JZOJ6757.【NOI2020模拟07.22】T3(singer)
Description
- n < 2 19 , m ( n − k + 1 ) < 2 20 n<2^{19},m(n-k+1)<2^{20} n<219,m(n−k+1)<220，模 998244353 998244353 998244353
Solution
- 首先容斥，恰好 k k k个容斥成至少 k k k个升调，那么对于长度 n n n来说至多有 n − k n-k n−k段。
- f ( n ) = ∑ i > = n C i n ( − 1 ) n − i g ( i ) f(n)=\sum_{i>=n}C_i^n(-1)^{n-i}g(i) f(n)=∑i>=nCin(−1)n−ig(i)
- 设 f ( x ) = ( x + 1 ) m − 1 f(x)=(x+1)^m-1 f(x)=(x+1)m−1
- 枚举 i i i段组合在了一起，长度为 n n n，枚举有 i > = k i>=k i>=k个升调。
- A n s = ∑ n = k N ∑ i = k n C i k ( − 1 ) i − k f n − i ( x ) [ x n ] Ans=\sum_{n=k}^N\sum_{i=k}^nC_i^k(-1)^{i-k}f^{n-i}(x)[x^n] Ans=∑n=kN∑i=knCik(−1)i−kfn−i(x)[xn]
- A n s = ∑ i = k N C i k ( − 1 ) i − k ∑ n = i N f n − i ( x ) [ x n ] Ans=\sum_{i=k}^NC_i^k(-1)^{i-k}\sum_{n=i}^Nf^{n-i}(x)[x^n] Ans=∑i=kNCik(−1)i−k∑n=iNfn−i(x)[xn]
- A n s = [ x N ] ∑ i = k N C i k ( − 1 ) i − k ∑ n = i N f n − i ( x ) x N − n Ans=[x^N]\sum_{i=k}^NC_i^k(-1)^{i-k}\sum_{n=i}^Nf^{n-i}(x)x^{N-n} Ans=[xN]∑i=kNCik(−1)i−k∑n=iNfn−i(x)xN−n
- A n s = [ x N ] ∑ i = k N C i k ( − 1 ) i − k f N − i + 1 − x N − i + 1 f − x Ans=[x^N]\sum_{i=k}^NC_i^k(-1)^{i-k}\frac{f^{N-i+1}-x^{N-i+1}}{f-x} Ans=[xN]∑i=kNCik(−1)i−kf−xfN−i+1−xN−i+1
- A n s = [ x N + 1 ] ∑ i = k N C i k ( − 1 ) i − k f N − i + 1 − x N − i + 1 f − x x Ans=[x^{N+1}]\sum_{i=k}^NC_i^k(-1)^{i-k}\frac{f^{N-i+1}-x^{N-i+1}}{{\frac{f-x}{x}}} Ans=[xN+1]∑i=kNCik(−1)i−kxf−xfN−i+1−xN−i+1
- 相当于求：
- F ( x ) = ∑ i = k N C i k ( − 1 ) i − k f N − i + 1 F(x)=\sum_{i=k}^NC_i^k(-1)^{i-k}f^{N-i+1} F(x)=∑i=kNCik(−1)i−kfN−i+1
- F ( x ) = ∑ i = k N C i k ( − 1 ) i − k ∑ j = 0 N − i + 1 ( x + 1 ) m j ( − 1 ) N − i + 1 − j C N − i + 1 j F(x)=\sum_{i=k}^NC_i^k(-1)^{i-k}\sum_{j=0}^{N-i+1}(x+1)^{mj}(-1)^{N-i+1-j}C_{N-i+1}^j F(x)=∑i=kNCik(−1)i−k∑j=0N−i+1(x+1)mj(−1)N−i+1−jCN−i+1j
- F ( x ) = ∑ i = k N ∑ j = 0 N − i + 1 ( x + 1 ) m j ( − 1 ) N − i + 1 − j + i − k C N − i + 1 j C i k F(x)=\sum_{i=k}^N\sum_{j=0}^{N-i+1}(x+1)^{mj}(-1)^{N-i+1-j+i-k}C_{N-i+1}^jC_i^k F(x)=∑i=kN∑j=0N−i+1(x+1)mj(−1)N−i+1−j+i−kCN−i+1jCik
- F ( x ) = ∑ j = 0 N − k + 1 ( x + 1 ) m j ( − 1 ) N + 1 + j + k ∑ i = k N − j + 1 C N − i + 1 j C i k F(x)=\sum_{j=0}^{N-k+1}(x+1)^{mj}(-1)^{N+1+j+k}\sum_{i=k}^{N-j+1}C_{N-i+1}^jC_i^k F(x)=∑j=0N−k+1(x+1)mj(−1)N+1+j+k∑i=kN−j+1CN−i+1jCik
- F ( x ) = ∑ j = 0 N − k + 1 ( x + 1 ) m j ( − 1 ) N + 1 + j + k C N + 2 j + k + 1 F(x)=\sum_{j=0}^{N-k+1}(x+1)^{mj}(-1)^{N+1+j+k}C_{N+2}^{j+k+1} F(x)=∑j=0N−k+1(x+1)mj(−1)N+1+j+kCN+2j+k+1
- F ( x ) = ∑ j = 0 N − k + 1 ∑ i = 0 m j C m j i ( − 1 ) N + 1 + j + k C N + 2 j + k + 1 x i F(x)=\sum_{j=0}^{N-k+1}\sum_{i=0}^{mj}C_{mj}^i(-1)^{N+1+j+k}C_{N+2}^{j+k+1}x^i F(x)=∑j=0N−k+1∑i=0mjCmji(−1)N+1+j+kCN+2j+k+1xi
- F ( x ) = ∑ i = 0 m ( N − k + 1 ) ∑ j = i / m N − k + 1 ( m j ) ! i ! ( m j − i ) ! ( − 1 ) N + 1 + j + k C N + 2 j + k + 1 x i F(x)=\sum_{i=0}^{m(N-k+1)}\sum_{j=i/m}^{N-k+1}\frac{(mj)!}{i!(mj-i)!}(-1)^{N+1+j+k}C_{N+2}^{j+k+1}x^i F(x)=∑i=0m(N−k+1)∑j=i/mN−k+1i!(mj−i)!(mj)!(−1)N+1+j+kCN+2j+k+1xi
- 由于题目保证 m > = 2 m>=2 m>=2，再特判一下 k = 0 k=0 k=0即可。
注意事项
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F ( x ) F(x) F(x)没有常数项！我也不知道为什么卷积之后突然出现了常数项，可能是我没有判掉吧。
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多项式求逆的时候如果分母常数项为0，就上下同时除以一个 x x x
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由于只需要求 x N x^{N} xN的值，所以只需要求出 m o d x N + 1 mod\ x^{N+1} mod xN+1后的 F ( x ) F(x) F(x)和 1 f − x \frac{1}{f-x} f−x1。
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注意乘法的时候要开两倍。
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多项式求逆的时候有值得下标不能超过 l i m / 2 lim/2 lim/2。
姿势
- 可以通过次数的移动使得求得东西变得简单，再用等比数列求解
- A n s = ∑ i = k N C i k ( − 1 ) i − k ∑ n = i N f n − i ( x ) [ x n ] Ans=\sum_{i=k}^NC_i^k(-1)^{i-k}\sum_{n=i}^Nf^{n-i}(x)[x^n] Ans=∑i=kNCik(−1)i−k∑n=iNfn−i(x)[xn]
- A n s = [ x N ] ∑ i = k N C i k ( − 1 ) i − k ∑ n = i N f n − i ( x ) x N − n Ans=[x^N]\sum_{i=k}^NC_i^k(-1)^{i-k}\sum_{n=i}^Nf^{n-i}(x)x^{N-n} Ans=[xN]∑i=kNCik(−1)i−k∑n=iNfn−i(x)xN−n
#include