Codeforces Round #448(Div.2)

6 лет назад, # |

No hacking, problems are too hard. Great contest...

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I_love_Palindromic_Tree

6 лет назад, # |

As predicted, problem C is much more difficult than usually.

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6 лет назад, # ^ |

+18

B as well.

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Naumtsev

6 лет назад, # |

Go no rating...

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sagarthecoder2

6 лет назад, # |

Good problem set.

Happy_minus_Rating

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Oreo_85

6 лет назад, # |

Is it Div-1 contest ?

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Oreo_85

6 лет назад, # |

+18

Is it Div-1 contest ?

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oversolver

6 лет назад, # ^ |

+78

typical Div 1.5

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coding_weeb

6 лет назад, # |

+28

The first A-B-C problems should be B-C-D.

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Gevorkyan_Konstantin

6 лет назад, # |

что-то второй уже раунд немного тяжеловат для див2

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Fulger

6 лет назад, # |

"High ratings to everybody!"

Problems are way harder than usual.

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6 лет назад, # ^ |

+12

I would agree. The problems were way too difficult, especially B and C. Goodbye rating :(

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6 лет назад, # |

← Rev. 2 →

How to solve D? I thought of fixing first position where a[i] < c[i] < b[i] but c[i] can be equal to b[i] and become less later(for example ab bc answer is 1(ba)).

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radoslav11

6 лет назад, # ^ |

DP[position in permutation][is current permutation already grater than a flag (0 or 1)][is current permutation already less than b flag (0 or 1)].

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CP_Learner_

2 года назад, # ^ |

How to calculate dp[position in permutation][1 (current permutation is already greater then a )][1(current permutation already less then b)] ???? If we somehow know the remaining count of each letter then this equal number of permutations of remaining letters, but we don't know the remaining count of each other, So how to calculate this??

Thanks in advance!

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Mahdi_Jfri

6 лет назад, # ^ |

← Rev. 4 →

Consider a function f(stringx, 26 - numbers) which returns number of strings s constructed with characters which is given in the input, that s < x

f(x, 26numbers) can be solved using the first position that you said and the answer is f(b, (a - characters)) - f(a, (a - characters)) - 1

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lolxxx

6 лет назад, # |

-58

the contest was very bad it was like div 1 please dont ever write a contest and please make it unrated because it was bullshit

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sheykh

6 лет назад, # ^ |

I'm agree with you

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kazuma_desu

6 лет назад, # |

Pretest 9 for C?

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szcoder

6 лет назад, # |

← Rev. 2 →

Why my code for E is so slow? http://codeforces.com/contest/895/submission/32688252

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Lucas97

6 лет назад, # ^ |

I think it is because you are always propagating and you only have to do it when it is necessary(i.e. lazy[nodo] != 0)

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szcoder

6 лет назад, # ^ |

Thank You, I will check this.

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Jynx

6 лет назад, # |

what is the 8th pretest for b ? What is wrong in this ? https://ideone.com/dIQKbm

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mtkaya

6 лет назад, # ^ |

You need to think about the case k = 0

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6 лет назад, # ^ |

I was also stuck on the 8th pretest. It has k == 0, which may lead to WA for some programs if you don't take care of the case.

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dragneel3131

6 лет назад, # ^ |

Even I got wrong answer on 8th pretest. I just handled the special case for k = 0 and the pretests passed

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Gigibyte

6 лет назад, # |

← Rev. 2 →

-52

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6 лет назад, # ^ |

+13

You need to choose subarray not subsequence.

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Jynx

6 лет назад, # ^ |

← Rev. 5 →

+17

Because segments have to be continuous

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HBK_Wasi

6 лет назад, # ^ |

It said "continuous" pieces......I didn't catch that "continuous" first and was hacked at 1:30 ..........

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6 лет назад, # ^ |

Why 2^N? I think N^2.

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Angeltapia

6 лет назад, # ^ |

yuShaf sets

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mtkaya

6 лет назад, # |

Is solution of E based on sqrt decomposition?

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6 лет назад, # ^ |

← Rev. 2 →

In E, notice how the update affects any particular value. Assume we store the expected value at each index at current time. Now we have to deal with an update.

Consider any value in the left subarray. Let size of left and right subarrays be S1 and S2 respectively.
final_value = (S1-1)/S1 * previous_value + (sum of values in right subarray)/(S1*S2)

Similar argument can be used for all values in the right subarray as well.
Simulate these operations using a segment tree and lazy propagation.

Code

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oversolver

6 лет назад, # ^ |

what about non independence of values?

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6 лет назад, # ^ |

+11

What do you mean by non-independence? I am using linearity of expectation which does not require any such condition.

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oversolver

6 лет назад, # ^ |

I am fool, i forgot about independence of E...

I thought E(x+y)!=E(x)+E(y) if x and y dependent of each other.

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szcoder

6 лет назад, # ^ |

It's expected value not variance.

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RohanRTiwari

6 лет назад, # |

How did people solve C? Unable to access other submissions as of now.

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6 лет назад, # ^ |

← Rev. 2 →

+14

The number is perfect square iff all prime degrees are even. So you can do dp with bitmask since there is only 19 primes until 70.

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6 лет назад, # ^ |

Didn't know this fact, thanks)

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RohanRTiwari

6 лет назад, # ^ |

I figured as much. Could you give a more detailed description of your dp (states and transition)?

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send_nodes

6 лет назад, # ^ |

← Rev. 3 →

edit:wait nm this is too complicated lol

If some number is prime and >35, then it's only prime is itself.

Otherwise, it can be decomposed into the first 11 primes (up to 31). Build a bitmask for each number where mask[i] = the parity of number of times the i'th prime divides the number.

Now do dp(first i numbers, xor value of j) dp (it's like knapsack). Use the sliding window thing to save memory. This takes care of first 11 primes. Take ans = dp(n,0)

For the rest of the primes, they need to occur an even number of times in a subset. So if there are count(p) occurrences of prime p, p>=37, then you can do this 2^(count(p)-1) ways (it's p choose 0 + p choose 2 + ...). So just multiply this together for each prime p, 37<=p<70, and multiply by res and subtract 1 for the empty case.

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I_Love_CardiB

6 лет назад, # ^ |

I did the exact same fuckin thing, still couldn't pass the 14th test case :(

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6 лет назад, # ^ |

I handled squares and non-squares differently. I even know where my bug is :( but too slow.

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sheykh

6 лет назад, # |

-41

this contest was awful! so sad!

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6 лет назад, # ^ |

+27

Nope! Great problems. I've never failed at B so hard tbh... but C was a lot of fun, only if I could finish my dp in time. Great problems.

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6 лет назад, # ^ |

What is an idea behind your DP approach?

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KieranHorgan

6 лет назад, # ^ |

What were your dp states might I ask?

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6 лет назад, # ^ |

← Rev. 2 →

i'll give you my incomplete code https://ideone.com/BP05Of

The idea is bitmask dp and knowing that choosing odd and even number of items from a collection of size n is just 2^n-1

Please look for "solve", which is the entry point. I know my template is big :|

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6 лет назад, # ^ |

Explanation from AdiZer0 : The number is perfect square if all prime degrees are even. So you can do dp with bitmask since there is only 19 primes until 70.

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6 лет назад, # ^ |

Correct. We want dp[n][mask=0]

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iMurat

6 лет назад, # |

← Rev. 2 →

Hack for the first problem: 5 72 72 73 74 69 Ans: 66

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achaitanya.sai

6 лет назад, # ^ |

Sad! I forgot that "sectors should be continuous" and submitted something useless :(

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Evro

6 лет назад, # |

Somebody please explain div2 A ,as it had the worst explanation ever :/

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KieranHorgan

6 лет назад, # ^ |

Select a continuous subsegment such that the difference between the sum of the selected slices and the sum of the unselected slices is as small as possible.

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Evro

6 лет назад, # ^ |

He Could've wrote subsegment instead of sectors, too misleading

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6 лет назад, # ^ |

The problem gives you N pieces of pizza, with sizes in array 'a'. The sum of all these sizes add up to 360. Now, the problem is asking for the minimum difference between 2 section if we split the pizza into 2 contiguous sections. We can start from position 1 and iterate through, keep track of the sum before called 'SUM'. Since we know the total sum is 360, we can find the sum of the other piece by doing 360 — SUM. We then take ans = min(ans, abs(SUM — (360 — SUM)). The pizza is circular, so we can start from all possible starting positions (1...n) and do the same thing. The answer is then the minimum over all starting positions.

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TooWeak_TooSlow

6 лет назад, # ^ |

There is no any case for this-> if one of sector is not selected ?

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6 лет назад, # ^ |

If one of the sectors is not selected, then we know the sum of one sector is 0 (because not selected), and the other one must be 360. So, the highest answer possible would be 360. We can set our answer to 360 initially to cover this case.

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Diegogrc

6 лет назад, # |

Problems were harder than usual ( specially A and B )

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TooWeak_TooSlow

6 лет назад, # |

← Rev. 2 →

How to solve div 2 problem A and B?

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iMurat

6 лет назад, # ^ |

In problem A you can use dp, or just calculate the minimum sum, by(pseudo) cyclic shift. More in my code: Click

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allllekssssa

6 лет назад, # |

+22

Very interesting tasks, I enjoy a lot of solving this problems !

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6 лет назад, # ^ |

How to solve C?

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6 лет назад, # ^ |

← Rev. 2 →

Note that there are atmost 19 primes less or equal to 70. And do something like dp[current_number][mask of primes you have to get rid of].

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flyman3046

6 лет назад, # ^ |

← Rev. 5 →

the time complexity is 2^19 * 100000 = 5e10. Can it fit the time limit? UPDATED: we can actually do in 2^19 * 70.

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6 лет назад, # ^ |

No. But you can notice that there are atmost 70 distinct numbers and reduce the complexity to 2¹⁹·70.

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6 лет назад, # ^ |

← Rev. 2 →

If you have a number repeated k times in input, then parity is 0 for it's prime factors if the number is chosen even number of times, and parity is odd otherwise.

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skpro19

6 лет назад, # ^ |

Please explain a it more about the dp formulation.

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6 лет назад, # ^ |

Imagine you grouped all numbers by their values (all except 1, we shall consider it separately).

Let mask[i] be a mask where ones denote primes that occur odd number of times in the factorization of i.

The main idea is that a number is a perfect square if all the degrees of its primes are even. So now denote dp[group][mask] as the number of ways to obtain such a mask of odd primes if we use numbers from groups 2 to group.

To make a transition note that if we take an even number of elements from group + 1, the mask doesn’t change. Otherwise the last taken element will change the parity of some primes. Hence out current mask gets xorred with mask[group + 1].

The only observation left is that the number of ways to take either an odd or an even amount of numbers if there are n of them is equal to 2^n - 1.

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allllekssssa

6 лет назад, # ^ |

← Rev. 2 →

For each integer in interval [1,70] you can save mask of prime divisiors ( 19 prime divisiors until 70). After it you can iterate over numbers from 1 to 70 and calculate next DP state : DP [ i ] [ mask ] = amount of subsets with numbers ( 1...i) and current mask of of prime divisiors.

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skpro19

6 лет назад, # ^ |

Can you elaborate it a bit more?

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Azret

6 лет назад, # ^ |

← Rev. 3 →

Let f_i = how many numbers i there are in array. Note that there are 19 primes up to 70. Also, note that in square number number of occurences of every prime must be even. Considering all of that do

dp_{i, parityMask} = how many subsets there are containing only numbers 1 ≤ x ≤ i such that parity of number of occurences of i-th prime is denoted by i-th bit in parityMask. Obviously, dp_0, 00...00 = 1. Figure out the remaining part of solution yourself. Answer to the problem will be dp_{70, 00...00}.

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skpro19

6 лет назад, # ^ |

Can you give some hints for the rest of the solution?

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6 лет назад, # |

Do you think B will pass If I used map instead of index compression? I had map<int,int> which tells me this: if map[a] = b, it means that b numbers have exactly a numbers strictly smaller than them, divisible by x.

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6 лет назад, # |

Do you think my solution for B will pass time limit, I used map<int,int> instead of index compression?

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Taran106

6 лет назад, # ^ |

i also used map.. but i used map<long long,long long>. i'm not sure it will pass or not :(

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6 лет назад, # ^ |

Well, it works in logN. So I think you both do not need to worry about TL

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6 лет назад, # ^ |

My mistake, I also used long long.

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Taran106

6 лет назад, # ^ |

got ac bro :D

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6 лет назад, # ^ |

me too :D

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WolfBlue

6 лет назад, # |

Gah problem B apparently had so many edge cases, I kept getting stuck at the later tests

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T1duS

6 лет назад, # |

What's the hack for B?

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6 лет назад, # ^ |

Handle k = 0 and duplicates in the array.

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RohanRTiwari

6 лет назад, # ^ |

I believe both of them were already there in the sample cases!

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6 лет назад, # ^ |

← Rev. 2 →

I meant duplicates when k = 0, for example the case
8 3 0
1 1 1 3 3 3 3 3

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fmqjpt

6 лет назад, # ^ |

What's the answer?

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6 лет назад, # ^ |

Ans = 9 as you can create 9 pairs using the three 1s.

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6 лет назад, # ^ |

I suppose case like

3 4 0
9 9 10

is better.

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tsuda1627406047

6 лет назад, # ^ |

is the answer 7?

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6 лет назад, # ^ |

Yes, it's 7 with indexes pairs: 1-1, 1-2, 1-3, 2-1, 2-2, 2-3, 3-3.

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code_hard123

6 лет назад, # ^ |

← Rev. 2 →

70 20 50 50 80 90

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T1duS

6 лет назад, # ^ |

not A. I am talking about B.

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code_hard123

6 лет назад, # ^ |

Oh sorry.

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getSchwifty

6 лет назад, # |

PEOPLE ON CODEFORCES ARE GOOD AT SARCASM

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zadrga

6 лет назад, # |

Very hard contest but problems were very interesting.

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HBK_Wasi

6 лет назад, # |

What's the answer of B:- 5 3 1 1 2 3 4 5

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mmmb80

6 лет назад, # ^ |

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mahim007

6 лет назад, # ^ |

Cam i ask how the answer is 9?

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mmmb80

6 лет назад, # ^ |

Good pairs are: (1,3),(1,4),(1,5),(2,3),(2,4),(2,5), (3,3), (3,4), (3,5)

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Sahashoo

6 лет назад, # |

← Rev. 2 →

-23

any idea about this code bug(bug'S)???-problem B-


#include<bits/stdc++.h>
#define F first
#define S second
using namespace std;
const int maxn=1e5+7;
int a[maxn];//,r[maxn],l[maxn];
map<int,int>mp,cnt;
pair<int,int>b[maxn];
vector<pair<int,int> >vec;
int32_t main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n,x,k;cin>>n>>x>>k;
	for(int i=1;i<=n;i++)cin>>a[i],cnt[a[i]]++;
	sort(a,a+n);
	int ans=0;
	for(int i=1;i<=n;i++){
		if(a[i]==a[i-1])vec.back().S++;
		else vec.push_back({a[i],1});
	}
	for(int i=1;i<vec.size();i++)vec[i].S+=vec[i-1].S,b[i]=vec[i];
	for(int i=0;i<vec.size();i++){
		int l=upper_bound(b,b+vec.size(),make_pair(b[i].F+k*x,0))-b-1;
		int r=upper_bound(b,b+vec.size(),make_pair(b[i].F+k*x,0))-b-1;
		ans+=(vec[r].S-(l-1?vec[l-1].S:0)+1)*vec[i].S;
		// cout<<r<<" "<<l<<" "<<cnt[a[i]]<<endl;
	}
	cout<<ans;
}

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HBK_Wasi

6 лет назад, # ^ |

5 3 1\n 1 2 3 4 5 ans is 9.your code prints 31.

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Cydiater

6 лет назад, # |

+18

excellent problem set!

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iMurat

6 лет назад, # |

div1.7-1.8

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6 лет назад, # |

Can anyone help explain why my solution to B times out on case 14? Thanks in advance!

32692235

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WolfBlue

6 лет назад, # ^ |

You have a double for loop from 1 to n, so the code runs in O(N^2) time. Since n = 10^5, you need O(NlogN) or similar. (In general, less than 10^8 or 10^9 operations is safe)

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6 лет назад, # ^ |

May I ask where I have a double for loop? I thought my time complexity was O(NlogN) due to multiset. I may have put a double for loop by accident somewhere though.

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WolfBlue

6 лет назад, # ^ |

Oh, sorry I read your code wrong. Idk

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6 лет назад, # ^ |

It's no problem, thanks for the help!

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6 лет назад, # ^ |

Afaik distance works in o(n) with std::set.

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GGOSinon

6 лет назад, # ^ |

std::distance works in O (N) . So your code works in O (N^2).

P.S. std::distance is same as using 'it++' several times.

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6 лет назад, # ^ |

oh I didn't realize that. That's unfortunate for me. Thanks for the help!

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vb7401

6 лет назад, # ^ |

Yep, std::distance is too costly for a multiset. If you use a vector, sort the values, and use lower_bound and upper_bound, getting the indices of iterators is O(1), rather than O(n) for the multiset!

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BeardAspirant

6 лет назад, # |

what is pretest 8 in prob A. solved nothing today. I calculated two segments with the min difference between them like partition problem but that is apparently not the case.

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gardusig

6 лет назад, # ^ |

I got WA in pretest 8 because I was taking a subsequence (not consecutive values).

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Gigibyte

6 лет назад, # |

+10

How does this contest fit into the pizza of life?

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Asgar

6 лет назад, # |

+34

In problem B statement, did you guys notice that previously it was written find the number of different UNordered pairs of indexes (i, j) which killed me. After the contest, it shows ordered. In the beginning, I realised that the solution would be with lower/upper bounds, but that very statement confused me and pushed me to thinking in terms of self-balancing trees & adding elements from end towards start, which I spent a lot of time and coded. After that in the test case 3, realised that it's not like that :(

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mochii

6 лет назад, # ^ |

+16

Why don't announce this change? NBAH

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6 лет назад, # ^ |

← Rev. 2 →

+13

That's a huge error, they have to address it.

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bira37

6 лет назад, # ^ |

+13

Spent the whole contest in this problem ...

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kabirsahni8

6 лет назад, # ^ |

← Rev. 4 →

+12

It should've been announced.Even I solved the earlier version until i reread it in the last few minutes.

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Dalisyron

6 лет назад, # ^ |

Solving for the number of unordered pairs is no different from ordered pairs since if (i,j) fulfills the condition Ai <= Aj then (j,i) doesn't, unless of course i is equal to j

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Asgar

6 лет назад, # ^ |

+11

No! For unordered pair, (i, j) = (j, i), however, for ordered pairs, (i, j) != (j, i). (of course for distinct i, j's).

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Vevadeus

6 лет назад, # |

Problem B: Is this

Spoiler

complexity optimal?

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ddevbrat

6 лет назад, # ^ |

Yes, That should pass.

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TwentyOneHundredOrBust

6 лет назад, # |

← Rev. 3 →

hmm

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sayedgkm

6 лет назад, # |

What is the idea of problem C?

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GGOSinon

6 лет назад, # ^ |

If we use naive bitmask dp for each prime, it is O (N)*2^19 (because there are 19 primes between 1 and 70) which results time exceed. The idea is if the number is a prime which exceed 35, it must be grouped with a same number. So it can be pre-processed. Since there are 11 primes between 1 and 35, this solution is O (N)*2^11 which gives AC. P.S.Be careful of using % while implmenting it -it gives TLE

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AshishYadav

6 лет назад, # ^ |

GGOSinon How to do bitmask transitions using iterative DP ??? I know only for the Recursive One and thus could not solve it in time.I thought the same :(

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GGOSinon

6 лет назад, # ^ |

Let d [i][j] as a number of way to make state j by using number a[1]~a[i].Let b [i] as a bitmask form of a[i]. We can divide the case : use the b [i] or not. So the recursive formula is d[i+1][j]=d [i][j^b [i]]+d[i][j] ('^' is xor operator) You can implement it by just iterating i and j using 2 'for's to 0~n-1, 0~(2^19)-1. (Initial value is d [0][0]=1, Answer is d [n][0])

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fmqjpt

6 лет назад, # ^ |

← Rev. 4 →

You can also correct this issue by noting that there are at most 70 distinct values in the array you are given, so you just use a frequency array for each of the numbers between 1 and 70 and noting that the number of ways to choose an odd number of numbers out of a set of K numbers is always 2^K - 1 for K > 0.

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AshishYadav

6 лет назад, # ^ |

fmqjpt 70 distinct value of what ???

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6 лет назад, # |

-6

Many people will get rating rise just for solving A :v

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6 лет назад, # ^ |

-7

And that's another reason why this contest was awful.

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6 лет назад, # ^ |

Well, since I'll probably get a rating rise, I'm not gonna say this is awful :)

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6 лет назад, # ^ |

← Rev. 2 →

I only solved A and my rating will go up but that was an awful contest.

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Asgar

6 лет назад, # ^ |

I only solved A and my rating went down so this was an awful contest! :)

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achaitanya.sai

6 лет назад, # ^ |

+17

I think they wanted to justify their statement "High ratings to everybody!" :P

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6 лет назад, # |

A contest where the one can be proud of solving A & B

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the_phoenixx

6 лет назад, # |

Good Questions , i really enjoyed it..hope others too ;-)

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KarimElSheikh

6 лет назад, # |

-11

Bitmasks + some knowledge of Combinatorics is Div.2 C now? What's the world coming to?

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Shihab_Hossain

6 лет назад, # |

For Problem A in

Test case 50:

41 38 41 31 22 41 146 Output: 14

Is it correct? and how Explain please

Thanks :)

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6 лет назад, # ^ |

take the first 41 with the last 146, their sum is 187, the remaining is 360-187 = 173, therefore the difference between the 2 shares is 187-173 = 14

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Shihab_Hossain

6 лет назад, # ^ |

But if i take

series 1: 41 41 41 38 22

series 2:146 31

than the difference between series1 and series2 is 6

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6 лет назад, # ^ |

you must take some numbers which are adjacent to each other, note that in this problem first and last numbers are adjacent too

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Shihab_Hossain

6 лет назад, # ^ |

Thanks :)

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DaysGone

6 лет назад, # |

← Rev. 2 →

can someone please explain for the pizza separation ques why the ans for the below sample test is 120 !!

5

30 60 180 60 30

what should be the ans for this ?? ans can be zero when the 1 continuous sector contains 60 +30 +30+ 60 = 180 (starting from the 2 last 30 to the first 30 in anticlockwise order) and the 2 continuous sector contains only 180 so 180-180=0 (minimal possible difference) But my code fails for this system test case ,it is showing the ans 120 but i think minimal diff 0 can be achieved .. plzz anyone tell me i cant find where i am wrong in logic or thinking ("ANY HELP WILL BE APPRECIATED")

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AshishYadav

6 лет назад, # ^ |

Your code gives 120 . The expected answer is 0 .

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6 лет назад, # ^ |

Your submission 32688201 checks:

A) Situations "FFFFSSSSSSS" for any numbers of Fs and Ss.

B) Situations "FSSSSSSFFFF" for any numbers of Ss and Fs only at the tail.

Your code can't correctly process something like "FFFSSSSFF".

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MaralK

6 лет назад, # |

+46

Good contest, Very good contest, the problems where good and where fun to solve, thank you NBAH for this good contest. But please keep in mind that the problems where too hard for a div2 contest

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Oreo_85

6 лет назад, # |

Can anyone explain why my solution to B gets WA on case 21? 32682296 Thanks.

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MaksymOboznyi

6 лет назад, # ^ |

Input:

5 3 0

2 4 5 7 9

Your answer: 4

Correct answer: 5. (1, 1), (2, 2), (3, 3), (4, 4), (2, 3)

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Romansp

6 лет назад, # |

why here UndefinedBehavior http://codeforces.com/contest/895/submission/32691004 ? but here everything is normal http://codeforces.com/contest/895/submission/32693522 . changed only this

auto x=lower_bound(a.begin(),a.end(),lp);
auto y=upper_bound(a.begin(),a.end(),rp);
ans=ans+y-x;

auto x=lower_bound(a.begin(),a.end(),lp);
auto y=upper_bound(a.begin(),a.end(),rp);
ll p=(ll)(y-x);
ans+=p;

he in int started to count all?

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mivael

6 лет назад, # ^ |

Note that y could be equal to a.end().
std::vector::end() is a special kind of iterator.
I think it is not specified what happens if you add an integer to it (as you do in ans+y part of 32691004).

Regardless, I would use parentheses to avoid unnecessary integer-iterator addition (as well as the subsequent subtraction): ans=ans+(y-x);

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6 лет назад, # |

Albeit I didn't manage to solve D in time, the contest was really cool. The only thing to bother me was the unclear statement of B. Keep it up :D

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gardusig

6 лет назад, # ^ |

I was thinking I was the only one who have read B many times until really understand what to do in this problem

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Graf_Rav

6 лет назад, # |

← Rev. 3 →

Писать автору задач не решился(я ведь просто нубас с 1.2к), поэтому спрошу тут. Разве важен в задаче А порядок кусков пиццы? К примеру инпут 5 90 89 88 87 6. 90+89 образуют ближайший к 180-и угол из возможных, ответ (180-179*)2, то бишь 2. Окей, все понятно, но вот такой инпут: 5 90 87 88 89 6. Тут уже поинтереснее, хотя куски такие же. Если порядок имеет значение, то ответ 6: abs(180-(6+90+87))*2. Но если порядок значения не имеет, то ответ тот же 2. Для интереса проверил несколько скриптов гуру с 1-й странице, все решают линейно, с тем же порядком. Так и должно быть? Если порядок не имеет значения, то все эти решения улетят в пропасть, нужно будет решать по-другому(рекурсивно, например, обходя неюзанные куски, пока сумма <180).

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6 лет назад, # ^ |

Сектора должны быть непрерывны, то есть порядок важен.

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shahidul_brur

6 лет назад, # |

+10

Wow !! This was one of the best contests I have ever participated in <3. All the problems were very , very nice ! I really enjoyed the round as well as the problems.

Take my heartiest love NBAH <3 <3 Hope we will get such awesome problems from you in the future. Keep up the good work man :)

And many many thanks for such nice round and all your efforts in it :)

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qareken

6 лет назад, # |

+12

I have trained for 5 days and became specialist. I am looking forward to next contest

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nonsense

6 лет назад, # |

http://codeforces.com/contest/895/submission/32679304 is judged an error, but works in my compiler?

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6 лет назад, # ^ |

the last line in your while loop should be if(i==n) i=0 not i==0

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Ehsan_sShuvo

6 лет назад, # |

← Rev. 3 →

[update: ACCEPTED] Where is my wrong approach for problem B 32694817

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Lets_chill_tonight

6 лет назад, # |

← Rev. 2 →

Contest : Omae wa mou shindeiru.
Me : Nani?

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meoconn

6 лет назад, # |

delete2/32687180, coutinho/32687203, OutSpace/32688038, blindspot/32689194

LOL, this is 4 photo machines. I was canceled for using the ideone for C. And these 4 people have found, then they copy my code. My fail but I hate their, LOL

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Anushi1998

6 лет назад, # |

Can someone please suggest some cases(possibly small n so that I can debug) for my code (Problem C) http://codeforces.com/contest/895/submission/32711673

It is giving correct answer till n=1000 (in System cases) but I don't know why it is giving WA for n around 10e5

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linkret

6 лет назад, # ^ |

← Rev. 2 →

You missed a prime number, you only have 19, and there are 20 in total.

Edit: nevermind.

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Anushi1998

6 лет назад, # ^ |

We have 19 primes before 70 https://primes.utm.edu/lists/small/100000.txt

Please tell me which one I am missing if I am wrong

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linkret

6 лет назад, # ^ |

My mistake, my code said 20 because it counted up to 75.

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Len

6 лет назад, # ^ |

pw2 array size should be 1e5 not 70.

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Anushi1998

6 лет назад, # ^ |

← Rev. 2 →

Thanks :)

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scorer

6 лет назад, # |

← Rev. 6 →

Can anyone tell me what's wrong with this code?

#include<bits/stdc++.h>
#include <algorithm>
using namespace std;

int main() {
	long long n;
	cin>>n;
	vector<long long> a(2*n);
	for(int i = 0 ; i < n; i++) {
		cin>>a[i];
		a[i+n] = a[i];
	}
	int mindiff = 360;
	for(int i = 0; i < 2*n; i++) {
		int curr = 0;
		for(int k = i; i < 2*n && curr < 180; k++) {
			curr+=a[k];
			mindiff = min(mindiff, 2*abs(180-curr));
		}
	}
	cout<<mindiff<<endl;
	return 0;
}

Running fine on local. :(

Test: #1, time: 0 ms., memory: 1844 KB, exit code: 0, checker exit code: 1, verdict: WRONG_ANSWER Input

4 90 90 90 90

Output -664469436

Answer 0

Checker Log wrong answer 1st numbers differ — expected: '0', found: '-664469436'

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haleyk100198

6 лет назад, # ^ |

← Rev. 2 →

Line 16 should be k < 2*n instead of i < 2*n

Check here: http://codeforces.com/contest/895/submission/32737382

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letmeC

6 лет назад, # |

THIS IS VERY IMPORTANT !!!

I submitted my solution for problem A during the contest and it passed the pretests then I got RTE on test 49 during system test phase, after the contest I resubmitted the exact same code and I got ACCEPTED !!!!

submission during contest : http://codeforces.com/contest/895/submission/32683171 submission after the contest : http://codeforces.com/contest/895/submission/32733925

PLEASE NBAH CHECK THIS PROBLEM !

THANKS.

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VLamarca

6 лет назад, # ^ |

I belive that you accessed invalid position on the array as you used

for (int j=i; j<N+i; j++)

N+i-1 will be outside of bounds since the array has only N positions and, by accessing not allocated memory you get an undefined behviour, which means that anything could happen, even geting AC even though it is wrong. That is actually a very common doubt for begginers, in his last post Mike talked about that. In his words:

"Many Codeforces visitors are already tired of the questions of less experienced participants: "Why does my solution not work on some test on the Codeforces servers, if I locally launch it and it works correctly? You have the wrong compiler/servers!" In 99% of cases this is an example of undefined behavior in a program."

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farmersrice

6 лет назад, # |

One of the most informative rounds ever. Thanks for your efforts!

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