数论傻谈
数论傻谈
颓废之余写的怕记不得随便写
CF870F Paths
首先分类
\[Case1:不互质的数对
\\
\sum\limits_{i=2}^{n}(i-1-\phi(i))
\\
Case2:互质的数对
\\
Subcase1:距离为2
\\
设两数为x,y，最小质因数分别为p_x,p_y
\\
p_x\times p_y\leq n满足
\\
不好算？？
\\
Subcase2:距离为3
\\
p_x\times p_y> n且2p_x\leq n,2p_y\leq n(可以理解为用2p_x,2p_y相连)
\\
考虑记录满足的质因数设P_i为i的最小质因数
\\
\sum\limits_{i=1}^{\dfrac{n}{2}}\sum_{j=i+1}^{\dfrac{n}{2}}[P_iP_j>n][2P_i\le n][2P_j\le n]
\\
设T_i为\sum\limits_{j=1}[P_j=i]
\\
=\sum_{p_i\in Prime}\sum_{p_j\in Prime}[p_i<p_j][p_ip_j>n][2p_i\le n][2p_j\le n]T_iT_j
\\
=\sum_{p_i\in Prime}T_i\sum_{p_j\in Prime}[max(p_i,\dfrac{n}{p_i})\leq p_j\leq\dfrac{n}{2}]T_j
\\
Subcase3:不连通
\\
max(2p_y,2p_x)> n
\\
=\sum_{p_i\in Prime}T_i\sum_{p_j\in Prime}[max(p_i,\dfrac{n}{2})\leq p_j\leq n]T_j
\\
注意到是区间加的形式
\]
\(关于a\bmod m\)
\(\gcd(a,m)=d,设a=a_0d,m=m_0d\)
\(a_0d\bmod m_0d=dk\)
\(\Rightarrow a\bmod m=d(a_0\bmod m_0)\)
什么智障问题
关于exlucas
求\(\binom{m}{n}\bmod m\),其中\(m\)不为质数
考虑对\(m\)质因数分解\(m=\prod\limits_{i=1}^{k}p_i^{a_i}\)，利用\(CRT\),即求\(\binom{m}{n}\bmod p_i^{a_i}\)
\(\binom{m}{n}\bmod p^{a}\)展开可化为\(\dfrac{n!}{m!(n-m)!}\bmod p_i^{a}\)
由于\(p^a\)不为质数且不与\(n!\)互质，所以依旧无法使用逆元，所以考虑消去阶乘中的\(p\)
即\(n!=\dfrac{n!}{p^x}p^x\),其中\(\dfrac{n!}{p^x}\)与\(p\)的互质，形象一点就是把\(n!\)中是\(p\)的倍数的\(i\)消去以达到逆元的效果，最后再乘回去
设\(f(n)为\dfrac{n!}{p^x},g(n)为x\)(\(f\)是去除\(p\)的结果,\(g\)是去除的个数让后面加回来)
考虑将阶乘展开\(\prod\limits_{i=1}^ni=p^{\dfrac{n}{p}}(\dfrac{n}{p})!\prod\limits_{i=1,i\bmod p\neq0}i\)
注意到后面的连乘具有循环节(取模时模数较小)
而\((\dfrac{n}{p})!\)中可能还包含\(p\)的因数
所以\(f(n)=f(\dfrac{n}{p})(\prod\limits_{i=1,i\bmod p\neq 0}^{p^k}i)^{\dfrac{n}{p^k}}\prod\limits_{i=1}^{n\bmod p^k}i,g(n)=\dfrac{n}{p}+g(\dfrac{n}{p})\)
注意如果多次询问可以预处理循环节
时间复杂度\(O(\sqrt{p}+\log_{p}n\sum p^k)\)
#include<bits/stdc++.h>
#define int long long
using namespace std;
int cnt_prime;
int prime[105];
int Num[105];
int Pow(int a,int b,int p)
{
int base=a;
int res=1;
while(b)
{
if(b&1){
res=((long long)res*base)%p;
}
base=((long long)base*base)%p;
b>>=1;
}
return res;
}
int Ex_x,Ex_y;
int Ex_gcd(int a,int b)
{
if(!b)
{
Ex_x=1;
Ex_y=0;
return a;
}
int Rex=Ex_gcd(b,a%b);
int t=Ex_x;
Ex_x=Ex_y;
Ex_y=t-(a/b)*Ex_y;
return Rex;
}
int inv(int a,int p)
{
int d=Ex_gcd(a,p);
Ex_x=(Ex_x+(p))%p;
return Ex_x;
}
pair<int,int>cal(int n,int p,int pk)
{
if(!n)
{
return make_pair(1,0);
}
int Ind=(n/pk);
int rest=(n%pk);
int Bas=1;
for(int i=1;i<=pk;i++)
{
if(i%p)
{
Bas=((long long)Bas*i)%pk;
}
}
Bas=(Pow(Bas,Ind,pk));
for(int i=1;i<=rest;i++)
{
if(i%p)
{
Bas=((long long)Bas*i)%pk;
}
}
pair<int,int>Nx=cal(n/p,p,pk);
return make_pair(((long long)Nx.first*Bas)%pk,Nx.second+(n/p));
}
int get(int n,int m,int p,int k)
{
int pk=Pow(p,k,1e9);
// printf("%lld\n",pk);
pair<int,int>Fn=cal(n,p,pk);
//printf("%d %d %d\n",n,Fn.first,Fn.second);
pair<int,int>Fm=cal(m,p,pk);
pair<int,int>Fx=cal(n-m,p,pk);
int Res=Fn.first;
Res=((long long)Res*inv(Fm.first,pk))%pk;
Res=((long long)Res*inv(Fx.first,pk))%pk;
int Ind=(Fn.second-Fm.second-Fx.second);
if(Ind>=k)
{
return 0;
}
Res=((long long)Res*Pow(p,Ind,pk))%pk;
// printf("%lld %lld %lld %lld\n",Fn.first,inv(Fm.first,pk),inv(Fx.first,pk),Pow(p,Ind,pk));
return Res;
}
int Exlucas(int n,int m,int p)
{
cnt_prime=0;
int X=p;
for(int i=2;i*i<=X;i++)
{
if(X%i==0)
{
int tot=0;
while(X%i==0)
{
X/=i;
tot++;
}
prime[++cnt_prime]=i;
Num[cnt_prime]=tot;
}
}
if(X>1)
{
prime[++cnt_prime]=X;
Num[cnt_prime]=1;
}
int M=p;
int res=0;
for(int i=1;i<=cnt_prime;i++)
{
int mi=Pow(prime[i],Num[i],M+1);
int a=get(n,m,prime[i],Num[i]);
//printf("%lld\n",a);
int Mi=(M/mi);
int Gcd=Ex_gcd(Mi,mi);
Ex_x=((long long)Ex_x+mi)%mi;
res=((long long)res+((((long long)Ex_x*Mi)%M)*a)%M)%M;
}
return res;
}
int n,m,p;
signed main()
{
scanf("%lld %lld %lld",&n,&m,&p);
printf("%lld",Exlucas(n,m,p));
}
关于类欧几里得算法
关于f
\[f(n,a,b,c)=\sum\limits_{i=0}^n(\lfloor\dfrac{ai+ b}{c}\rfloor)=\sum\limits_{i=0}^n(\lfloor\dfrac{a}{c}\rfloor i+\lfloor\dfrac{b}{c}\rfloor+(\lfloor\dfrac{(a\bmod c)i+ (b\bmod c)}{c}\rfloor))\\=(\dfrac{n(n+1)}{2}\lfloor\dfrac{a}{c}\rfloor+(n+1)\lfloor\dfrac{b}{c}\rfloor+\sum\limits_{i=0}^n(\lfloor\dfrac{(a\bmod c)i+ (b\bmod c)}{c}\rfloor))
\\
=\dfrac{n(n+1)}{2}\lfloor\dfrac{a}{c}\rfloor+(n+1)\lfloor\dfrac{b}{c}\rfloor+f(n,a\bmod c,b\bmod c,c)
\\
现在只考虑a<c,b<c的情况,设M=\lfloor\dfrac{an+b}{c}\rfloor
\\
\sum\limits_{i=0}^n(\lfloor\dfrac{ai+ b}{c}\rfloor)=\sum\limits_{i=0}^n\sum_{j=1}^M[j\le\lfloor\dfrac{ai+ b}{c}\rfloor]=\sum\limits_{i=0}^n\sum_{j=0}^{M-1}[cj+c\le ai+b]
\\
=\sum\limits_{i=0}^n\sum_{j=0}^{M-1}[cj+c< ai+b+1]=\sum\limits_{i=0}^n\sum_{j=0}^{M-1}[cj+c-b-1< ai]=\sum_{j=0}^{M-1}(n-\sum\limits_{i=0}^n[cj+c-b-1\geq ai])
\\
=\sum_{j=0}^{M-1}(n-\sum\limits_{i=1}^n[\lfloor\dfrac{cj+c-b-1}{a}\rfloor\geq i])=\sum_{j=0}^{M-1}(n-\lfloor\dfrac{cj+c-b-1}{a}\rfloor)=nM-\sum_{j=0}^{M-1}\lfloor\dfrac{cj+c-b-1}{a}\rfloor
\\
=nM-f(M-1,c,c-b-1,a)
\\
边界a=0
\]
关于g,h
\[g(n,a,b,c)=\sum\limits_{i=0}^n(\lfloor\dfrac{ai+ b}{c}\rfloor)^2
\\
h(n,a,b,c)=\sum\limits_{i=0}^ni(\lfloor\dfrac{ai+ b}{c}\rfloor)
\\
Case1:a\ge c,b\ge c
\\
g(n,a,b,c)=\sum\limits_{i=0}^n(\lfloor\dfrac{(a\bmod c)i+ b\bmod c}{c}\rfloor+\lfloor\dfrac{b}{c}\rfloor+\lfloor\dfrac{a}{c}\rfloor i)^2
\\
=\sum\limits_{i=0}^n(\lfloor\dfrac{(a\bmod c)i+ b\bmod c}{c}\rfloor)^2 +(\lfloor\dfrac{b}{c}\rfloor+\lfloor\dfrac{a}{c}\rfloor i)^2+2(\lfloor\dfrac{(a\bmod c)i+ b\bmod c}{c}\rfloor)(\lfloor\dfrac{b}{c}\rfloor+\lfloor\dfrac{a}{c}\rfloor i)
\\
=g(n,a\bmod c,b\bmod c,c)+\sum_{i=0}^n(\lfloor\dfrac{b}{c}\rfloor+\lfloor\dfrac{a}{c}\rfloor i)^2+2(\lfloor\dfrac{(a\bmod c)i+ b\bmod c}{c}\rfloor)(\lfloor\dfrac{b}{c}\rfloor+\lfloor\dfrac{a}{c}\rfloor i)
\\
=g(n,a\bmod c,b\bmod c,c)+ \sum_{i=0}^n(\lfloor\dfrac{b}{c}\rfloor)^2+(\lfloor\dfrac{a}{c}\rfloor i)^2+2(\lfloor\dfrac{b}{c}\rfloor)(\lfloor\dfrac{a}{c}\rfloor)i\\+2(\lfloor\dfrac{(a\bmod c)i+ b\bmod c}{c}\rfloor)\lfloor\dfrac{b}{c}\rfloor+2(\lfloor\dfrac{(a\bmod c)i+ b\bmod c}{c}\rfloor)\lfloor\dfrac{a}{c}\rfloor i
\\
=g(n,a\bmod c,b\bmod c,c) + (\lfloor\dfrac{b}{c}\rfloor)^2(n+1) +(\lfloor\dfrac{a}{c}\rfloor)^2\dfrac{n(n+1)(2n+1)}{6} + 2(\lfloor\dfrac{b}{c}\rfloor)(\lfloor\dfrac{a}{c}\rfloor)\dfrac{(n+1)n}{2}
\\
+2\lfloor\dfrac{b}{c}\rfloor f(n,a\bmod c,b\bmod c,c)+2\lfloor\dfrac{a}{c}\rfloor h(n,a\bmod c,b\bmod c,c)
\\
h(n,a,b,c)=\sum\limits_{i=0}^ni(\lfloor\dfrac{(a\bmod c)i+ b\bmod c}{c}\rfloor+\lfloor\dfrac{b}{c}\rfloor+\lfloor\dfrac{a}{c}\rfloor i)\\
=h(n,a\bmod c,b\bmod c,c)+\sum\limits_{i=0}^ni(\lfloor\dfrac{b}{c}\rfloor+\lfloor\dfrac{a}{c}\rfloor i)
\\
=h(n,a\bmod c,b\bmod c,c)+\lfloor\dfrac{b}{c}\rfloor\dfrac{n(n+1)}{2}+(\lfloor\dfrac{a}{c}\rfloor)\dfrac{n(n+1)(2n+1)}{6}
\]
在做下面的情况之前首先推个结论
\[X^2=X(X)+1-X
\\
=2\sum\limits_{i=0}^Xi-X
\]
利用上面的公式展开平方项
\[Case2:a<c,b<c
\\
设M=\lfloor\dfrac{an+b}{c}\rfloor
\\
g(n,a,b,c)=\sum\limits_{i=1}^n(2\sum\limits_{j=1}^{\lfloor\dfrac{ai+ b}{c}\rfloor}j-\lfloor\dfrac{ai+ b}{c}\rfloor)=2\sum\limits_{i=1}^n \sum\limits_{j=1}^{\lfloor\dfrac{ai+ b}{c}\rfloor}j - f(n,a,b,c)\\
=2\sum\limits_{i=1}^n \sum\limits_{j=1}^{M}j[j\le\lfloor\dfrac{ai+ b}{c}\rfloor] - f(n,a,b,c)=2\sum\limits_{i=1}^n \sum\limits_{j=0}^{M-1}(j+1)[cj+c\le ai+ b] - f(n,a,b,c)
\\
=2\sum\limits_{j=0}^{M-1}(j+1)\sum\limits_{i=1}^n [cj-b-c\le ai] - f(n,a,b,c)=2\sum\limits_{j=0}^{M-1}(j+1)(n-\sum\limits_{i=1}^n [cj-b-c-1\ge ai]) - f(n,a,b,c)
\\
=2\sum\limits_{j=0}^{M-1}(j+1)(n-\lfloor\dfrac{cj+c-b-1}{a}\rfloor) - f(n,a,b,c)
\\
=2\sum\limits_{j=0}^{M-1}nj+n-j\lfloor\dfrac{cj+c-b-1}{a}\rfloor-\lfloor\dfrac{cj+c-b-1}{a}\rfloor - f(n,a,b,c)
\\
=2[n\dfrac{M(M-1)}{2}+nM-h(M-1,c,c-b-1,a)-f(M-1,c,c-b-1,a)]-f(n,a,b,c)
\\
=nM(M+1)-2h(M-1,c,c-b-1,a)-2f(M-1,c,c-b-1,a)-f(n,a,b,c)
\\
h(n,a,b,c)=\sum\limits_{i=0}^ni\sum_{j=1}^M[j\le\lfloor\dfrac{ai+ b}{c}\rfloor]=\sum\limits_{i=0}^ni\sum_{j=0}^{M-1}[cj+c-b-1< ai]
\\
=\sum_{j=0}^{M-1}(\dfrac{n(n+1)}{2}-\dfrac{\lfloor\dfrac{cj+c-b-1}{a}\rfloor(\lfloor\dfrac{cj+c-b-1}{a}\rfloor+1)}{2})\\=\sum_{j=0}^{M-1}(\dfrac{n(n+1)}{2}-\dfrac{(\lfloor\dfrac{cj+c-b-1}{a}\rfloor)^2}{2}-\dfrac{(\lfloor\dfrac{cj+c-b-1}{a}\rfloor)}{2})\\=\dfrac{Mn(n+1)}{2}-\dfrac{g(M-1,c,c-b-1,a)}{2}-\dfrac{f(M-1,c,c-b-1,a)}{2}
\]
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int MOD=998244353;
int Ex_x,Ex_y;
int Pow(int a,int b,int p)
{
int base=a;
int res=1;
while(b)
{
if(b&1){
res=((long long)res*base)%p;
}
base=((long long)base*base)%p;
b>>=1;
}
return res;
}
int inv(int a,int p)
{
return Pow(a,p-2,p);
}
int inv2,inv6;
struct Euclid_Like{
int h,f,g;
Euclid_Like()
{
h=0;
f=0;
g=0;
}
void print()
{
printf("%lld %lld %lld\n",f,g,h);
}
};
Euclid_Like cal(int n,int a,int b,int c)
{
Euclid_Like Res;
int Sumf=0;
int Sumh=0;
int Sumg=0;
if(a==0)
{
Sumf=(n+1)%MOD;
Sumf=(Sumf*(b/c))%MOD;
Sumg=(n+1)%MOD;
Sumg=(Sumg*(b/c))%MOD;
Sumg=(Sumg*(b/c))%MOD;
Sumh=(b/c)%MOD;
Sumh=(Sumh*n)%MOD;
Sumh=(Sumh*(n+1))%MOD;
Sumh=(Sumh*inv2)%MOD;
Res.f=Sumf;
Res.g=Sumg;
Res.h=Sumh;
// printf("%lld %lld %lld %lld\n" ,n,a,b,c);
// Res.print();
return Res;
}
if((a>=c)||(b>=c))
{
Euclid_Like Lax=cal(n,a%c,b%c,c);
Sumf=((a/c)*(n%MOD))%MOD;
Sumf=(Sumf*((n+1)%MOD))%MOD;
Sumf=(Sumf*inv2)%MOD;
Sumf=(Sumf+(((n+1)%MOD)*(b/c))%MOD)%MOD;
Sumf=(Sumf+Lax.f)%MOD;
Sumg=((b/c)*(b/c))%MOD;
Sumg=(Sumg*(n+1)%MOD)%MOD;
// printf("%lld-\n",Sumg);
int Nssx;
Nssx=((a/c)*(a/c))%MOD;
Nssx=(Nssx*(n)%MOD)%MOD;
Nssx=(Nssx*(n+1)%MOD)%MOD;
Nssx=(Nssx*(2*n+1)%MOD)%MOD;
Nssx=(Nssx*inv6)%MOD;
Sumg=(Sumg+Nssx)%MOD;
// printf("%lld--\n",Sumg);
Nssx=((n+1)%MOD)%MOD;
Nssx=(Nssx*n)%MOD;
Nssx=(Nssx*(b/c))%MOD;
Nssx=(Nssx*(a/c))%MOD;
Sumg=(Sumg+Nssx)%MOD;
// printf("%lld----\n",Sumg);
Nssx=Lax.f;
Nssx=(Nssx*2)%MOD;
Nssx=(Nssx*(b/c))%MOD;
Sumg=(Sumg+Nssx)%MOD;
Nssx=Lax.h;
Nssx=(Nssx*2)%MOD;
Nssx=(Nssx*(a/c))%MOD;
Sumg=(Sumg+Nssx)%MOD;
Nssx=Lax.g;
Sumg=(Sumg+Nssx)%MOD;
Nssx=Lax.h;
Sumh=Nssx;
// printf("%lld-\n",Sumh);
Nssx=(b/c);
Nssx=(Nssx*n)%MOD;
Nssx=(Nssx*(n+1)%MOD)%MOD;
Nssx=(Nssx*inv2)%MOD;
Sumh=(Sumh+Nssx)%MOD;
// printf("%lld--\n",Sumh);
Nssx=(a/c);
Nssx=(Nssx*(n)%MOD)%MOD;
Nssx=(Nssx*(n+1)%MOD)%MOD;
Nssx=(Nssx*(2*n+1)%MOD)%MOD;
Nssx=(Nssx*inv6)%MOD;
Sumh=(Sumh+Nssx)%MOD;
// printf("%lld---\n",Sumh);
Res.f=Sumf;
Res.g=Sumg;
Res.h=Sumh;
// printf("%lld %lld %lld %lld\n" ,n,a,b,c);
// Res.print();
return Res;
}
else
{
int M=(a*n+b)/c;
// printf("%lldXSHWOSHINIDEGOU\n",M);
Euclid_Like Lax=cal(M-1,c,c-b-1,a);
Sumf=(n*(M)%MOD)%MOD;
Sumf=(Sumf-Lax.f+MOD)%MOD;
Sumg=(n*(M)%MOD)%MOD;
Sumg=(Sumg*(M+1)%MOD)%MOD;
// printf("%lld-\n",Sumg);
int Nssx;
Nssx=Lax.h;
Nssx=(Nssx*2)%MOD;
Sumg=(Sumg-Nssx+MOD)%MOD;
// printf("%lld--\n",Sumg);
Nssx=Lax.f;
Nssx=(Nssx*2)%MOD;
Sumg=(Sumg-Nssx+MOD)%MOD;
// printf("%lld---\n",Sumg);
Nssx=Sumf;
Sumg=(Sumg-Nssx+MOD)%MOD;
// printf("%lld---\n",Sumg);
Sumh=(n*(M)%MOD)%MOD;
Sumh=(Sumh*(n+1)%MOD)%MOD;
Nssx=Lax.g;
Sumh=(Sumh-Nssx+MOD)%MOD;
Nssx=Lax.f;
Sumh=(Sumh-Nssx+MOD)%MOD;
Sumh=(Sumh*inv2)%MOD;
Res.f=Sumf;
Res.g=Sumg;
Res.h=Sumh;
// printf("%lld %lld %lld %lld\n" ,n,a,b,c);
// Res.print();
return Res;
}
}
int t;
int n,a,b,c;
signed main()
{
inv2=inv(2,MOD);
inv6=inv(6,MOD);
scanf("%lld",&t);
while(t--)
{
scanf("%lld %lld %lld %lld",&n,&a,&b,&c);
Euclid_Like RP=cal(n,a,b,c);
RP.print();
}
}
//1
//1 6 2 3
关于阶与原根
阶
设\(x\)为使得\(a^x\equiv1(\bmod m)\)最小的\(x\)，且\(\gcd(a,m)=1\),称\(x\)为\(a\)在\(\bmod m\)意义下的阶,记作\(Ord_ma=x\)
\(Quality1:\)\(a^x\equiv1(\bmod m)\)的充要条件是\(Ord_ma|x\)
充分性:\(a^x\equiv1\equiv a^{Ord_ma}(\bmod m)\)
\(m|a^{Ord_ma}-a^{x}\Rightarrow m|a^x(a^{Ord_ma-x}-1)\)
\(\because\gcd(a,m)=1,\gcd(a^x,m)=1\)
\(\therefore m|a^{Ord_ma-x}-1,a^{Ord_ma-x}\equiv1(\bmod m)(???)\)
设\(x=Ord_ma\times p+r(0\leq r<Ord_ma)\)
\(m|a^{x}-a^{{Ord_ma}}\Rightarrow m|a^{{Ord_ma}}(a^{Ord_ma\times (p-1)+r}-1)\)
\(m|a^{Ord_ma\times (p-1)+r}-1\)
\(a^{Ord_ma\times (p-1)+r}\equiv1(\bmod m)\Rightarrow a^r\equiv1(\bmod m)\Rightarrow r=0\)
必要性:设\(x=Ord_ma\times p\)
\(a^x\equiv a^{Ord_ma\times p}\equiv1(\bmod m)\)
推论:\(Ord_ma|\phi(m)\),显然
\(Quality2:若a\equiv b(\bmod m),则Ord_ma=Ord_mb\)
\(m|b-a\)设\(km=b-a,b=km+a\)
\((km+a)^{Ord_mb}\equiv1(\bmod m)\)
考虑展开二项式\(\Rightarrow a^{Ord_mb}\equiv1(\bmod m)\Rightarrow Ord_ma|Ord_mb\)
同理\(Ord_mb|Ord_ma\),合并即\(Ord_ma=Ord_mb\)
\(Quality3:a^p\equiv a^q(\bmod m)的充要条件是p\equiv q(\bmod Ord_ma)\)
充分性:\(a^{Ord_ma}\equiv1(\bmod m)\)
设\(p=Ord_ma\times k_p+r_p,q=Ord_ma\times k_q+r_q\)
\(a^{r_p}\equiv a^{r_q}(\bmod m)\)
\(r_p,r_q<Ord_ma\),根据\(Ord\)的定义\(r_p=r_q\)
必要性:\(a^{p}\equiv a^{p\bmod Ord_ma}\equiv a^{q\bmod Ord_ma}\equiv a^{q}\)
\(Quality4:a^0,a^1,a^2...a^{Ord_ma-1}\)构成一个关于\(m\)的剩余系
\(即证\forall i,j< Ord_ma,a^i\not\equiv a^j(\bmod m)\)
反证法
考虑\(\exists i<j< Ord_ma,a^i\equiv a^j(\bmod m)\)
\(m|a^j-a^i\Rightarrow m|a^i(a^{j-i}-1)\)
\(\gcd(a,m)=1\Rightarrow m|(a^{j-i}-1)\Rightarrow a^{j-i}\equiv1(\bmod m)\)
\(j-i<Ord_ma\),与\(Ord\)定义不符
\(Quality5:Ord_ma=Ord_m{a^{-1}}\)
\((aa^{-1})^{Ord_ma}\equiv (a^{-1})^{Ord_ma}\Rightarrow Ord_m{-a}|Ord_ma\)
同理\(Ord_m{a}|Ord_m{-a}\Rightarrow Ord_m{a}=Ord_m{-a}\)
\(Quality6:设x=Ord_ma,则Ord_ma^t=\dfrac{x}{gcd(x,t)}\)
设\(k=Ord_ma^t\),\(a^{kt}\equiv1(\bmod m)\)
\(a^x\equiv1(\bmod m),x|kt\)
即最小的\(k\),使得\(x|kt\),\(k=\dfrac{x}{gcd(x,t)}\)
\(Quality7:若w|m,则Ord_wa|Ord_ma\)
\(a^{Ord_ma}\equiv1(\bmod m)\)
\(a^{Ord_ma}=km+1=kdw+1\)
\(a^{Ord_ma}\equiv 1(\bmod w)\Rightarrow Ord_wa|Ord_ma\)
\(Quality8:若m\perp n,a\perp n,m\perp n,Ord_{nm}a=lcm(Ord_ma,Ord_na)\)
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^m(\gcd(i,j)^k)=\sum_{p}p^k\sum\limits_{i=1}^n\sum\limits_{j=1}^m([gcd(i,j)=p])\\
=\sum_{p}p^k\sum\limits_{i=1}^{\lfloor\dfrac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\dfrac{n}{p}\rfloor}([gcd(i,j)=1])=\sum_{p}p^k\sum\limits_{i=1}^{\lfloor\dfrac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\dfrac{m}{p}\rfloor}\sum\limits_{d|gcd(i,j)}\mu(d)
\\
=\sum_pp^k\sum_d\mu(d)\lfloor\dfrac{n}{dp}\rfloor\lfloor\dfrac{m}{dp}\rfloor
\\
设T=dp
\\
=\sum\limits_{T}\lfloor\dfrac{n}{T}\rfloor\lfloor\dfrac{m}{T}\rfloor\sum\limits_{d|T}\mu(d)(\dfrac{T}{d})^k
\\
设f(n)=\sum\limits_{d|n}\mu(d)(\dfrac{n}{d})^k,n=\prod_{i=1}^kp_i^{a_i}
\\
考虑x=\prod_{i=1}^kpx_i^{ax_i},y=\prod_{i=1}^kpy_i^{ay_i},x\perp y
\\
f(xy)=\sum\limits_{d|xy}\mu(d)(\dfrac{xy}{d})^k
\\
f(y)=\sum\limits_{d|y}\mu(d)(\dfrac{y}{d})^k
\\
f(x)=\sum\limits_{d|x}\mu(d)(\dfrac{x}{d})^k
\\
考虑x\perp y,x,y的质因子互不相同，所以D_{xy}=\left\{d|(d|xy)\right\}=\left\{d|(d=d_xd_y,d_x|x,d_y|y)\right\}
\\
由于\mu是积性函数
\\
所以f为积性函数(就是狄利克雷卷积的形式)
\\
f(p)=p^k-1,f(p^x)=p^{kx}-p^{(x-1)k}
\\
x\in prime,p_t=x,f(nx)=\prod_{i} f(p_i^{a_i})\dfrac{p_t^{ka_t+k}-p_t^{ka_t}}{p_t^{ka_t}-p_t^{ka_t-k}}=\prod_{i} f(p_i^{a_i})p_t^k
\]
\[d(xy)=\sum\limits_{d|xy}1=\sum_d[d|xy]\\
=\sum_d[\dfrac{d}{gcd(d,y)}|x]
\\
令p=\dfrac{d}{gcd(d,y)}
\\
=\sum\limits_{p|x}\sum_{d}[\dfrac{d}{\gcd(d,y)}=p]
\\
令k=gcd(d,y)
\\
=\sum\limits_{p|x}\sum\limits_{k|y}(\gcd(pk,y)=k)=\sum\limits_{p|x}\sum\limits_{k|y}(\gcd(p,\dfrac{y}{k})=1)=\sum\limits_{p|x}\sum\limits_{k|y}(\gcd(p,k)=1)
\\
d(xyz),令w=xy
\\
d(xyz)=\sum\limits_{p|w}\sum\limits_{k|z}[\gcd(p,k)=1]=\sum\limits_{p|x}\sum\limits_{q|y}\sum\limits_{k|z}[\gcd(p,q )=1][\gcd(p,k )=1][\gcd(k,q)=1]
\\
\sum\limits_{x=1}^a\sum\limits_{y=1}^b\sum\limits_{z=1}^cd(xyz)=\sum\limits_{x=1}^a\sum\limits_{y=1}^b\sum\limits_{z=1}^c\sum\limits_{p|x}\sum\limits_{q|y}\sum\limits_{k|z}[\gcd(p,q )=1][\gcd(p,k )=1][\gcd(k,q)=1]\\
=\sum\limits_{x=1}^a\sum\limits_{y=1}^b\sum\limits_{z=1}^c\sum\limits_{p|x}\sum\limits_{q|y}\sum\limits_{k|z}[\gcd(p,q )=1][\gcd(p,k )=1]\sum\limits_{d|k,d|q}\mu(d)
\\
=\sum\limits_{x=1}^a\sum\limits_{y=1}^b\sum\limits_{z=1}^c\sum\limits_{p|x}\sum\limits_{d}\mu(d)\sum\limits_{d|q}\sum\limits_{d|k}[\gcd(p,q )=1][\gcd(p,k )=1][q|y][k|z]
\\
=\lfloor\dfrac{a}{p}\rfloor\lfloor\dfrac{b}{q}\rfloor\lfloor\dfrac{c}{k}\rfloor\sum\limits_{p}\sum\limits_{d}\mu(d)\sum\limits_{d|q}\sum\limits_{d|k}[\gcd(p,q )=1][\gcd(p,k )=1]
\\
=\lfloor\dfrac{a}{p}\rfloor\sum_{p}\sum\limits_{d}\mu(d)\sum_{x=1}\lfloor\dfrac{b}{xd}\rfloor[gcd(p,xd)=1]\sum_{y=1}\lfloor\dfrac{c}{yd}\rfloor[gcd(p,yd)=1]
\\
\sum_{x=1}\lfloor\dfrac{b}{xd}\rfloor[gcd(p,xd)=1],\sum_{y=1}\lfloor\dfrac{c}{yd}\rfloor[gcd(p,yd)=1]可以暴力求解(推了这麽久优化了一个根号。。。)
\]
