http://en.wikipedia.org/wiki/Digital_root
As S(n)=n-9 * ⌊n/9⌋
=>
S(n) mod 9 = n mod 9
And S(n) = (N1+N2+…Ni+…)  (Ni means the digit of n) 
=>  
dr(n)=n mod 9=S(n) mod 9=(N1+N2+…Ni+…) mod 9 (S(n) mod 9 ≠0), 
dr(n) = 9 (S(n) mod 9 = 0)
Then we can get the algorithm (n ≠0)
For Ni:N
	tmp = (Ni + tmp) mod 9
if tmp equals 0
	dr(n) = 9
else
	dr(n)=tmp