Difference between revisions of "ECNU XCPC Team Training Round"

From EOJ Wiki
Jump to navigation Jump to search
(Created page with "== Problem A == Unsolved. == Problem B == Solved by yanghong. 04:34 (+) 类似数位dp的想法: 从高位到低位枚举, 如果小于边界则直接计算答案, 等...")
 
 
(15 intermediate revisions by the same user not shown)
Line 5: Line 5:
 
== Problem B ==
 
== Problem B ==
  
Solved by yanghong. 04:34 (+)
+
Solved by Once. 00:09 (+)
 
 
类似数位dp的想法: 从高位到低位枚举, 如果小于边界则直接计算答案, 等于边界则固定此位置的值然后去低位计算
 
  
 
== Problem C ==
 
== Problem C ==
  
Solved by bingoier. 00:35 (+)
+
Unsolved.
 
 
质因数分解后发现答案为每一个质数的指数 $k,((k+2)*(k+1)/2$ 的乘积之和
 
  
 
== Problem D ==
 
== Problem D ==
  
Unsolved.
+
Solved by Once. 01:31 (+)
  
 
== Problem E ==
 
== Problem E ==
  
Solved by bingoier. 01:17 (+1)
+
Solved by Once. 00:42 (+)
 
 
模拟题,将时间转化为秒数可以简便运算
 
  
 
== Problem F ==
 
== Problem F ==
  
Solved by yanghong. 01:06 (+)
+
Unsolved.
 
 
计算树上路径交
 
  
 
== Problem G ==
 
== Problem G ==
  
Unsolved.
+
Solved by bingoier. 03:52 (+4)
  
 
== Problem H ==
 
== Problem H ==
  
Solved by Once. 03:08 (+2)
+
Solved by yanghong. 02:16 (+)
 
 
大概可以猜测一下,一个积木盘只有绕一圈之后才会出现一个轮换。所以我们就暴力把原来的积木标号之后转一圈,得到所有轮换,求出轮换的最小公倍数就是答案。然而模数不是素数,所以答案需要处理出所有的质因子后暴力相乘。
 
 
 
不过还有一个结论感觉很对但是不会证明,就是经过一个轮换之后积木形状相同。
 
 
 
时间复杂度 $O(n^2)$。
 
  
 
== Problem I ==
 
== Problem I ==
Line 51: Line 37:
 
== Problem J ==
 
== Problem J ==
  
Solved by bingoier. 00:07 (+)
+
Solved by yanghong. 01:09 (+)
 
 
签到,将题面上的代码复制过来即可
 

Latest revision as of 08:08, 23 December 2020

Problem A

Unsolved.

Problem B

Solved by Once. 00:09 (+)

Problem C

Unsolved.

Problem D

Solved by Once. 01:31 (+)

Problem E

Solved by Once. 00:42 (+)

Problem F

Unsolved.

Problem G

Solved by bingoier. 03:52 (+4)

Problem H

Solved by yanghong. 02:16 (+)

Problem I

Unsolved.

Problem J

Solved by yanghong. 01:09 (+)