In the next ABC contest, I will try the following strategy.
1. If the remaining time is more than 10/15 min, solve D or E.
2. Otherwise solve ABC

ABC problems are so easy yet, they consume more than 30 min and get me out of mood. I think it will be fun solving ABC under 10 min.

It's been a long time since I last did the problem, now I have time to do it

GL && HF!

Hope to solve more problems.

Harder D Easier E

Hope to get to 1200 in this contest.

I am 1190 now.

Have fun & Good luck!

What are you guys f**king? Easy ABCDEF and hard G , 180+ competitors solved F in 40 min.Harder F,Pls!!!!!!

Any hints regrading E and F ? (Contests have Ended!)

how to write this code optimally for problem E:-

#define vi vector<ll>
#define rep(i, a, b) for(ll i = (a); i < (b); i++)
void solve() {
    ll n,m;cin>>n>>m;
    vi p(n+1);
    rep(i,1,n+1) {
        cin>>p[i];
        p[i]+=p[i-1];
    }
    ll ans =0;
    ll prev =0;
    rep(i,1,n+1){
        prev = p[i-1];
        rep(j,i,n+1){
            //cout<<p[j]-ki<<endl;
            ans+=(p[j]-prev)%m;
        }
    }
    cout<<ans<<endl;

}

how to solve last problem?

Can you guys find the WA problem with my submission for E?

I followed a different approach than the editorial to solve F. First, I merge adjacent nodes that have degree 3 in the original graph. Let $$$A$$$ denote the subset of nodes of the new graph that correspond to a group of adjacent nodes of degree 3 in the original graph. On the new graph the solution is $$$\sum_{v \in A} {cnt_v \choose 2}$$$ where $$$cnt_v$$$ is the number of neighbors of degree 2 of $$$v$$$.

I got TLE from atcoder, can someone help me? I think my complexity of time is right.

#include <bits/stdc++.h>
#define ll long long
#define pii pair<int, int>
#define fi first
#define se second

using namespace std;
int a[200010], sum[200010] = {0}, n, m;

struct fw_tree
{
	int tree[200010] = {0};

	int lowbit(int x)
	{
		return x & (-x);
	}

	void modify(int x, int k)
	{
		for (int i = x; i <= m; i += lowbit(i))
			tree[i] += k;
	}

	int query(int l, int r)
	{
		int sum1 = 0, sum2 = 0;

		for (int i = l - 1; i; i -= lowbit(i))
			sum1 += tree[i];
		for (int i = r; i; i -= lowbit(i))
			sum2 += tree[i];

		return sum2 - sum1;
	}
}tr;

void solve()
{
	cin >> n >> m;
	int res = 0, SUM = 0;

	for (int i = 1; i <= n; i++)	cin >> a[i], sum[i] = (sum[i - 1] + a[i]) % m;

	for (int i = 1; i <= n; i++)
	{
		res += i * sum[i] - SUM + m * tr.query(sum[i] + 1, m);
		SUM += sum[i];
		tr.modify(sum[i], 1);
	}

	cout << res << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);	cout.tie(0);

	int t = 1;	//cin >> t;
	while (t--)	solve();

	return 0;
}

How do you find the number of AC submissions for a problem during a contest?

I think F is much easier than E.

https://atcoder.jp/contests/abc378/submissions/59414779 Can someone help me check why my solution for problem E is wrong? I can't figure it out. Thank you!

Can anyone help me out with why my code gives wrong answer for some tc in question F? Here is my submisiion link https://atcoder.jp/contests/abc378/submissions/59440996

I did not get the solution of problem E. Can anyone please explain me the solution of problem E in details ? ( details explanation )

Anyone solved E using divide and conquer?