[Tutor] Is there a better way to do this? (original) (raw)
jb jb at riseup.net
Thu Jul 22 03:45:15 CEST 2004
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On Wed, Jul 21, 2004 at 09:22:26PM -0400, Hee-Seng Kye wrote:
I'm trying to write a program that computes six-digit numbers, in which the left digit is always smaller than its following digit (i.e., it's always ascending). The output of the program starts with "0 1 2 3 4 5" and ends on "6 7 8 9 A B." The best I could do was to have many embedded 'for' statements:
c = 1 for p0 in range(0, 7): for p1 in range(1, 12): for p2 in range(2, 12): for p3 in range(3, 12): for p4 in range(4, 12): for p5 in range(5, 12): if p0 < p1 < p2 < p3 < p4 < p5: print repr(c).rjust(3), "\t", print "%X %X %X %X %X %X" % (p0, p1, p2, p3, p4, p5) c += 1 print "...Done" This works, except that it's very slow. I need to get it up to nine-digit numbers, in which case it's significantly slower. I was wondering if there is a more efficient way to do this.
this version is a bit quicker:
c = 1 for p0 in range(0, 7): for p1 in range(p0+1, 12): for p2 in range(p1+1, 12): for p3 in range(p2+1, 12): for p4 in range(p3+1, 12): for p5 in range(p4+1, 12): print repr(c).rjust(3), "\t", print "%X %X %X %X %X %X" % (p0, p1, p2, p3, p4, p5) c += 1 print "...Done"
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