I remember solving this @ spoj Here is an O(1) solution
#!/usr/bin/python
def solve(n):
val=1
for i in range(1,9):
val*=(n+i)
return float((n+9)/9.0-(40320.0/val))
cases=int(raw_input())
while(cases):
cases-=1
n=int(raw_input())
print '%.6f'%(solve(n))
On Tue, Jan 25, 2011 at 6:28 PM, juver++ <avpostni...@gmail.com> wrote:
> @Skywalker your solution is ok. But is works only for the small value of n.
> Cause amount of desired numbers with n=10^6 digits is very big ))
> After n=27 there is a regularity for the ratio.
> However, here is more simplified dp - http://codepad.org/9bzFzDtV
>
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