二项式反演
公式 1
\[g(n) = \sum_{i = 0}^n \binom n if(i)
\]
\[f(n) = \sum_{i = 0}^n (-1)^{n - i} \binom n ig(i)
\]
证明
\[f(n) = \sum_{i = 0}^n(-1)^{n - i} \binom n i g(i)
\]
\[= \sum_{i = 0}^n (-1)^{n - i} \binom ni \sum_{j = 0}^i \binom i j f(j)
\]
\[=\sum_{j = 0}^nf(j)\sum_{i = j}^n\binom n i \binom i j (-1)^{n - i}
\]
\[= \sum_{j = 0}^n f(j) \sum_{i = j}^n\binom n j \binom {n - j} {i - j} (-1)^{n - i}
\]
\[= \sum_{j = 0}^n\binom n j f(j) \sum_{i = 0}^{n- j} \binom {n - j} i (-1)^{n - j - i}
\]
\[= \sum_{j = 0}^n f(j) (\binom n j\sum_{i = 0} ^{n - j} (1 - 1)^{n - j} )
\]
\[=f(n)
\]
公式 2
\[g(k) = \sum\limits_{i = k} ^ n \binom i k f(i)
\]
\[f(k) = \sum\limits_{i = k} ^ n \binom i k g(i)* (-1) ^ {i - k}
\]
证明
\[g(k) = \sum\limits_{i = k} ^ n \binom i k \sum\limits_{j = i} ^ n \binom j i g(j) * (-1) ^ {j - i}
\]
\[= \sum_{j = k} ^n g(j) \sum_{i = k} ^j \binom j i \binom i k (-1) ^ {j - i}
\]
\[= \sum_{j = k} ^n g(j)\sum_{i = k} ^ j \binom j k \binom {j - k} {i - k} (-1)^ {j - i}
\]
\[= \sum_{j = k}^n g(j) \binom j k \sum_{i = 0} ^ {j - k} \binom {j - k} i (-1) ^{j - k - i}
\]
\[= \sum_{j = k}^n g(j) \binom j k (1 - 1) ^ {j - k}
\]
\[= g(k)
\]
例题
题意
有 \(n\) 个相邻格子， \(m\) 种颜色， 每种颜色有价值 \(v_i\) 把 \(m\) 种颜色喷涂到格子上，要求相邻格子颜色不相同，每种颜色的价值只算一遍，求格子总价值的期望，由于答案过大，你只需要求出答案 \(mod \ 998244353\) 即可
\(n \leq 1e16, m \leq 2000\)
解法
设 \(g(m)\) 为 \(m\) 种颜色随便喷涂的方案数
显然
\[g(m) = m * (m - 1) ^ {n - 1}
\]
设 \(f(m)\) 为恰好用 \(m\) 种颜色的方案数
\[g(m) = \sum_{i = 0}^m \binom m i f(i)
\]
\[f(m) = \sum_{i = 0}^m (-1) ^ {m - i} \binom m i g(i)
\]
\[total = \sum_{i = 1}^m v_i
\]
\[E = \sum_{i = 0}^m \frac {f(i)} {g(m)} * \binom m i i * \frac { total} m
\]
