容斥原理
容斥原理
多步容斥的一般形式如下
∣ A 1 ∪ A 2 ∪ ⋯ ∪ A n ∣ = ∑ 1 ≤ i ≤ n ∣ A i ∣ − ∑ 1 ≤ i < j ≤ n ∣ A i ∩ A j ∣ + ⋯ + ( − 1 ) n − 1 × ∣ A 1 ∩ A 2 ∩ ⋯ ∩ A n ∣ |A_1\cup A_2\cup\cdots\cup A_n|=\sum\limits_{1\le i\le n}|A_i|-\sum\limits_{1\le i<j\le n}|A_i\cap A_j|+\cdots+(-1)^{n-1}\times |A_1\cap A_2\cap \cdots\cap A_n| ∣A1∪A2∪⋯∪An∣=1≤i≤n∑∣Ai∣−1≤i<j≤n∑∣Ai∩Aj∣+⋯+(−1)n−1×∣A1∩A2∩⋯∩An∣
也可以写成
∣ ⋃ i = 1 n A i ∣ = ∑ ∅ ≠ S ⊆ { 1 , 2 , … , n } ( − 1 ) ∣ S ∣ − 1 ∣ ⋂ j ∈ S A j ∣ \left|\bigcup_{i=1}^{n} A_{i}\right|=\sum_{\emptyset \neq S \subseteq\{1,2, \ldots, n\}}(-1)^{|S|-1} \left|\bigcap_{j \in S} A_{j} \right| ∣∣∣∣∣i=1⋃nAi∣∣∣∣∣=∅=S⊆{1,2,…,n}∑(−1)∣S∣−1∣∣∣∣∣∣j∈S⋂Aj∣∣∣∣∣∣
二项式反演
f ( n ) = ∑ i = 0 n ( n i ) g ( i ) ⇔ g ( n ) = ∑ i = 0 n ( − 1 ) n − i ( n i ) f ( i ) f(n)=\sum\limits_{i=0}^n{n\choose i}g(i)\Leftrightarrow g(n)=\sum\limits_{i=0}^n(-1)^{n-i}{n\choose i}f(i) f(n)=i=0∑n(in)g(i)⇔g(n)=i=0∑n(−1)n−i(in)f(i)
略证
将左式代入右式
g ( n ) = ∑ i = 0 n ( − 1 ) n − i ( n i ) ∑ j = 0 i ( i j ) g ( j ) = ∑ i = 0 n ( − 1 ) n − i ∑ j = 0 i ( n i ) ( i j ) g ( j ) g(n)=\sum\limits_{i=0}^n(-1)^{n-i}{n\choose i}\sum\limits_{j=0}^i{i\choose j}g(j) \\ =\sum\limits_{i=0}^n(-1)^{n-i} \sum\limits_{j=0}^i {n\choose i}{i\choose j}g(j) \\ g(n)=i=0∑n(−1)n−i(in)j=0∑i(ji)g(j)=i=0∑n(−1)n−ij=0∑i(in)(ji)g(j)
将求和顺序变换，因为 ( n m ) n \choose m (mn) 在 m > n m > n m>n 时值为 0 0 0
g ( n ) = ∑ j = 0 n ∑ i = j n ( − 1 ) n − i ( n i ) ( i j ) g ( j ) g(n)=\sum\limits_{j=0}^n\sum\limits_{i=j}^n(-1)^{n-i} {n\choose i}{i\choose j}g(j) g(n)=j=0∑ni=j∑n(−1)n−i(in)(ji)g(j)
因为
( n i ) ( i j ) = ( n j ) ( n − j i − j ) \binom{n}{i}\binom{i}{j}=\binom{n}{j}\binom{n-j}{i-j} (in)(ji)=(jn)(i−jn−j)
所以
g ( n ) = ∑ j = 0 n ∑ i = j n ( − 1 ) n − i ( n j ) ( n − j i − j ) g ( j ) = ∑ j = 0 n ( n j ) g ( j ) ∑ i = j n ( n − j i − j ) ( − 1 ) n − i = ∑ j = 0 n ( n j ) g ( j ) ∑ k = 0 n − j ( n − j k ) ( − 1 ) n − k − j = ∑ j = 0 n ( n j ) g ( j ) ∑ k = 0 n − j ( n − j k ) ( − 1 ) n − k − j × 1 n − j g(n)=\sum\limits_{j=0}^n\sum\limits_{i=j}^n(-1)^{n-i} {n\choose j}{n-j\choose i-j}g(j) \\ = \sum\limits_{j=0}^n {n\choose j} g(j) \sum\limits_{i=j}^n {n-j\choose i-j} (-1)^{n-i} \\ = \sum\limits_{j=0}^n {n\choose j} g(j) \sum\limits_{k=0}^{n-j} {n-j\choose k} (-1)^{n-k-j} \\ = \sum\limits_{j=0}^n {n\choose j} g(j) \sum\limits_{k=0}^{n-j} {n-j\choose k} (-1)^{n-k-j} \times 1^{n-j} g(n)=j=0∑ni=j∑n(−1)n−i(jn)(i−jn−j)g(j)=j=0∑n(jn)g(j)i=j∑n(i−jn−j)(−1)n−i=j=0∑n(jn)g(j)k=0∑n−j(kn−j)(−1)n−k−j=j=0∑n(jn)g(j)k=0∑n−j(kn−j)(−1)n−k−j×1n−j
根据二项式定理
( x + y ) k = ∑ i = 0 k x i y k − i × C i k (x+y)^{k}=\sum_{i=0}^{k} x^{i} y^{k-i} \times C_{i}^{k} (x+y)k=i=0∑kxiyk−i×Cik
得
g ( n ) = { ∑ j = 0 n ( n j ) g ( j ) ∑ k = 0 n − j ( n − j k ) ( − 1 ) n − k − j = ∑ j = 0 n ( n j ) g ( j ) ( n − j = 0 ) ∑ j = 0 n ( n j ) g ( j ) × ( 1 − 1 ) n − j = 0 ( n − j ≠ 0 ) g(n) =\begin{cases} \sum\limits_{j=0}^n {n\choose j} g(j) \sum\limits_{k=0}^{n-j} {n-j\choose k} (-1)^{n-k-j} =\sum\limits_{j=0}^n {n\choose j} g(j) & (n - j =0) \\ \sum\limits_{j=0}^n {n\choose j} g(j) \times (1-1)^{n-j} = 0& (n - j \neq 0) \\ \end{cases} g(n)=⎩⎪⎪⎨⎪⎪⎧j=0∑n(jn)g(j)k=0∑n−j(kn−j)(−1)n−k−j=j=0∑n(jn)g(j)j=0∑n(jn)g(j)×(1−1)n−j=0(n−j=0)(n−j=0)
综上所述
g ( n ) = ∑ j = 0 n ( n j ) [ j = n ] = ( n n ) g ( n ) = g ( n ) g(n) = \sum\limits_{j=0}^n {n\choose j} [j=n]\\ ={n\choose n}g(n) \\ =g(n) g(n)=j=0∑n(jn)[j=n]=(nn)g(n)=g(n)
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