随笔分类 - 数学-容斥
摘要:比赛链接 "cf" A 最后一位判定 C 所有方案 全是红边的方案 并查集维护 cpp include include include include include include include include include using namespace std; typedef long
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摘要:第一道容斥 $ans[i] = \sum_{j = 0}^{min(cnt, n / 2)} ( 1)^j \tbinom{cnt}{j} \tbinom{n 2 j + k 1}{k 1}$ i 为 [2, 2k] cnt是满足x, y小于等于k 且 x + y = i 的对的种类数 后面那个组合
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