Numbers on a Chain
Bash+coreutils, 197 bytes
for i in $(eval printf '%s\\n' ${2//_/{0..9\}}|grep -vP '(\d).*\1');{
for((f=d=0;d<${#i}-$1;d++));{
((${i:d+1:1}==0||10#${i:d+1:$1}%${i:d:1}))&&f=
}
[ $f ]&&echo $i&&((c++))
}
((c))||echo no answer
Output:
$ ./chain.sh 3 714_
7140
$ ./chain.sh 2 7141
no answer
$ ./chain.sh 2 14208
no answer
$ ./chain.sh 3 14208
14208
$ ./chain.sh 2 1_208
no answer
$ ./chain.sh 3 1_208
14208
$ ./chain.sh 2 6__2__4508
no answer
$ ./chain.sh 3 6__2__4508
6132794508
$
Explanation
- The parameter expansion
${2//_/{0..9\}}
replaces all underscores with{0..9}
. - The resulting string is
eval
ed to expand all these brace expressions. - The
grep
weeds out all possibilities where there are any repeated digits. - Then each remaining number is checked, digit-by-digit for conditions 1 and 2.
Mathematica Ruby, 349 224 229 bytes
n=$*[0].to_i
r='no answer'
(?0..?9).to_a.permutation($*[1].count'_'){|q|s=$*[1]
q.map{|d|s=s.sub'_',d}
c=s.chars
(t=1
c.each_cons(n+1){|c|e=c.shift.to_i
(t=!t
break)if e<1||c[0]==?0||c.join.to_i%e>0}
(r=s)if t)if c==c.uniq}
$><<r
This is a very naive implementation. I count the number of underscores, and then simply create a list of all digit-permutations of that length, to brute force every possible combination. This will perform horribly for larger numbers of underscores, but this is code golf and not fastest-code. :)
Edit: Ported this from Mathematica. See the edit history for the original version.
Edit: Fixed leading underscore cases.
Python - 239 267
from itertools import*
T=raw_input()
n=int(T[0])
N=len(T)-2
J=''.join
for i in permutations('0123456789',N):
if all([S in[I,'_']for S,I in zip(T[2:],i)])*all([i[j]>'0'<i[j+1]and int(J(i[j+1:j+n+1]))%int(i[j])<1for j in range(N-n)]):print J(i);exit()
print'no answer'
Slow, but short. Simply compare every possible N-digit permutation with the given pattern and check all requirements. I've tested it only with 7 or 8 digits. Should work for 9 or 10 as well, but will take quite a while.
Edit: I added the missing default output "no answer".