Difference between revisions of "2020 CCPC Online Contest"
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== Problem C == | == Problem C == | ||
Solved by bingoier. 00:27:30 | Solved by bingoier. 00:27:30 | ||
| + | |||
| + | 贪心选择最后一个信封柜即可 | ||
== Problem D == | == Problem D == | ||
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== Problem E == | == Problem E == | ||
Solved by yanghong. 00:36:01(+2) | Solved by yanghong. 00:36:01(+2) | ||
| + | |||
| + | 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 == | == Problem F == | ||
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== Problem K == | == Problem K == | ||
| − | Solved by Once. 00:45:19( | + | 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。 | 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。 | ||
| Line 48: | Line 54: | ||
== Problem M == | == Problem M == | ||
| + | Unsolved. | ||
Latest revision as of 14:32, 7 October 2020
Problem A
Unsolved.
Problem B
Solved by all. 02:24:09(+1)
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
贪心选择最后一个信封柜即可
Problem D
Unsolved.
Problem E
Solved by yanghong. 00:36:01(+2)
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)
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
大胆猜测答案就是出现最多的字母的出现次数。
Problem H
Unsolved.
Problem I
Unsolved.
Problem J
Solved by Once. 00:08:01
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.