摘要:
dpdp( x , k ) = max( dp( x - 1 , k - 1 ) + *** , dp( x - 1 , k ) + *** ) *** = 0 or 1 ,根据情况(BZOJ 1750双倍经验)--------------------------------------------... 阅读全文
posted @ 2015-06-25 20:20
JSZX11556
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摘要:
LCT..--------------------------------------------------------------------------------#include#include#include#include#define rep( i , n ) for( int i =... 阅读全文
posted @ 2015-06-25 14:15
JSZX11556
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