Difference between revisions of "2020 ICPC Xiaomi Invitational Second Round Online Contest"

Latest revision as of 08:53, 4 November 2020

Solved by yanghong. 04:24:26(-4)

Unsolved.

Solved by Once. 03:19:36(-2)

Solved by bingoier & Once. 00:53:26(+)

线性代数推导可得结论.

Unsolved.

Unsolved.

Solved by Once. 01:48:41(-1)

Solved by yanghong. 03:02:04(-3)

Solved by bingoier & Once. 00:51:30(+3)

Unsolved.

Unsolved.

@@ Line 1: / Line 1: @@
 == Problem A ==
-Unsolved.
+Solved by yanghong. 04:24:26(-4)
 == Problem B ==
-Solved by all. 02:24:09(+1)
+Unsolved.
-每个节点向其某个质因子连边，之后所有素数向 $2$ 连边。答案即为区间自然数和加上区间素数和，用 min25 筛出即可。
 == Problem C ==
-Solved by bingoier. 00:27:30
+Solved by Once. 03:19:36(-2)
-贪心选择最后一个信封柜即可
+== Problem D ==
+Solved by bingoier & Once. 00:53:26(+)
-== Problem D ==
+线性代数推导可得结论.
-Unsolved.
 == Problem E ==
-Solved by yanghong. 00:36:01(+2)
+Unsolved.
-结论为 $sg(x) = [c_2] + \sum_{i > 2} c_i$
-其中 $x = 2^{c_2} 3 ^ {c_3} 5 ^ {c_5} \dots $
 == Problem F ==
-Solved by Once. 04:43:11(+4)
+Unsolved.
-把所有边界点带入后方的连续函数验证连续性即可。
-取消同步要人命。
 == Problem G ==
-Solved by Once. 00:16:43
+Solved by Once. 01:48:41(-1)
-大胆猜测答案就是出现最多的字母的出现次数。
 == Problem H ==
-Unsolved.
+Solved by yanghong. 03:02:04(-3)
 == Problem I ==
-Unsolved.
+Solved by bingoier & Once. 00:51:30(+3)
 == Problem J ==
-Solved by Once. 00:08:01
+Unsolved.
-检查是否有相邻的 $1$ 即可。
 == Problem K ==
-Solved by Once. 00:45:19(+1)
-大胆猜测，当且仅当只有左上角非零时答案为原矩阵，感觉证明也挺简单的。
-== Problem L ==
-Unsolved.
-== Problem M ==
 Unsolved.