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简单证明

Posted by whd0810 at 2008-07-31 21:05:40 on Problem 1989

// | ... a1| | ... a2| ... |  ac|| ... | 
// 后面的| ... | 是没有完整的 1 - k  
// 统计前面出现完整的 1 - k 的个数 c , c + 1 就是答案
// 设每一个完整的1 - k 以 ai结束( 1 <= i <= c )
// 假设最后一个缺少  aj , 则无法构造 a1 a2 ... ac aj

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Re:简单证明 (12) scaneeling 2008-10-03 13:52:15

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