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.

(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

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

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!

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?

Very good problem :)
Short and beauty.