I have a non linear recurrence relation a(n)=(n-1)*(a(n-1)+a(n-2)).How can I calculate a(n)%MOD in O(log(n)) time? Base conditions:-a(0)=0,a(1)=1
I have a non linear recurrence relation a(n)=(n-1)*(a(n-1)+a(n-2)).How can I calculate a(n)%MOD in O(log(n)) time? Base conditions:-a(0)=0,a(1)=1
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This is basically a recurrence for n!. (The base conditions are not mentioned but even with those, the solution will not be very different). And I dont think there is a simple way to do it in sub-linear time. :(
Actually, the base conditions matter a lot. For example, if a0 = 1, a1 = 0, then it's a recurrence for !n (number of derangements).
Xellos,since you are saying that base condition matters a lot,I take a(0)=1 and a(1)=0 .Can I calculate the recurrence relation in O(log(n))?
Maybe there's some obscure algorithm that does it, I don't know. But for the purpose of solving contest problems, assume you can't.
Sorry! By solution I meant "solving the task in logN time", because there is at least one base condition in which solution takes linear time. I guess I messed up with the wordings :P