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其实这题也可以解模方程求解，如果题目的要求变成<10000的所有数那用暴力会TLEIn Reply To:正点, 11...11(n个1)的余数=11..1(n-1个1)的余数*10+1 Posted by:tzkq at 2008-12-18 20:50:36 Followed by: Post your reply here: |
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