We will hold Toyota Programming Contest 2024#11（AtCoder Beginner Contest 379）.
- Contest URL: https://atcoder.jp/contests/abc379
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20241109T2100&p1=248
- Duration: 100 minutes
- Writer: MtSaka, toam, physics0523
- Tester: yuto1115, cn449
- Rated range: ~ 1999
- The point values: 100-200-300-400-475-550-575
We are looking forward to your participation!
I want to ask how can come up with the solution in a short time. Sorry for my weak English.
practice
GL && HF!
Today is my birthday!
happy birthday ! cook this round
Seems a lot easier than previous contests. G is only 575.
GL&&HF!
Problem C is too difficult!
C is way much harder than a 300-score problem, I think it worth at least 400.
EDIT: only ~3.1k solves o_O
interesting; why do you think that? it seems like it only required sorting and speeding up the simulation with triangular numbers... (and it doesn't seem like basically bruteforcing the thing is hard to come up with but i'm probably biased here)
though i did get stuck on C for some reason
F was beautiful!
For E, my only observation is:
For some number, says, 349685
Ans = 333333 + 2 * 44444 + 3 * 9999 + 4 * 666 + 5 * 88 + 6 * 5
But this will definitely get TLE, so how the hell did u guys manage to solve it?
I unaccepted C at about 20:30,and I dont accept so far,but I dont know why I unaccepted.
Why did it WA?my submission
Perhaps overflow caused by line 50 and 55 because even if your ans stored in ll, but you perform multiplication on integers with l, r, and b
Thanks.
I assumed the input for boxes containing stones was sorted. Was wondering why getting WA the entire time ;)
Ikrrr! so infuriating
Why bruteforce works for D? Solution
I didnt try hoping it wud get TLE
During he contest, I thought that was the intended solution. It works because a plant only be harvest(count/delete) for once.
I think it should TLE. Eg a case where the first 1e5 queries are type 1, and the next 1e5 queries are type 2. Even if you compress Type 2 queries, should still TLE if you alternate between type 1 and type 2 queries...
sorry, I missunderstood the OP's solution.
But if the second operation was changed to something similar to lazy tag(then you insert -tag, and delete >= H-tag), then it works.
How to solve E? My approach is to consider the contribution of each digit, but used High Precision which caused TLE. :(
Thank you very much!
My bad for not checking conditions, but still not to sort input in C was totally unnecessary :(
Anybody can explain me the F problem ;-;
C was hard
Could not optimize to final state
felt F was a next greatest ele problem but got confused from the testcases?
Example:
2 1 4 3 5
query: 1 and 2
expected ans: 2 (buildings with heights 3 and 5)
how is 1 able to see 3 and 5 ? isn't it blocked by 4 (building 3)
You misread the example. It is building index 3 and 5 (not height) that can be seen. Building index 3's height is 4 and building index 5's height is 5.
ohh it was a previous greater + next greater combo
damn could've solved it
E made me laugh because I thought it was a easy simulation problem haha. Surprisingly, I got a 2210 runtime running at $$$O(n^2)$$$. I read the editorial last night and it had such a beautiful solution. 10/10 contest.
Can someone help with finding what's wrong with my submission for problem C https://atcoder.jp/contests/abc379/submissions/59600147
Try this test case;
Thanks, need to practice reading statements more carefully :)
when the Testcases will be updated in dropbox and then i can download the wrong test?
my English is not very good.excusmeT T
Could someone tell me where am wrong in C ? I've tried many times, but didn't accepted yet.... Below is my code. ~~~~~
define _CRT_SECURE_NO_WARNINGS
include <bits/stdc++.h>
using ll = long long int; using namespace std;
struct ch { ll x, y; } a[200005];
bool cmp(ch q, ch p) { return q.x < p.x; }
int main() { ll n, m; cin >> n >> m;
for (int i = 1; i <= m; i++) cin >> a[i].x; for (int i = 1; i <= m; i++) cin >> a[i].y; sort(a + 1, a + m + 1, cmp); ll ans = 0; for (int i = 1; i < m; i++) { ll gap = a[i + 1].x - a[i].x - 1; ll d = gap * (gap + 1) / 2; if (gap == 0) { a[i + 1].y += a[i].y - 1; ans += a[i].y - 1; } if (gap <= a[i].y - 1&&gap!=0) { a[i + 1].y += a[i].y - 1 - gap; ans += d; } if(gap>a[i].y-1) { cout << -1 << endl; return 0; } } ll last_gap = n - a[m].x; ll last_cost = last_gap * (last_gap + 1) / 2; if (last_gap < a[m].y - 1) { cout << -1 << endl; return 0; } else { ans += last_cost; } cout << ans << endl; return 0;} ~~~~~
Why no blog of ABC380 yet?