I tried many times and I got TLE7, on code with cin cout, and the visual studio compiler. BUT with this input only 5 million numbers. It should not give TLE (I used ios and tie), which is more strange in g++, ABSOLUTELY the same code gives AC, with a time of 0.187, the code is MORE THAN 5 TIMES FASTER! I understand that g++ and visual differ a lot in terms of compiler settings, but to have such a big difference, I'm just shocked. I am very interested to find out why this is happening. Usually visual is faster, it is very interesting to find out which settings and in which cases make the compiler faster.

you don't understand what your code does at all
try to think more

try to reproduce the out in sample

I just use the scanf to read the input data. Why do I also receive a TLE response?

Edited by author 11.06.2015 18:48

Edited by author 11.06.2015 18:48

#define I(x) int x; cin >> x
#define F(i,n) for(int i = 0; i < (n); ++i)
int main(){
  cin.sync_with_stdio(false);
  I(n); I(k); I(m); cout << "YES" << endl;
  F(i,k){
    I(s); F(i,s){ I(x); //...

Уважаемые администраторы сайта!

Подскажите, пожалуйста, что Вы собираетесь делать с посылками, которые получали AC на компиляторе Intel C++ 7, и получают Time Limit на компиляторах G++ 4.7.2 и Visual C++ 2010?

У меня, например, около 4-5 таких задач было, пришлось все их перерешивать по-другому.
Ведь теперь нет возможности отправить задачу на старом компиляторе.

Извиняйте, но тогда нужно перетестировать посылки под Intel C++ 7 на новых компиляторах, а то результаты и рейтинг получаются НЕЧЕСТНЫМИ !!! Ведь не все перерешивают уже "сданные" задачи.

 If there whould be no cicles then a topological sort would do the
job but I can't figure out how to solve the problem ...

(a+b)%c!=a%c where c>b
(a+b)%c!=(a+p)%c where c>b&&c>p

cute problem. but how could someone ac in 0.046 sec? mine was 0.625
pls share faster idea of solution

Very good problem :)
Short and beauty.

As i've understood none of the sets with equal number of good can be similar.

But with this assumption, i've got WA1 all the time.

Help!

1. First of all it seems that YES alltimes;
2. Secondly evident that we should find partition
of set of all possiible multisets of products not only given K~50000;
3. Let P(i,A)- predicate for i-th subdivision of the partition. 1<=i<=N+1.
Eqution P(i,A)==1 must become wrong if we add any element to A or raplace any element in A.
It seems that |A| mod(N+1) has this stable property.

Nice problem...

Edited by author 08.07.2007 21:56

Why "1 2 3 4" is similar to "1 2 3 4 2" ans is not similar to "1 1 2 3 4". The first set is obtained from these two by deleting "2" or "1" respectively. What do I misunderstand?

This problem seems to be very strange. The graph theory sais that it always has a solution. But I think, that any algorithm works slower than O(K^2) and for K=50000 it is TLE. The O(K) algorithm (like Locomotive's one) is obviously incorrect, because gives the wrong answer even in a sample input.

(The first two sets in sample input: 1 3 5 6 4 and 1 3 5 6 3 are similar, but his program sais that I must send these sets to shop №1.)

The problem sais that two sets are "similar" if one of them is obtained by deleting one good form the second set or by replacing one good to another. But if the similar sets can be obtained only by deleting one good, the Locomotive's solution is correct.

Maybe the problem text is incorrect, or there is no tests with K=50000 (so I can solve it, finding the minimal number of colors to paint this graph in O(K^2)).

Who can tell me, why I am wrong?

Edited by author 07.08.2004 11:57