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| Показать все ветки Спрятать все ветки Показать все сообщения Спрятать все сообщения | | TO ADMIN (Problem 1456) | cash | 1456. Джедайский ребус 2 | 4 авг 2016 14:54 | 5 | I have "Time Limit" in test #4.
Can you help me.
[code deleted]
Edited by moderator 01.06.2006 16:14 Of course you got TLE...
you algo is O(n）and certainly will get TLE Real Help!
First of all we must diminish number of candidate
to optimal.
MathHelp:if (A^k)%N==1 => fi(N)%k==0 where
fi(N)-Eiler function of N
Try find in Internet effective algorith for fi(N).
we can see that number of candidates diminished from
1000000000 to 2*sqrt(1000000000)~64000
Also if Nod(A,N)>0 print 0.
Final trick is using divide recursion
when calculating (A^K)%N=>O(log K)-time
and don't forget use __int64 when form A*B
Edited by author 17.08.2007 02:26 Good point svr! :) But number of ways diminishes to amount of divisors of phi(N) which is way less than 64000, and it's enough to factorize phi(N) itself to get them all recursively.
Edited by author 20.08.2008 17:46 NOD is russian for GCD :-) | | Some Tests | Lev_Sky | 1456. Джедайский ребус 2 | 7 июл 2012 22:30 | 1 | Give some some tests ,please
Edited by author 07.07.2012 22:30 | | WA 8 | wiza | 1456. Джедайский ребус 2 | 11 окт 2011 13:17 | 1 | WA 8 wiza 11 окт 2011 13:17 Please help. What is the input? | | to Admin | 12_liz | 1456. Джедайский ребус 2 | 18 авг 2011 01:30 | 1 | I have "Wrong answer" in test 4,
but I don`t understand where my
mistake.Can you help me? Please,
show me test 4 or something like.
Edited by author 18.08.2011 23:48
Edited by author 19.08.2011 15:36 | | Test | Denis Koshman | 1456. Джедайский ребус 2 | 25 сен 2008 16:45 | 3 | Test Denis Koshman 20 авг 2008 18:00 131313213 453535
Result: 14860 by both way (O(ln n) and O(n) ) result is 112562728 453535^14860 (131313213) !=1 !!! | | Help me, what algorithms against a limit time the Test ¹ 4 operates??? | Barselona | 1456. Джедайский ребус 2 | 13 апр 2007 11:27 | 1 | Help me, what algorithms against a limit time the Test №4 operates???
PLEASSSSSSSSS!!!! | | Something reminded me finding a square root modulo N | SPIRiT | 1456. Джедайский ребус 2 | 9 июн 2006 17:27 | 1 | A^X=1 mod N;
if X is even, we can find quickly A^(X/2) (watch problem 1134).
If X is odd, we have to solve A*A^(X-1)/2=1 mod N. Therefore we have to find inverse number for A. And so on. Well, now, it's a fast quick recursion checking two cases...
Gonna check that soon... |
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