Monoxer Programming Contest 2024（AtCoder Beginner Contest 345） Announcement

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Автор atcoder_official, история, 2 месяца назад, По-английски

We will hold Monoxer Programming Contest 2024（AtCoder Beginner Contest 345）.

Contest URL: https://atcoder.jp/contests/abc345
Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240316T2100&p1=248
Duration: 100 minutes
Writer: Nyaan, math957963, evima, PCTprobability
Tester: MtSaka, Aus21
Rated range: ~ 1999
The point values: 100-200-350-450-525-550-675

We are looking forward to your participation!

atcoder

atcoder_official
2 месяца назад
43

Комментарии (43)

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xiaobai7363

2 месяца назад, # |

Good luck to everyone!

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rafaelbg

2 месяца назад, # |

The contest cover page is in Japanese language even though my atcoder account is in English. Is that a contest registration issue or is it going to be in Japanese? Thanks, Rafael

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llc5pg

2 месяца назад, # |

+16

ABC 345... Time for me to practice arithmetic series to prepare for this contest.

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yuanzihan

2 месяца назад, # |

ABC345? Increasing sequence?

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abhishekharkar

2 месяца назад, # |

contest questions are in japanese language?

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shentianlin

2 месяца назад, # |

Does the contest provide problem statements in English or only in Japanese?

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shiny_shine

2 месяца назад, # |

Please tell us if you only provide Japanese problem statements.

Although I'm a Chinese, I know that there're some differences between Chinese and Japanese, so it may cause ambiguity.

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Rubikun

2 месяца назад, # |

← Rev. 2 →

+26

To those who are confused

All previous rated ABC have Japanese and English statements. I don't know this time, but I guess it's the same since no special announcement was made.(I'm not an official)

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AllenAK

2 месяца назад, # |

-8

How C

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alpha1215

2 месяца назад, # ^ |

← Rev. 2 →

hint

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Wasif_Shahzad

2 месяца назад, # ^ |

I actually tried implementing the same thing but I got WA x13

#include <bits/stdc++.h>

using namespace std;
#define int long long
typedef long double ld;
#define all(x) x.begin(), x.end()
#define pb push_back

signed main(){
	string s;
	cin >> s;
	set<char> a;
	int n = s.size();
	vector<int> suffix(n);
	suffix[n-1] = 0;
	a.insert(s[n-1]);
	for(int i = (n - 2); i >= 0; i--){
		if(!a.count(s[i])) suffix[i] = a.size();
		else suffix[i] = a.size() - 1;
		a.insert(s[i]);
	}
	int ans = 0;
	for(int i = 0; i < n; i++){
		ans += suffix[i];
	}
	cout << max(ans, 1LL) << endl;
}

What could I be missing in my logic?

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Sheldor.The.Conqueror

2 месяца назад, # ^ |

I could not understand your logic, but I think you are doing the same mistake I made yesterday.

I figured it out now, if we can generate the original string again then it would count as well in the final result, that would happen only if the frequency of any character is $$$>1$$$, so make $$$ans = 1$$$ when you see that, then iterate over the string to calculate the remaining answer.

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batsyboy

2 месяца назад, # ^ |

why ain't this working , pls can you help me out

string s; cin>>s; int n=s.size(); map<char,int>mp;

for(int i=0; i<n;++i){ ++mp[s[i]]; } vectorv; for (auto it:mp ) { /* code */ v.push_back(it.second); }

int sum=0;

for (int i = 0; i < v.size(); i++) { /* code */

for (int j = i+1; j < v.size(); j++)
{
    /* code */
    sum+=(v[i]*v[j]);
}

} for (auto it:v) { /* code */ if(it>1){ ++sum; break; } }

cout<<sum<<endl;

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pb_messi

2 месяца назад, # |

+11

Problems were more difficult than usual today.

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xcx0902

2 месяца назад, # |

← Rev. 2 →

Atcoder Rand Contest.

This code passed Problem D. (Even though I tried 5 times)

...
bool dfs(int x) {
    if (clock() >= CLOCKS_PER_SEC * 1.95) {
        mt19937 rnd(random_device{}());
        cout << ((rnd() & 1) ? "Yes" : "No") << endl;
        exit(0);
    }
    ...
}
...

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VinCenzo___

2 месяца назад, # ^ |

Was E dp + 2-d segment tree?

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milind0110

2 месяца назад, # ^ |

It was just dp. Optimization could be done without segment tree using sets or without those as well.

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VinCenzo___

2 месяца назад, # ^ |

← Rev. 2 →

I just messed up everything.

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milind0110

2 месяца назад, # ^ |

← Rev. 2 →

It's ok I solved it 10 mins after contest.

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chuanyu.fan

2 месяца назад, # ^ |

aMaZiNg

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J.A.K.D.Z

2 месяца назад, # |

-22

Top 200 EZ

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qiuzx

2 месяца назад, # |

+18

Don't know if it was intended or not, but I guess the time limit is too loose for G. My $$$O(n^2/k)$$$ solution with a fast mod managed to squeeze into the time limit. XD

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LilyWhite

2 месяца назад, # |

How to prune D?

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J.A.K.D.Z

2 месяца назад, # ^ |

← Rev. 2 →

-6

with ‌Backtracking.

You can first solve this problem for learning backtrack. Eight queens puzzle

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LilyWhite

2 месяца назад, # ^ |

Well it turned out I shouldn't sort the tiles by increasing area. I should sort them in decrease

Never mind

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ExplodingFreeze

2 месяца назад, # ^ |

← Rev. 5 →

You don't really need to, there is a $$$O(n! \cdot 2^n \cdot H \cdot W)$$$ brute which fits into the time limit pretty easily.

Basically, for all possible orders of the tiles, try all combinations of both possible orientations (initial or rotate 90) among all tiles and for each of them try to fit the grid using the following pseudocode:

for i in [1, H]:
    for j in [1, W]:
        if (i, j) is not covered:
            try to place the next tile with its top left corner at (i, j)

Code — 51308848

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ExplodingFreeze

2 месяца назад, # |

Toughest ABC I've seen in a while, problem E feels tougher to me than most CF Div2E problems @_@.

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VinCenzo___

2 месяца назад, # ^ |

was E dp+segment tree?

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ExplodingFreeze

2 месяца назад, # ^ |

To be honest, I don't know. I have an $$$O(n \cdot k)$$$ dp but its getting 5xWA and I'm still not sure where my logic is incorrect.

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variety-jones

2 месяца назад, # ^ |

Try

The expected output is 3.

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aryanc403

2 месяца назад, # ^ |

← Rev. 2 →

I ended up creating a video editorial for this problem. Here

Do take some pauses before jumping between chapters. Each chapter adds one optimisation to move one step closer.

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VinCenzo___

2 месяца назад, # ^ |

thanks,it was helpful!!

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J.A.K.D.Z

2 месяца назад, # |

-8

F was easier than D :)

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OMAR_JR

2 месяца назад, # |

← Rev. 5 →

if you get wrong in c this because if there is a repetition of the letter, when you make a switch between the same letter, the resulting string calculates a new string just once. aab -> aab , baa , aba so the answer is 3 not 2 just do it before your code

for(int i = 0;i < s.size();i++){
    if(mp[s[i]] > 1){
        sum++;
        break;
     }
}

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yazhouli0607

2 месяца назад, # |

for B ，why int(x/10) is wrong for negative ?

def solve(): x=int(input()) if x<=0: print(int(x/10)) else: if x%10!=0: print(x//10+1) else: print(x // 10)

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zhongyu

2 месяца назад, # |

I think the test case of A should add "<=<=>=>".Almost all the solutions didn't conside this.

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darklord002

2 месяца назад, # |

← Rev. 4 →

why can't we print Yes for this string in A "><==>"?

upd:Got it

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SummerSky

2 месяца назад, # |

Problem E is fantastic, and I spend about two hours and finally get it accepted. My idea is to use dpmax[j][0] = pair<color0, value0>, and dpmax[j][1] = pair<color1, value1> to denote that, until now, if we remove j balls, the optimal two values with color0 and color1 as the rightmost ball. Besides, we use dp[i][j] to denote the optimal value that we can get, if we consider the first i balls and remove j of them. dp[i][j] can be computed simply by using dpmax[j][0] and dpmax[j][1], and then we should update dpmax[j][0] and dpmax[j][1].

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deydhananjoy5

2 месяца назад, # |

Why A is giving WA?

my code

string s; cin>>s;

bool a = false, b = false, c = false;
for(auto &ch : s)
{
    if(ch == '<')
    {
        a = true;
        b = false;
        c = false;
    }
    if(ch == '=')
    {
        b = true;
        c = false;
    }
    if(ch == '>')
    {
        c = true;
    }
    if(a && b && c)
    {
        cout<<"Yes"<<endl;
        return;
    }
}
cout<<"No"<<endl;

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