"Recycled numbers" challenge
f=open('C-large.in')


Don't sue single letter variable names, its really hard to read

    noi=int(f.readline().rstrip('\r\n'))


Since the default is already to remove whitespace, why both with the \r\n parameter?

    i=1
    ou=open('jam.out','w')
    while i&lt;=noi:


Use for loops, not while loops to count

      re=f.readline().rstrip('\r\n')


re is a really bad name because its the same as the regular expression module

      rs=re.split()
      a=int(rs[0])
      b=int(rs[1])


I'd use a,b = map(int, re.split())

      ans=0
      c=[x for x in range(a,b+1)]


Use c = list(range(a,b+1))

      for ind,x in enumerate(c):
          for y in c[ind+1:]:


That's gonna be inefficient. Firstly, you can easily express it using simple ranges

for x in range(a, b+ 1):
    for y in range(x + 1, b + 1):


and therefore avoid the slices. 

But you can do much better. You are only interested in values of y which are recyclings of x. Instead of generating the rather large number of possible second numbers, try just generating the fairly small number of possible ways to recycle x. 

              n=str(x)
              m=str(y)


Converting all your numbers to strings will be expensive. See if you can keep them as integers and just do math operations on them

              if len(n)==len(m) and len(n)&gt;1 and len(m)&gt;1:
                 ki=1 
                 while ki &lt;len(n):


Again, use for loops to count

                     temp=n[-1:-(ki)-1:-1][::-1]+n[:-(ki)]


temp should be banned variable name. All variables are temporary, use a more helpful name.

                     if temp[0]!='0':
                      if temp==m:
                         ans+=1
                         break
                     ki+=1   
              else :
                  break



      print("Case #%s:"%(i),ans,file=ou)


Normally, we use %d for numbers

      i+=1
    ou.close()
    f.close()