Actually, that was China too. :|

When I was in America, I used Facebook, Twitter, and many other social networks that Americans are using.

Usually the first thing I feel when I open either online web pages or their mobile apps is that the UI is simply ugly, untidy and not easy to read. And when I look deeper into them, I found the logic of menus are usually not user-friendly. Company as large as Facebook spent too much space of their app menus to show the privacy settings or other. Are these settings necessary? Maybe. But Facebook puts them their to avoid law suit rather than help the users to achieve some useful function. (Sometimes few people like to sue a company who provides them free service for the service is not good enough lol)...

Regarding the mobile app. I noticed that Facebook and Twitter consume a lot of data but providing no more information than WeChat and QQ. I think they just don't care about optimizing app performances. The iOS version of Quora is another example. Is it really necessary to refresh the entire frame every time I move into the detail page of a question and move back to main feed? The app works like a web broswer rather than an app. It is slow and waste too much data.

As a person who uses Chinese apps more often, I have to say that the user-experience of these popular International social apps s-u-c-k...Forgive me for bad words. But before I explore the content, the web design and app performance are already pushing me away.

Once I look into the content. I feel that there is nothing more interesting or more important as well. I can not even come up with an example about the important and valuable information I acquired from Facebook (Trump is winning? China does not have democracy? lol). Even my American classmates do not really like using Facebook any more...

In terms of functions. Alipay and Wechat Pay in China has achieved I believe everything Apple Pay are still trying to achieve. We scan our phones to check out in restaurants, super markets. We order tickets, hotel rooms and we call cabs with one finger, in one app...

As a whole, I do not see any significant advantages Facebook or other social networks have over convenient and beautiful local apps...If there is one, maybe arrogance and ignorance? Because they even believe that the reason why they do not have too many Chinese users is that China Government bans them...

Maybe they will experience a user number explosion at the first several weeks of cancelling the banning order. But everything probably just returns to the beginning after all.

R E K T

Lucky jqdai0815... #:-s

Unlucky me , didn't get in the same room maybe i'll take my Great Revenge later ,,,, time pass and revenge grow up

In case you have not locked. Otherwise, you feel bad due to locking in case of hack.

Is there any reason to believe this person is a 'she'?

angel is called farishta

Exactly, India was once part of Persian Empire and thus origin of a lot of Indian words is Iran, not India.

you're right , that's weird

in div2 rounds I hack others , but in div1 I'm the one who will be hacked hahaha

maybe :p

will be to be accurate.

Two numbers x and y have the same remainder modulo c[i]'s if and only if abs(x-y) = lcm(c[1], c[2], ... c[n])

Now if x mod k and y mod k are not the same, then there will always be ambiguity for Arya. x mod k and y mod k will be the same iff lcm(c[1], c[2], ... c[n]) % k is 0 since the differ by this amount (or a multiple of this amount)

If LCM of c1,c2..cn is not divisible by K, then (x+ m*LCM) (m= any +ve integer), would give different remainder for K, but same for c1,c2..cn as m is increased. Hence, not possible to uniquely determine remainder with K

Here is another view on the problem and explanation.

Any set of distinct numbers C (e.g. c = {3, 4, 6} as below) "generates" unique remainder sets for increasing positive numbers X with Period = LCM(c1, c2, .. cn). Those unique remainder-sets allow to uniquely distinguish remainders x mod k. E.g. remainders row "2 1 5" always correspond to (x mod k) == 5, or "1 2 4" to (x mod k) == 10. So to have things unique CycleLength = LCM(C) needs to be "syncronized" with K, more presicely it needed CycleLength % K == 0.

Out of that two solutions come:

1. Find LCM and check if it is divisible by K. If true "Yes", otherwise "No". Solution #1

2. Factorize K to prime powers. E.g. factorize K=108 to 4 * 27, or K=12 to 3 * 4. If EACH of that prime powers divides any number from C-set — the answer is "Yes", otherwise "No". In fact it has the same meaning as in "1": if each prime power from K can be found in some subset of C numbers then => (c1*c2*c3*..cn) % K == 0, which leads to LCM(c1,c2,c3..cn) % K == 0. Solution #2

k = 12, c : {3, 4, 6}
  x = 01, rems: 1 1 1 , i mod k = 1
  x = 02, rems: 2 2 2 , i mod k = 2
  x = 03, rems: 0 3 3 , i mod k = 3
  x = 04, rems: 1 0 4 , i mod k = 4
  x = 05, rems: 2 1 5 , i mod k = 5
  x = 06, rems: 0 2 0 , i mod k = 6
  x = 07, rems: 1 3 1 , i mod k = 7
  x = 08, rems: 2 0 2 , i mod k = 8
  x = 09, rems: 0 1 3 , i mod k = 9
  x = 10, rems: 1 2 4 , i mod k = 10
  x = 11, rems: 2 3 5 , i mod k = 11 !!! Cycle end.
  x = 12, rems: 0 0 0 , i mod k = 0  !!! New cycle iteration.
  x = 13, rems: 1 1 1 , i mod k = 1
  ...
  x = 22, rems: 1 2 4 , i mod k = 10
  x = 23, rems: 2 3 5 , i mod k = 11
  x = 24, rems: 0 0 0 , i mod k = 0

We are interested only in divizors of K. Other numbers give no information about X mod K (by Chinese Remainder Theorem). Therefore with GCD we will not interesting divizors.

Will solution with bool array get TLE?

Thanks!