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摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/132123173?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22% 阅读全文
posted @ 2023-08-05 18:46 Morning_Glory 阅读(169) 评论(0) 推荐(0) 编辑
摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/131953904?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22% 阅读全文
posted @ 2023-07-27 10:28 Morning_Glory 阅读(115) 评论(0) 推荐(0) 编辑
摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/131583841?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22% 阅读全文
posted @ 2023-07-06 19:37 Morning_Glory 阅读(34) 评论(0) 推荐(0) 编辑
摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/131561922?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22% 阅读全文
posted @ 2023-07-05 19:05 Morning_Glory 阅读(73) 评论(0) 推荐(0) 编辑
摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/131522235?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22% 阅读全文
posted @ 2023-07-04 10:46 Morning_Glory 阅读(15) 评论(0) 推荐(0) 编辑
摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/131462805?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22% 阅读全文
posted @ 2023-06-29 19:44 Morning_Glory 阅读(21) 评论(0) 推荐(0) 编辑
摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/131456608?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22% 阅读全文
posted @ 2023-06-29 15:18 Morning_Glory 阅读(16) 评论(0) 推荐(0) 编辑
摘要: [也许更好的阅读体验](https://blog.csdn.net/Morning_Glory_JR/article/details/131415680) # 简述 cdq分治是一种思想:将问题从中间分成两个部分,并且满足只有一部分(设为左)对另一部分(设为右)有贡献,右部分对左部分是没有贡献的,然 阅读全文
posted @ 2023-06-27 17:00 Morning_Glory 阅读(161) 评论(0) 推荐(0) 编辑
摘要: 也许更好的阅读体验 #include <stdio.h> #include <string.h> /* ****************************************************************** ALTERNATING BIT AND GO-BACK-N N 阅读全文
posted @ 2023-04-06 16:07 Morning_Glory 阅读(125) 评论(0) 推荐(0) 编辑
摘要: 也许更好的阅读体验 \(\mathcal{Description}\) 能否构造出一棵 \(n\) 个节点的树,使得以每个点为根的子树的大小加起来等于$s$,如果能,输出使得儿子最多的点的儿子数目最少的那种。 \(\mathcal{Solution}\) 边界:菊花是子树和最小的构造方法,链是子树和 阅读全文
posted @ 2022-05-07 15:35 Morning_Glory 阅读(54) 评论(0) 推荐(0) 编辑
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