h) then I will go to

nxt[h-arr[i]]=max(dp[h] -- means dont fight , nxt[h-arr[i]] --> fight and have h-arr[i] health for next state with m magic)

b--> or I can use my magic

if(dp[h]>=brr[i] )

nxt[h]=max(dp[h]-brr[i] --> use magic in this state , nxt[h]--> dont use )

Initialize nxt with -1 at starting of all health choices If after all the choices if my maximum element in nxt is <0 it means I cannot have any valid choice return the cnt or value of level (i) .

Here is the code

for (int i = 0; i < N; ++i) {
    fill(nxt.begin(), nxt.end(), NEG);
    for (int h = 0; h <= H; ++h) if (dp[h] != NEG) {
        if (h >= A[i])
            nxt[h - A[i]] = max(nxt[h - A[i]], dp[h]);
        if (dp[h] >= B[i])
            nxt[h] = max(nxt[h], dp[h] - B[i]);
    }
    if (*max_element(nxt.begin(), nxt.end()) == NEG) break;
    dp.swap(nxt);
    ++defeated;
}

This is for the first time I am explaining in such huge length of words . Sorry If I failed

in D, why won't dp[v] = std::min(dp[v], dp[u] ^ w) work. We do it every time even if a node is visited since cycles can be present. Please explain.

void dfs(ll u)
{
    vis[u] = 1;
    for(auto &it : g[u])
    {
        ll v = it.ff;
        ll w = it.ss;
        dp[v] = std::min(dp[v], dp[u] ^ w);
        if(!vis[v])
        {
            dfs(v);
        }
    }
}
void solve()
{

    ll n,m;
    cin >> n >> m;
    forn(i,m)
    {
        ll u, v, w;
        cin >> u >> v >> w;
        g[u].pb(mpr(v, w));
    }

    forn(i,n+5) dp[i] = inf;
    dp[1] = 0; 
    dfs(1);
    if(dp[n] == inf) cout<<-1<<endl;
    else cout<<dp[n]<<endl;

}

can anyone help to figure out the mistake in ques 3

#include <bits/stdc++.h>
using namespace std ;
void sitaram(){
    int n , q ; cin >> n >> q;
    vector<int>v(n+1);
    for(int i =0;i<=n;i++)v[i]=i;
    int r = 0 , l = 0 ;
    while(q--){
        int t ; cin >> t;
        if(t==1){
            int p , x ; 
            cin >> p >> x ;
            if((n+p-l)%n==0)v[n]=x ;
            else v[(n+p-l)%n]=x ;
        }
        else if(t==2){
            int p ; cin >> p ;
            if((n+p-l)%n==0)cout << v[n] << endl;
            else cout << v[(n+p-l)%n] << endl;
        }
        else{
            int k ; cin >> k ;
            r+=k ;
            l = n - r%n ;
        }
    }
}
int main()
{
    sitaram();

    return 0;
}

Are we possible to solve F in a complexity of less than $$$O(N \sqrt N)$$$?