UNIQUE VISION Programming Contest 2024 Spring(AtCoder Beginner Contest 346) Announcement

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

We will hold UNIQUE VISION Programming Contest 2024 Spring(AtCoder Beginner Contest 346).

Contest URL: https://atcoder.jp/contests/abc346
Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240323T2100&p1=248
Duration: 100 minutes
Writer: yuto1115, cn449, physics0523, nok0, kyopro_friends, leaf1415, ynymxiaolongbao
Tester: sotanishy, toam
Rated range: ~ 1999
The point values: 100-200-250-400-450-525-575

We are looking forward to your participation!

atcoder

atcoder_official
2 месяца назад
73

Комментарии (73)

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Ashnu

2 месяца назад, # |

-26

Hope to solve a,b,c,d,e.

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Ashnu

2 месяца назад, # |

-74

I hope all problems are Div3 type only.

Atcoder Beginner contest == Codeforces Div 3

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xiaobai7363

2 месяца назад, # |

-20

Hope to solve ABCDEF.

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Eternal_crystal

2 месяца назад, # |

-20

Hope to solve ABCDE. I hope I won't be hacked this time

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railgunjjj

2 месяца назад, # |

Hope to solve ABCD

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Tangninghaha

2 месяца назад, # |

Hope to solve ABCDE

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rui_er

2 месяца назад, # |

+64

Problem E is almost the same as 2023 Chinese NOI Spring Test Problem A. 涂色游戏 (paint), and is an easy version of LGR-162-Div.3 Problem D. 头 which I coauthored.

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nichke

2 месяца назад, # |

← Rev. 2 →

Please, can somebody tell me why this fails on F it literally binary searches the first left position to place "K" characters for every character in T. These contests are literally implementation only and I hate every minute of it.

16xAC 37xWA Link

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Rideris

2 месяца назад, # |

I deducted the n^2 brute force solution mentioned at the end of G's editorial in 5 min, but just can't find a way to use a data structure to optimize it:( maybe i just ain't Chinese enough:P

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SpecialOperations0

2 месяца назад, # ^ |

-10

registered 5 months ago and already master, i am pretty sure that you are chinese D:

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nskybytskyi

2 месяца назад, # ^ |

In sweepline you add and remove segments. Use a lazy segment tree to maintain the number of segments that cover each point. Adding is += 1 on a range and removing is -= 1 on a range. Query union length after each operation. It is n minus the count of zeros. The count of zeros is the count of minimums if the minimum is zero and zero otherwise code

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Rideris

2 месяца назад, # ^ |

yeah i figured out the sweepline part with ease but couldnt think of the sgt part...

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Chimpanzee

2 месяца назад, # |

wtf is F, I cannot understand why this fails: binsearch on K, for checking each character, binsearch again on how far we have to go

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cleanser

2 месяца назад, # ^ |

I did the same thing but idk where I'm getting 2 WAs from.

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kHarsh3715

2 месяца назад, # ^ |

← Rev. 2 →

+14

For me, the 2 WAs was because of limits of binary search. Initially I put high to be 1e18 but I don't know why this was causing 2 WAs. Changing it to n*(size of s) worked. 1e18 probably caused overflow somewhere i don't know why. But even 1e17 limit which is 1e12*1e5 was causing 1 WA.

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UP84

2 месяца назад, # ^ |

← Rev. 2 →

The upper bound of length is (n*(size of s))/(size of t),if you take a take k such that k*(size of t) > 2^63,it can overflow.

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kHarsh3715

2 месяца назад, # ^ |

ohh yeah, wierd 1e17 failed, but n*(size of s) passed and I don't think it is possible that they did not put n = 1e12 and |s| = 1e5 testcase. Maybe i got lucky lol.

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redcircle

2 месяца назад, # ^ |

Try this test: n = 1

s = a

t = aaaaa

ans = 0

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harvey_wzy

2 месяца назад, # ^ |

Me:

1
aa
aa

The answer is 1 and my code output 0. Now I fixed the code by this test.

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fclvbfm934

2 месяца назад, # ^ |

Had the same issue where I put 1e18 and got 2 WAs. Turns out this is because of overflows on long longs when you do the binary search. Switching to bigint made it AC.

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lw22019

2 месяца назад, # ^ |

Me too,I used "long long" at first and got "WA",and I used "unsigned long long" and got "AC".

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UP84

2 месяца назад, # ^ |

This works, i did the same thing, code -Link

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nichke

2 месяца назад, # ^ |

-8

Did the same and failed 37 WA lol. It's a shit problem tbh just implementationforces.

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nskybytskyi

2 месяца назад, # ^ |

+11

It may be beneficial for you to focus on improving your implementation or at least debugging skills. It's unfortunate to miss out on some good points because of silly bugs but it's on you.

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TG-STUDIO

2 месяца назад, # |

first time with Corona Virus （just to D sadness）say no more go to sleep

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cleanser

2 месяца назад, # |

+10

Is F binary search on answer and for each search, you binary search to find when you have enough letters? My solution was n(logn^2) but it TLEs on a couple cases.

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Farhan20

2 месяца назад, # ^ |

+17

Binary search for problem F will work perfectly. I used binary search for problem F and got accepted.

Here is my submission link : https://atcoder.jp/contests/abc346/submissions/51615694

Let, p = s.size() and q = t.size()

Then, time complexity : O(q * log(1e17) * log(p))

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650iq

2 месяца назад, # |

Today was a bit on easier side..could solve upto E :)

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amoad

2 месяца назад, # |

How to solve D using DP ?

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alpha1215

2 месяца назад, # ^ |

you can do it with prefix sum

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kishor236jha

2 месяца назад, # ^ |

can u explain prefix sum approach?

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alpha1215

2 месяца назад, # ^ |

idea

code

#include<bits/stdc++.h>
using namespace std; 
char nl = '\n';
using i64 = long long;

void solve(int t) {
    // cout << "test #" << t << << nl;
    int n; cin >> n;
    string str; cin >> str;
    vector<i64> C(n);
    for(auto& a : C) cin >> a;
    vector<i64> match(n), not_match(n);
    i64 a = 0, b = 0;
    for(int i = 0; i < n; i++) {
        if((i & 1) == str[i] - '0') a += C[i];
        else b += C[i];
        match[i] = a;
        not_match[i] = b;
    }
    a = 0, b = 0;
    i64 ans = LLONG_MAX;
    for(int i = n - 1; i > 0; i--) {
        if((i & 1) == str[i] - '0') a += C[i];
        else b += C[i];
        ans = min(ans, min(match[i - 1] + b, not_match[i - 1] + a));
    }
    cout << ans << nl;
} 

int main() {
    int tt = 1;
    // cin >> tt;
    for(int t = 1; t <= tt; t++) solve(t);
}

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kishor236jha

2 месяца назад, # ^ |

sorry i really not getting your ideas can you explain more?

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alpha1215

2 месяца назад, # ^ |

← Rev. 3 →

you have to make string such that s[i] == s[i + 1] for exactly one i and the const should be minimum the idea is calculate cost for all possible i for both cases s[i] == 1 and s[i] == 0

suppose we choose s[i] == 0 and i is odd
now on the left <= i
all 1's should occur at even position
all 0's should occur at odd position

on right >= i + 1
all 1's should occur at odd position
all 0's should occur at even position

total cost = cost of making left + cost of making right

similarly do it for s[i] == 1

and the ans will be min cost considering all possible i for both cases

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yadavDipesh

2 месяца назад, # ^ |

code

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_Aryan_

2 месяца назад, # |

how to do D with dp?

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brave-kid

2 месяца назад, # ^ |

← Rev. 4 →

$$$dp[pos][prev][count]$$$

pos = current position

prev= value of previous character

count = min( 2 ,count of $$$1 \le i \le n-1$$$ where $$$s_i=s_i+1$$$)

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alpha1215

2 месяца назад, # ^ |

← Rev. 3 →

for s[i] == s[i + 1] you just have to make sure either on the left for all k <= i parity of k == s[k] and on the right k >= i + 1 parity of k != s[k] or vice versa

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_Aryan_

2 месяца назад, # ^ |

← Rev. 3 →

i am getting wrong on 3 test cases. What have i missed?

Code

void _segfault_()
{
    ll n; cin >> n;
    string s; cin >> s;
    s = '@' + s;
    vector<ll>vec(n + 1);
    for (ll i = 1; i <= n; i++)
        cin >> vec[i];
    ll dp[n + 5][3][3];
    for (ll i = 0; i < n + 4; i++) {
        for (ll j = 0; j < 3; j++) {
            for (ll k = 0; k < 3; k++) {
                dp[i][j][k] = LLONG_MAX / 2;
            }
        }
    }
    for (ll i = 0; i < 3; i++) {
        for (ll j = 0; j < 3; j++) {
            dp[0][i][j] = 0;
        }
    }
    for (ll i = 1; i <= n; i++) {
        for (ll j = 0; j <= 1; j++) { //lst character
            if (s[i] - '0' == j) {
                dp[i][1][j] = min({dp[i][1][j], dp[i - 1][0][j]}); //already 1 pair done
                dp[i][1][j ^ 1] = min({dp[i][1][j ^ 1], dp[i - 1][1][j] + vec[i]}); // doing at curr indiex
                dp[i][0][j ^ 1] = min({dp[i][0][j ^ 1], dp[i - 1][0][j] + vec[i]});
            }
            else {
                int k = s[i] - '0';
                dp[i][1][k] = min({dp[i][1][k], dp[i - 1][1][j]});
                dp[i][1][k ^ 1] = min({dp[i][1][k ^ 1], dp[i - 1][0][j] + vec[i]});
                dp[i][0][k] = min({dp[i][0][k], dp[i - 1][0][j]});
            }
            // cout << dp[i][0][0] << " " << dp[i][0][1] << " " << dp[i][1][0] << " " << dp[i][1][1] << endl;
        }
    }
    cout << min(dp[n][1][1], dp[n][1][0]) << endl;
}

https://atcoder.jp/contests/abc346/submissions/51621500

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Yugandhar_Master

2 месяца назад, # |

Solved all problems after so many contests, thank you for the contest

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Farhan20

2 месяца назад, # |

← Rev. 5 →

+10

I have an edge case for problem F.

testcase :

aaa

expected output : 3

But output of my accepted code for this testcase is 1. So, my code should got WA verdict. But my code got accepted.

Here is my submission link : https://atcoder.jp/contests/abc346/submissions/51615694

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nichke

2 месяца назад, # ^ |

Mine gives 3 lmao but still fails 37 cases. Can you try finding a counter for mine please?

Submission

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variety-jones

2 месяца назад, # ^ |

Take a look at Ticket 17424 from CF Stress for a counter example.

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nichke

2 месяца назад, # ^ |

Thanks so much, I rounded one division down instead of up and messed up whole problem. :(

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_Shimla_05_

2 месяца назад, # ^ |

+15

10 yyqml y answer should be 20 , your code gives 21

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nichke

2 месяца назад, # ^ |

I got it, I rounded down division in one place instead of up. Thanks.

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rishichowdhury711

2 месяца назад, # |

I have a question for problem B. I solved it in this manner: https://atcoder.jp/contests/abc346/submissions/51573486.

I was curious though, what if the values of W and B are very large(upto 1e18 or 1e9). In that case how will I be able to approach the problem. Is that even solvable? I am not able to think of any solution .Any help would be appreciated. Thank you

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blaze

2 месяца назад, # ^ |

Iterate over all starting points on the string and calculate how many strings you need for b since there are less occurences of b in the strings and then iterate over the remainder of the string.

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alpha1215

2 месяца назад, # ^ |

yes

how

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shivam_sikotra

2 месяца назад, # |

In B, how answer is "No" for W=7,B=5???

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ilovenandini

2 месяца назад, # ^ |

the answer is Yes for w=7 and b=5

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Wasif_Shahzad

2 месяца назад, # ^ |

that's the original string itself so the answer is Yes

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trytoqualify2025voi

2 месяца назад, # |

absolutely interesting problem, G is hard but the tutorial is very clear.

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RYOMEN__SUKUNA

2 месяца назад, # |

-13

This Round was really amazing. I really enjoyed this round. The problems are very good

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as_09

2 месяца назад, # |

← Rev. 2 →

Can anybody tell me why this code is giving RE in problem D. Link to solution: https://atcoder.jp/contests/abc346/submissions/51621300

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nskybytskyi

2 месяца назад, # ^ |

Because your memory usage (3628612 KB) is greater than the memory limit (1024 MB). You probably run into bad_alloc

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as_09

2 месяца назад, # ^ |

Can you tell why my code memory usage exceeds, as it is a simple dp solution.

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llc5pg

2 месяца назад, # ^ |

I modified your submission into this Now it passes. The idea is to change string to string& in the function declaration.

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as_09

2 месяца назад, # ^ |

Ohh shit man silly error. Thanks for your help.

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sergiu128

2 месяца назад, # |

← Rev. 2 →

For D, I don't really get why this (http://atcoder.jp/contests/abc346/editorial/9651) is a correct solution. I don't understand how the answer is computed. I thought that if n is even, then the series first character must be different than the last character, so ans = min(ans, l0[i] + g1[i]) and ans = min(ans, l1[i] + g0[i]). If n is odd then the first and last characters are the same so ans = min(ans, l0[i] + g0[i]) and ans = min(ans, l1[i], g1[i]).

However I see that doesn't matter for the solution. I'm confused — can anybody explain it to me?

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aryanc403

2 месяца назад, # ^ |

Here is the crux of the idea

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sergiu128

2 месяца назад, # ^ |

Thank you! Everything is clear to me until the last step. For the last point, isn't the validity of a pair out of the 4 based on whether n is even or odd? What i am saying is that if n is even, then (pref0, suf0) and (pref1, suf1) are valid, and if n is odd, then (pref0, suf1) and (pref1, suf0) are valid. What am i missing here?

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aryanc403

2 месяца назад, # ^ |

Yes, thats the correct interpretation of the last line.

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monkey_d_viper

2 месяца назад, # |

can i do problem G by taking number of elements to right until not finding current element and left and add to result like res+=l*r is this wrong approach please help me

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