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 inom{n}{i}s^ia_{imod 4}
]
对(998244353)取模
范围
(1le Tle 10^5,1le nle 10^{18},1le s,a_0,a_1,a_2,a_3le 10^8)
单位根反演有个套路
[[kequiv l ( ext{ mod } n) ]=frac{1}{n}sum_{i=0}^{n-1}w_n^{(k-l)}
]
然后用套路推柿子
[egin{aligned}
&sum_{i=0}^ninom{n}{i}s^ia_{i mod 4}\
=&sum_{d=0}^3a_dsum_{i=0}^ninom{n}{i}s_i[iequiv dmod 4]\
=&sum_{d=0}^3a_dsum_{i=0}^ninom{n}{i}s_i(frac{1}{4}sum_{j=0}^3w_4^{(i-d)j})\
=&frac{1}{4}sum_{d=0}^3a_dsum_{j=0}^3sum_{i=0}^nw_4^{(i-d)j}inom{n}{i}s^i\
=&frac{1}{4}sum_{d=0}^3a_dsum_{j=0}^3w_4^{-dj}sum_{i=0}^ninom{n}{s}(sw_4^j)^i\
=&frac{1}{4}sum_{d=0}^3a_dsum_{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
",ans);
}
return 0;
}
2019.5.7