题目描述:
题解:
单位根反演。
$[n|x]=\frac{1}{n} \sum _{i=0}^{n-1} (ω_n^x)^i$
证明?显然啊,要么停在$(1,0)$要么转一圈。
所以说题目要求的是$\sum _{i=0}^{n} C(n,i) * s^i * a_{i\;mod\;4}$
把$a$提前,变成$\sum_{k=0}^{3}a_k \sum _{i=0} ^{n} C(n,i) *s^i [4|i-k]$
然后把上面单位根反演式子套进去。后面变成$\sum _{i=0} ^n C(n,i) * s^i * \frac{1}{4} \sum _{j=0} ^{3} (ω_4 ^{i-1})^j$
把后面提前面:$\frac{1}{4} \sum_{j=0}^3 ω_4^{-j} \sum_{i=0}^{n} C(n,i)*s^i*ω_4^{ij}$
发现二项式定理:$\frac{1}{4} \sum_{j=0}^3 ω_4^{-j} * (sω_4^j+1)^n$
最后就剩快速幂了?
代码:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MOD = ;
template<typename T>
inline void read(T&x)
{
T f = ,c = ;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){c=c*+ch-'';ch=getchar();}
x = f*c;
}
ll fastpow(ll x,ll y)
{
ll ret = ;
while(y)
{
if(y&)ret=ret*x%MOD;
x=x*x%MOD;y>>=;
}
return ret;
}
int T;
ll n,s,a0,a1,a2,a3,w0,w1,w2,w3,W0,W1,W2,W3,ans,inv;
void work()
{
read(n),read(s),read(a0),read(a1),read(a2),read(a3);n%=(MOD-),ans=;
W0 = fastpow(s*w0%MOD+,n),W1 = fastpow(s*w1%MOD+,n);
W2 = fastpow(s*w2%MOD+,n),W3 = fastpow(s*w3%MOD+,n);
ans=(ans+a0*(w0*W0%MOD+w0*W1%MOD+w0*W2%MOD+w0*W3%MOD)%MOD)%MOD;
ans=(ans+a1*(w0*W0%MOD+w3*W1%MOD+w2*W2%MOD+w1*W3%MOD)%MOD)%MOD;
ans=(ans+a2*(w0*W0%MOD+w2*W1%MOD+w0*W2%MOD+w2*W3%MOD)%MOD)%MOD;
ans=(ans+a3*(w0*W0%MOD+w1*W1%MOD+w2*W2%MOD+w3*W3%MOD)%MOD)%MOD;
printf("%lld\n",ans*inv%MOD);
}
int main()
{
// freopen("tt.in","r",stdin);
read(T);inv = fastpow(,MOD-);
w0=,w1=fastpow(,(MOD-)/),w2=w1*w1%MOD,w3=w1*w2%MOD;
while(T--)work();
return ;
}