Difference between revisions of "2012-2013 ACM-ICPC Northeastern European Regional Contest (NEERC 12)"

From EOJ Wiki
Jump to navigation Jump to search
Line 13: Line 13:
 
== Problem B ==
 
== Problem B ==
  
Unsolved.(-16)
+
Upsolved by oxx.(-16)
  
 
题意:
 
题意:

Revision as of 08:24, 26 August 2018

Neerc 12

  • 某三题题解[[题解][1]]

Problem A

Solved by oxx1108. 00:34

题意:

题解:

Problem B

Upsolved by oxx.(-16)

题意:

题解:

Problem C

Solved by Xiejiadong.02:53(+9)

题意:给每个人分配一个相同的长度,使得每个人可以在自己的区间里面找出互不重叠的区间。

题解:用long double来二分答案,然后暴力枚举分母,找出分数的表示。

需要注意精度问题,wa了好几发。

Problem D

Unsolved.

题意:

题解:

Problem E

Solved by oxx1108.02:29(+1)

题意:

题解:

Problem F

Unsolved.

题意:

题解:

Problem G

Solved by dreamcloud.02:01( +1(by oxx) )

题意:[0,n]中有多少个数满足k进制下,和-k进制下表达一样

题解:转换一下题意,就变成了一个数取k进制之后,$a_m,a_{m-1},……a_3,a_2,a_1,a_0$,要满足$a_m,,……,0,a_2,0,a_0$,即奇数位一定是0,数位dp一下。

Problem H

Solved by dreamcloud.01:34

题意:给你一个字符串s,求它有多少个子序列,满足重构之后是回文串。(重构指的是将子序列中的字符任意交换位置)

题解:子序列重构是回文,其实就是最多只有一个字母出现次数是奇数。利用hash异或和来维护前缀中每个字母出现的次数为奇还是为偶。从前往后扫,要么每个字母都是偶数,要么枚举一个字母出现奇数,只需要看前面该状态有多少个。类似于EOJ七月月赛。

Problem I

Unsolved.

题意:

题解:

Problem J

Solved by Xiejiadong. 03:59(+1)

题意:从0出发,用$a$次$\pm 1$,$b$次$\pm 2$,$c$次$\pm 3$遍历所有的$0$到$a+b+c$点。

题解:先把$c$次$\pm 3$的处理掉。我们通过向右、向左、再向右三部分,把前面的一部分填满,并且最后填的一个的左边已经全部填满,右边全部没有填过

再向右一步一步填,剩下一个$\pm 1$用来在最后改变方向,剩下的$\pm 2$我们用过向右、转弯、向左来填满右边剩下的部分

注意各类情况的处理与细节问题

Problem K

Unsolved.

题意:

题解:

Problem L

Unsolved.

题意:

题解: