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Re:计算((p^(eB+1)-1)/(p-1))%MODIn Reply To:Re:计算((p^(eB+1)-1)/(p-1))%MOD Posted by:mxd543 at 2009-09-16 21:58:04 > > 我感觉(rem-1+MOD*(p-1))%(p-1)不一定为0. 一定的。p^(eB+1)-1=kMOD(p-1)+rem;两边对p-1取模。这里的rem=(p^(eB+1)-1)%MOD(p-1) Followed by: Post your reply here: |
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