Hdoj HDU 1847 good luck in CET-4 everybody! ACM 1847 in HDU

Miyu original, post Please note: Reprinted from __________ White House

Question address:
Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1847
Description: Problem description
The CET-4 is coming soon. Are you busy reviewing? Maybe even the short term ACM has no time to practice. I know Kiki and Cici. Of course, Kiki and Cici, as contemporary college students infiltrated more than a dozen years in the test room, better understand the relaxation before the test. The so-called "Zhang Chi youdao" means this. No, Kiki and Cici have to play cards for a while before taking a rest every night to relax their nerves.
"Upgrade "? "Double buckle "? "Red Five "? Or "Landlords "?
Of course not! That's cool. ~
As a computer school student, Kiki and Cici did not forget their major when playing cards. The rules for playing cards are as follows:
1 , A total of N cards;
2 The two sides take turns to capture cards;
3 The number of cards each time can only be 2 power times (I .e: 1 , 2 , 4 , 8 , 16 ...)
4 After the cards are captured, the winning and losing results also come out: the people who finally caught the cards are the winners;
Suppose Kiki and Cici are both smart enough (in fact, you don't have to assume that there are unintelligent students ~ ), And Kiki is the first player each time. Who can win?
Of course, no matter who wins the cards, it's not a problem. What's important is the coming CET. - 4. Good status.

Good luckInCet-4Everybody!
 

Input
The input data contains multiple test cases. Each test case occupies one row and contains an integer N (1<=N<=1000).
 

Output
If Kiki can win, output "Kiki"; otherwise, output "Cici", which occupies one line for each instance.
 

Sample Input
1
3
 

Sample output
Kiki
Cici

Question Analysis:Miyu original, post Please note: Reprinted from__________ White House

First of all, we can think of a defeat in the face of 3. Whoever faces 3 will lose no matter how much! <--- This is the key
Then try to cause such a situation to the other party, because any number plus 1 or 2 that is not a multiple of 3 can become a multiple of 3,
Similarly, subtraction of 1 or 2 can also be a multiple of 3. That is to say, if the current number is not a multiple of 3, then I can certainly take it.
Take it as a multiple of 3. For example, if it is 11 now, I will take 2 and change it to 9, which leads to the situation where the opposite party is a multiple of 3.
If the other party takes m, I can take 1 or 2 to make the total number of rounds into a multiple of 3, so there are two situations:
1. Just finished.
2. The remaining three are multiples. Continue;
So it will win

CodeAs follows:Miyu original, post Please note: Reprinted from__________ White House

# Include < Iostream >
Using   Namespace STD;
Int Main ()
{
Int N;
While (CIN > N)
{
Puts (n %   3   ! =   0   ?   " Kiki " : " Cici " );
}
Return   0 ;
}