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发信人: kaman (天外飞仙), 信区: ACMICPC
标 题: 一道题
发信站: 荔园晨风BBS站 (Sat Apr 17 12:03:42 2004), 站内信件
的确是简单题，但要看所用的策略，怎样在最少的时间里做出来
Runaround Numbers
Runaround numbers are integers with unique digits, none of which is zero
(e.g., 81362) that also have an interesting property, exemplified by
this demonstration:
If you start at the left digit (8 in our number) and count that number
of digits to the right (wrapping back to the first digit when no
digits on the right are available), you'll end up at a new digit (a
number which does not end up at a new digit is not a Runaround Number).
Consider: 8 1 3 6 2 which cycles through eight digits: 1 3 6 2 8 1 3
6 so the next digit is 6.
Repeat this cycle (this time for the six counts designed by the `6') and
you should end on a new digit: 2 8 1 3 6 2, namely 2.
Repeat again (two digits this time): 8 1
Continue again (one digit this time): 3
One more time: 6 2 8 and you have ended up back where you started, after
touching each digit once. If you don't end up back where you started
after touching each digit once, your number is not a Runaround number.
Given a number M (that has anywhere from 1 through 9 digits), find and
print the next runaround number higher than M, which will always fit
into an unsigned long integer for the given test data.
PROGRAM NAME: runround
INPUT FORMAT
A single line with a single integer, M
SAMPLE INPUT (file runround.in)
81361
OUTPUT FORMAT
A single line containing the next runaround number higher than the input
value, M.
SAMPLE OUTPUT (file runround.out)
81362
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