Using Wladimir's formulas, you have:
F(0) + F(1) + F(2) + F(3) + F(4) + F(5) + F(6) + ... + F(3n-2) + F(3n-1) +
F(3n) = F(3n+2) - 1
F(0) +(F(1) + F(2))+ F(3) +(F(4) + F(5))+ F(6) + ... +(F(3n-2) + F(3n-1))+
F(3n) = F(3n+2) - 1
F(0) + 2F(3) + 2F(6) + ... + 2F(3n) = F(3n+2) - 1
Since F(0) = 0,
I think you're getting TLE because your solution calculate the same stuff
over and over again.
Look at this part:
else{
*ans[s]=find(s+1,len-1);*
if(s[0]'3's[1]'7')
*ans[s] = ans[s]+find(s+2,len-2);*
}
Calculating find(s+1,len-1) imply that
Notice that changing the order of button presses doesn't change anything.
Notice also that pressing a button twice is the same as not pressing it.
So, in the end, given N, there are at most 2^4 = 16 final configurations
(each button is either pressed once or not).
I hope this will help you.
--
The problem statement doesn't make it clear, but I guess one can rotate the
envelopes (i.e. exchange x_i and y_i).
I mean, in the example case of {(1,2),(2,13),(5,10),(7,9),(9,1)}, isn't
(1,2),(1,9),(5,10) a valid solution (with length 3), instead of the ones
shown (with length 2)?
On Sat, Oct
