Sum of Fibonacci numbers

GolfScript 28 26 characters

Since the problem is here we might as well golf it. Anyone who plan on participating on spoj.pl better close their eyes while they read this.

~](;{11.@5+{.@+}*;.10%+n}%

It basically says: For all input, start with the numbers 11 and 11, do input+5 Fibonacci iterations, discard the highest of the two results, add itself mod 10 to the lowest result, done. As a formula this can be describes as 11*Fib(input+6) + (11*Fib(input+6)) mod 10.

Why does this work? It is simply a condensed way of calculating two second identities in the same run, one could start at [0 1] and [55 89], do a run on both of the same length and subtract the first result from the second to get the sum of a series of 10 Fibonacci numbers, but one may as well do the subtraction on the initial sets, thus getting the set [55 88], that can be stepped back a few steps to [11 11] which is short to write.

The Fibonacci series mod 10 has a period of 60, so Fib(i+246) mod 10 = Fib(i+6) mod 10.

Python, 99 98 93 characters

F=[0,1]
exec'F+=[F[-2]+F[-1]];'*300
exec'G=F[input():];print sum(G[:10])+G[246]%10;'*input()

Perl, 78

sub F{$c=shift;$c>1?F($c-2)+F($c-1):$c}$_=<>;$x=F($_+11)-F($_+1);print$x+$x%10

This makes use of my observation that

• the sum of F(i) to F(i+9) is equal to F(i+11) − F(i+1) — proof:

     F(i) + F(i+1) + F(i+2) + F(i+3) + F(i+4) + F(i+5) + F(i+6) + F(i+7) + F(i+8) + F(i+9)
  =      F(i+2)    +     F(i+4)      +     F(i+6)      +     F(i+8)      +    F(i+10)
  =      F(i+2)    - F(i+3) + F(i+5) +     F(i+6)      +     F(i+8)      +    F(i+10)
  =           -F(i+1)       +      F(i+7)              +     F(i+8)      +    F(i+10)
  =           -F(i+1)       +              F(i+9)                        +    F(i+10)
  =           -F(i+1)       +                           F(i+11)
  

• F(i+246) mod 10 is equal to (F(i+11) − F(i+1)) mod 10 (discovered by experimentation; no idea how to prove this)