CF1342E. Placing Rooks

Solution

有一个显然的性质：
要么每行都有至少一个，要么每列至少有一个。

只看每行有至少一个的情况：
显然必有n−kn-kn−k列有棋子。
考虑容斥，枚举空列的数量iii。
Ans=∑(−1)i∗(n−ki)∗(n−k−i)n∗(nk)Ans=\sum{(-1)^i*\binom{n-k}{i}*(n-k-i)^n*\binom{n}{k}}Ans=∑(−1)i∗(in−k)∗(n−k−i)n∗(kn)

列的情况与行的情况等价，特判n=mn=mn=m的情况即可。

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CF1342E. Placing Rooks

CF1342E. Placing Rooks

Solution

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