考虑
A(k+1)-A(k)
A(k) 去掉第k位后2008位数的奇数位减去偶数位的和。。
可以化为全部由0或1组成
本帖于 2009-07-08 17:46:53 时间, 由普通用户 康MM 编辑
考虑
A(k+1)-A(k)
A(k) 去掉第k位后2008位数的奇数位减去偶数位的和。。
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