反演原理

已知: f(n)=∑i=0nani⋅g(i) f ( n ) = ∑ i = 0 n a n i ⋅ g ( i )

即: ⎧⎩⎨⎪⎪⎪⎪⎪⎪f(0)=a00⋅g(0)f(1)=a10⋅g(0)+a11⋅g(1)⋯f(n)=an0⋅g(0)+an1⋅g(1)+⋯+ann⋅g(n) { f ( 0 ) = a 00 ⋅ g ( 0 ) f ( 1 ) = a 10 ⋅ g ( 0 ) + a 11 ⋅ g ( 1 ) ⋯ f ( n ) = a n 0 ⋅ g ( 0 ) + a n 1 ⋅ g ( 1 ) + ⋯ + a n n ⋅ g ( n )

化成矩阵形式:

⎛⎝⎜⎜⎜⎜⎜⎜⎜f(0)f(1)f(2)⋮f(n)⎞⎠⎟⎟⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜⎜⎜a00a10a20⋮an00a11a21⋮an100a22⋮an2000⋮an3⋯⋯⋯⋱⋯000⋮ann⎞⎠⎟⎟⎟⎟⎟⎟⎟×⎛⎝⎜⎜⎜⎜⎜⎜⎜g(0) g(1) g(2) ⋮ g(n)⎞⎠⎟⎟⎟⎟⎟⎟⎟ ( f ( 0 ) f ( 1 ) f ( 2 ) ⋮ f ( n ) ) = ( a 00 0 0 0 ⋯ 0 a 10 a 11 0 0 ⋯ 0 a 20 a 21 a 22 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ a n 0 a n 1 a n 2 a n 3 ⋯ a n n ) × ( g ( 0 )   g ( 1 )   g ( 2 )   ⋮   g ( n ) )

---

反演求: g(n)=∑i=0nbni⋅f(i) g ( n ) = ∑ i = 0 n b n i ⋅ f ( i )

即: ⎧⎩⎨⎪⎪⎪⎪⎪⎪g(0)=b00⋅f(0)g(1)=b10⋅f(0)+b11⋅f(1)⋯g(n)=bn0⋅f(0)+bn1⋅f(1)+⋯+bnn⋅f(n) { g ( 0 ) = b 00 ⋅ f ( 0 ) g ( 1 ) = b 10 ⋅ f ( 0 ) + b 11 ⋅ f ( 1 ) ⋯ g ( n ) = b n 0 ⋅ f ( 0 ) + b n 1 ⋅ f ( 1 ) + ⋯ + b n n ⋅ f ( n )

考虑: f(n)=∑i=0nani⋅g(i)=∑i=0nani⋅∑j=0ibij⋅f(j) f ( n ) = ∑ i = 0 n a n i ⋅ g ( i ) = ∑ i = 0 n a n i ⋅ ∑ j = 0 i b i j ⋅ f ( j )

即下列矩阵所有数字之和:

⎛⎝⎜⎜⎜⎜⎜⎜⎜an0b00f(0)an1b10f(0)an2b20f(0)⋮annbn0f(0)an1b11f(1)an2b21f(1)⋮annbn1f(1)an2b22f(2)⋮annbn2f(2)⋱annbn3f(3)⋯annbnnf(n)⎞⎠⎟⎟⎟⎟⎟⎟⎟ ( a n 0 b 00 f ( 0 ) a n 1 b 10 f ( 0 ) a n 1 b 11 f ( 1 ) a n 2 b 20 f ( 0 ) a n 2 b 21 f ( 1 ) a n 2 b 22 f ( 2 ) ⋮ ⋮ ⋮ ⋱ a n n b n 0 f ( 0 ) a n n b n 1 f ( 1 ) a n n b n 2 f ( 2 ) a n n b n 3 f ( 3 ) ⋯ a n n b n n f ( n ) )

即: f(n)=∑j=0nf(j)∑i=jnani⋅bij f ( n ) = ∑ j = 0 n f ( j ) ∑ i = j n a n i ⋅ b i j

所以: ∑i=jnani⋅bij={0(j≠n)1(j=n) ∑ i = j n a n i ⋅ b i j = { 0 ( j ≠ n ) 1 ( j = n )

对于二项式反演: ani=Cin a n i = C n i

对于莫比乌斯反演: ani={1(n%i=0)0(n%i≠0) a n i = { 1 ( n % i = 0 ) 0 ( n % i ≠ 0 )