位运算。。。。RevengeofNimIITimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):229   &nbs...

题目： http://acm.hdu.edu.cn/showproblem.php?pid=1863注意有可能出现无法生成树的情况。 #include<iostream>#include<cstring>usingnamespacestd;constintinf=0xffff...

【HDOJ2089】不要62第一个数位dp的题做的老困难了。。。只是好歹是做出来了迈出了第一步。。对大牛来说这样的题都是小caseps:新上一个记忆化方法一些绕弯的题里用dfs好想些代码例如以下:#include<iostream>#include<cstdio>#include<cstr...

ProblemDescriptionVictorhasamachine.Whenthemachinestartsup,itwillpopoutaballimmediately.Afterthat,themachinewillpopoutaballevery w seconds.However,the...

/*hdu1999不可摸数TimeLimit:2000/1000MS(Java/Others)   MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):8274   AcceptedSubmissio...

  刚看题...觉得这不是棵树...可能有回路...仔细一想..这还真是棵树(森林)...这是由于每个城堡所需要提前击破的城堡至多一个..对于一个城堡.其所需提前击破的城堡作为其父亲构图....  dp[k][i]代表以k为跟的子树击破i个城堡所能获得的最大收益...泛化背包问题...

 TimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)ProblemDescription完数的定义：如果一个大于1的正整数的所有因子之和等于它的本身，则称这个数是完数，比...

ProblemDescriptionAsubsequenceofagivensequenceisthegivensequencewithsomeelements(possiblenone)leftout.GivenasequenceX=<x1,x2,...,xm>anothersequenceZ=<z...

/*大菲波数TimeLimit:1000/1000MS(Java/Others)   MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):11168   AcceptedSubmission(s):3...

现场赛的时候错了十四次。。。。。ProblemDescriptionAsanAndroiddeveloper,itˇsreallynoteasytofigureoutanewerversionoftwokernels,becauseAndroidisupdatedsofrequentlyandhasmanybranch...

TimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):17986    AcceptedSubmissio...

TimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):6015AcceptedSubmission(s):4205ProblemDescriptionAsweknow,BigNumberisal...

TimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1373    AcceptedSubmission...

ProblemDescriptionDZYlovesplayingballs.Hehas n ballsinabigbox.Oneachballthereisanintegerwritten.Onedayhedecidestopicktwoballsfromthebox.Firstherandoml...

http://acm.hdu.edu.cn/showproblem.php?pid=2546ProblemDescription电子科大本部食堂的饭卡有一种很诡异的设计，即在购买之前判断余额。如果购买一个商品之前，卡上的剩余金额大于或等于5元，就一定可以购买成功（即使购买后卡上余额为负），否则无法购买（即使金额足够）。...