[LOJ 6485]LJJ学二项式定理(单位根反演)

也许更好的阅读体验

\(\mathcal{Description}\)

原题链接

\(T\)组询问，每次给\(n,s,a_0,a_1,a_2,a_3\)求

\(\begin{aligned}\left(\sum ^{n}_{i=0}\begin{pmatrix} n \\ i \end{pmatrix}\cdot s^{i}\cdot a_{i\ mod\ 4}\right)mod\ 998244353\end{aligned}\)

\(\mathcal{Solution}\)

这道题要用单位根反演

\(\begin{aligned}\left[ n\ |\ k\right] =\dfrac {1}{n}\sum ^{n-1}_{i=0}w^{ik}_{n}\end{aligned}\)

我们先打表，或者查表找出在模\(998244353\)下的原根\(\omega_4^0\)并处理好\(\omega_4^1,\omega_4^2,\omega_4^3\)

考虑对每个\(a_i\)单独计算其答案

以\(a_0\)为例

\(\begin{aligned}
ans_{a_0}&=\frac{1}{4}a_0\sum_{i=0}^n [4\ |\ i]{n\choose i}s^i\\
&=\frac{1}{4}a_0\sum_{i=0}^n{n\choose i}s^i\sum_{j=0}^3 (\omega_4^{j})^i\\
&=\frac{1}{4}a_0\sum_{j=0}^3\sum_{i=0}^n {n\choose i}s^i(\omega_4^j)^i\\
&=\frac{1}{4}a_0\sum_{j=0}^3(s\omega_4^j+1)^n
\end{aligned}\)

对于其他\(a_i\)，我们可以将其写成\([4\ |\ i+4-k]\)的形式，并提一个\(\omega^x\)出来

下面完整的推一遍

\(\begin{aligned}ans&=\sum ^{3}_{k=0}a_{k}\sum ^{n}_{i=0}\left[ 4\ |\ i+4-k\right] s^{i}\cdot \begin{pmatrix} n \\ i \end{pmatrix}\\
&=\sum ^{3}_{k=0}\dfrac {1}{4}a_{k}\sum ^{n}_{i=0}\sum ^{3}_{j=0}\omega^{j\cdot\left( i+4-k\right) }_{4}\cdot s^{i}\cdot \begin{pmatrix} n \\ i \end{pmatrix}\\
&=\sum ^{3}_{k=0}\dfrac {1}{4}a_{k}\sum ^{n}_{i=0}\sum ^{3}_{j=0}\omega^{\left(4j-jk\right) }_{4}\cdot \omega^{j^i}_4\cdot s^{i}\cdot \begin{pmatrix} n \\ i \end{pmatrix}\\
&=\sum ^{3}_{k=0}\dfrac {1}{4}a_{k}\sum ^{3}_{j=0}\omega^{\left(4j-jk\right) }_{4}\sum ^{n}_{i=0} \omega^{j^i}\cdot s^{i}\cdot \begin{pmatrix} n \\ i \end{pmatrix}\\
&=\sum ^{3}_{k=0}\dfrac {1}{4}a_{k}\sum ^{3}_{j=0}\omega^{\left(4j-jk\right) }_{4}\left(s\cdot \omega^j_4+1\right)^n \\
\end{aligned}\)

拿出来，再写一遍

\(\begin{aligned}ans=\sum ^{3}_{i=0}\dfrac {1}{4}a_{i}\sum ^{3}_{j=0}\omega^{\left(4j-ij\right) }_{4}\left(s\cdot \omega^j_4+1\right)^n \\
\end{aligned}\)

\(\mathcal{Code}\)

#include <cstdio>
#define ll long long
using namespace std;
const int mod = 998244353;
const int w [] = {1,911660635,998244352,86583718};
ll T,n,s,ans,res,inv;
ll a[4];
ll ksm (ll a,ll b)
{
	a%=mod;
	ll s=1;
	for (;b;b>>=1,a=a*a%mod)
		if (b&1)	s=s*a%mod;
	return s;
}
int main ()
{
	scanf("%lld",&T);
	inv=ksm(4,mod-2);
	while (T--){
		scanf("%lld%lld%lld%lld%lld%lld",&n,&s,&a[0],&a[1],&a[2],&a[3]);
		ans=0;
		for (int i=0;i<=3;++i){
			res=0;
			for (int j=0;j<=3;++j)
				res=(res+w[(4*j-i*j)%4]*ksm(s*w[j]+1,n)%mod)%mod;
			ans=(ans+a[i]*res%mod)%mod;
		}
		printf("%lld\n",ans*inv%mod);
	}
	return 0;
}

如有哪里讲得不是很明白或是有错误，欢迎指正

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