LOJ6485 LJJ 学二项式定理 解题报告

LJJ 学二项式定理

题意

\(T\)组数据，每组给定\(n,s,a_0,a_1,a_2,a_3\)，求

\[\sum_{i=0}^n \binom{n}{i}s^ia_{i\bmod 4}
\]

对\(998244353\)取模

范围

\(1\le T\le 10^5,1\le n\le 10^{18},1\le s,a_0,a_1,a_2,a_3\le 10^8\)

---

单位根反演有个套路

\[[k\equiv l \ (\text{ mod } n)\ ]=\frac{1}{n}\sum_{i=0}^{n-1}w_n^{(k-l)}
\]

然后用套路推柿子

\[\begin{aligned}
&\sum_{i=0}^n\binom{n}{i}s^ia_{i \bmod 4}\\
=&\sum_{d=0}^3a_d\sum_{i=0}^n\binom{n}{i}s_i[i\equiv d\bmod 4]\\
=&\sum_{d=0}^3a_d\sum_{i=0}^n\binom{n}{i}s_i(\frac{1}{4}\sum_{j=0}^3w_4^{(i-d)j})\\
=&\frac{1}{4}\sum_{d=0}^3a_d\sum_{j=0}^3\sum_{i=0}^nw_4^{(i-d)j}\binom{n}{i}s^i\\
=&\frac{1}{4}\sum_{d=0}^3a_d\sum_{j=0}^3w_4^{-dj}\sum_{i=0}^n\binom{n}{s}(sw_4^j)^i\\
=&\frac{1}{4}\sum_{d=0}^3a_d\sum_{j=0}^3w_4^{-dj}(sw_4^j+1)^n
\end{aligned}
\]

---

Code:

#include <cstdio>
#include <cctype>
#define ll long long
const int SIZE=1<<21;
char ibuf[SIZE],*iS,*iT;
#define gc() (iS==iT?(iT=(iS=ibuf)+fread(ibuf,1,SIZE,stdin),iS==iT?EOF:*iS++):*iS++)
template <class T>
void read(T &x)
{
	x=0;char c=gc();
	while(!isdigit(c)) c=gc();
	while(isdigit(c)) x=x*10+c-'0',c=gc();
}
const int mod=998244353;
const int inv4=748683265;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
#define mul(a,b) (1ll*(a)*(b)%mod)
int qp(int d,int k)
{
	int f=1;
	while(k)
	{
		if(k&1) f=mul(f,d);
		d=mul(d,d);
		k>>=1;
	}
	return f;
}
int w[4],a[4],s;
ll n;
int main()
{
	int T;read(T);
	w[0]=1,w[1]=qp(3,mod-1>>2),w[2]=mod-1,w[3]=mul(w[1],w[2]);
	while(T--)
	{
		read(n),read(s);
		int ans=0;
		for(int a,i=0;i<4;i++)
		{
			read(a);int sum=0;
			for(int j=0;j<4;j++)
			{
				int f=add(mul(s,w[j]),1);
				sum=add(sum,mul(w[(20-i*j)%4],qp(f,n%(mod-1))));
			}
			ans=add(ans,mul(a,sum));
		}
		ans=mul(ans,inv4);
		printf("%d\n",ans);
	}
	return 0;
}

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2019.5.7