这篇博客探讨了数论中的一些常见问题，如莫比乌斯函数之和、欧拉函数之和、最大公约数与最小公倍数的求和等。通过介绍杜教筛、分块方法等技巧，阐述了如何利用数学方法，如反演和积性函数，解决这些数论问题。同时，文章提供了51Nod和洛谷等平台的相关题目作为实践案例。
上套路
-
反演
F ( n ) = ∑ d ∣ n f ( d ) = > f ( n ) = ∑ d ∣ n μ ( d ) F ( n d ) F(n)=\sum\limits_{d|n}f(d)=>f(n)=\sum\limits_{d|n}\mu(d)F(\frac{n}d) F(n)=d∣n∑f(d)=>f(n)=d∣n∑μ(d)F(dn)
F ( n ) = ∑ n ∣ d f ( d ) = > f ( n ) = ∑ n ∣ d μ ( d n ) F ( d ) F(n)=\sum\limits_{n|d}f(d)=>f(n)=\sum\limits_{n|d}\mu(\frac{d}n)F(d) F(n)=n∣d∑f(d)=>f(n)=n∣d∑μ(nd)F(d) -
μ 与 ϕ \mu与\phi μ与ϕ
∑ d ∣ n μ ( d ) = [ n = = 1 ] \sum\limits_{d|n}\mu(d)=[n==1] d∣n∑μ(d)=[n==1]
∑ d ∣ n ϕ ( d ) = n \sum\limits_{d|n}\phi(d)=n d∣n∑ϕ(d)=n
ϕ ( n ) n = ∑ d ∣ n μ ( d ) d \frac{\phi(n)}{n}=\sum\limits_{d|n}\frac{\mu(d)}{d} nϕ(n)=d∣n∑dμ(d)
ϕ ( n ) = ∑ d ∣ n μ ( d ) n d \phi(n)=\sum\limits_{d|n}\mu(d)\frac{n}{d} ϕ(n)=d∣n∑μ(d)dn -
σ 约 数 和 \sigma约数和 σ约数和
σ ( n ) = ∑ d ∣ n d \sigma(n)=\sum\limits_{d|n}d σ(n)=d∣n∑d
∑ i = 1 n σ ( i ) = ∑ i = 1 n i ∗ ⌊ n i ⌋ \sum\limits_{i=1}^{n}\sigma(i)=\sum\limits_{i=1}^ni*\left \lfloor \frac{n}{i} \right \rfloor i=1∑nσ(i)=i=1∑ni∗⌊in⌋
σ k ( i ∗ j ) = ∑ x ∣ i n ∑ y ∣ j n [ g c d ( x , y ) = = 1 ] x k ∗ j k / y k \sigma_k(i*j)=\sum\limits_{x|i}^n\sum\limits_{y|j}^n[gcd(x,y)==1]x^k*j^k/y^k σk(i∗j)=x∣i∑ny∣j∑n[gcd(x,y)==1]xk∗jk/yk
k 等 于 0 就 是 约 数 个 数 和 k等于0就是约数个数和 k等于0就是约数个数和 -
杜 教 筛 杜教筛 杜教筛
g 1 S ( n ) = ∑ i = 1 n h ( i ) − ∑ i = 2 n g ( i ) S ( ⌊ n i ⌋ ) g_1S(n)=\sum\limits_{i=1}^nh(i)-\sum\limits_{i=2}^ng(i)S(\left \lfloor \frac{n}{i} \right \rfloor) g1S(n)=i=1∑nh(i)−i=2∑ng(i)S(⌊in⌋)
h = f ∗ g h=f*g h=f∗g
∑ i = 1 n i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum\limits_{i=1}^n{i^2}=\frac{n(n+1)(2n+1)}{6} i=1∑ni2=6n(n+1)(2n+1)
∑ i = 1 n i 3 = ( n ( n + 1 ) 2 ) 2 \sum\limits_{i=1}^n{i^3}=(\frac{n(n+1)}{2})^2 i=1∑ni3=(2n(n+1))2 -
g c d ( a n − b n , a m − b m ) = a g c d ( n , m ) − b g c d ( n , m ) gcd(a^n-b^n,a^m-b^m)=a^{gcd(n,m)}-b^{gcd(n,m)} gcd(an−bn,am−bm)=agcd(n,m)−bgcd(n,m)
g c d ( a n − 1 , a m − 1 ) = a g c d ( n , m ) − 1 gcd(a^n-1,a^m-1)=a^{gcd(n,m)}-1 gcd(an−1,am−1)=agcd(n,m)−1 -
g c d ( F i b m , F i b n ) = F i b g c d ( n , m ) gcd(Fib_m,Fib_n)=Fib_{gcd(n,m)} gcd(Fibm,Fibn)=Fibgcd(n,m)
-
∑ i = 1 n 1 i ∑ j = 1 i l c m ( i , j ) = 1 2 ( ∑ d = 1 n ∑ i = 1 n d i ϕ ( i ) + n ) \sum\limits_{i=1}^n\frac{1}{i}\sum\limits_{j=1}^{i}lcm(i,j)=\frac{1}{2}(\sum\limits_{d=1}^n\sum\limits_{i=1}^{\frac{n}{d}}i\phi(i)+n) i=1∑ni1j=1∑ilcm(i,j)=21(d=1∑ni=1∑dniϕ(i)+n)
51Nod 1244 - 莫比乌斯函数之和
51Nod 1239 - 欧拉函数之和
51Nod 1237 - 最大公约数之和 V3
51Nod 1238 - 最小公倍数之和 V3
51Nod 1227 - 平均最小公倍数
51Nod 1220 - 约数之和
HDU 6706 - huntian oy
洛谷 P2257 - YY的GCD
51Nod 1244 - 莫比乌斯函数之和
回目录
杜教筛,
h
=
μ
∗
g
h=\mu*g
h=μ∗g卷上恒等函数
I
(
n
)
,
即
g
=
1
I(n),即g=1
I(n),即g=1
则
∑
i
=
1
n
h
(
i
)
=
1
\sum\limits_{i=1}^nh(i)=1
i=1∑nh(i)=1
S
(
n
)
=
1
−
∑
i
=
2
n
S
(
⌊
n
i
⌋
)
S(n)=1-\sum\limits_{i=2}^nS(\left \lfloor \frac{n}{i} \right \rfloor)
S(n)=1−i=2∑nS(⌊in⌋)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=5e6+9;
int su[N],mu[N],num,vis[N];
unordered_map<LL,LL> mp;
void init()
{
mu[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,mu[i]=-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0)break;
mu[i*su[j]]=-mu[i];
}
}
for(int i=1;i<N;++i)mu[i]+=mu[i-1];
}
LL djs(LL n)
{
if(n<N)return mu[n];
if(mp[n])return mp[n];
LL ans=1;
for(LL L=2,R;L<=n;L=R+1){
R=n/(n/L);
ans-=(R-L+1)*djs(n/L);
}
return mp[n]=ans;
}
int main()
{
init();
LL L,R;
scanf("%lld%lld",&L,&R);
printf("%lld\n",djs(R)-djs(L-1));
return 0;
}
51Nod 1239 - 欧拉函数之和
回目录
杜教筛,
h
=
ϕ
∗
g
h=\phi*g
h=ϕ∗g卷上恒等函数
I
(
n
)
,
即
g
=
1
,
则
h
(
n
)
=
n
I(n),即g=1,则h(n)=n
I(n),即g=1,则h(n)=n
则
∑
i
=
1
n
h
(
i
)
=
(
n
∗
n
+
1
)
/
2
\sum\limits_{i=1}^nh(i)=(n*n+1)/2
i=1∑nh(i)=(n∗n+1)/2
S
(
n
)
=
(
n
∗
n
+
1
)
/
2
−
∑
i
=
2
n
S
(
⌊
n
i
⌋
)
S(n)=(n*n+1)/2-\sum\limits_{i=2}^nS(\left \lfloor \frac{n}{i} \right \rfloor)
S(n)=(n∗n+1)/2−i=2∑nS(⌊in⌋)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=5e6+9;
int su[N],phi[N],num,vis[N];
unordered_map<LL,LL> mp;
const LL mod=1e9+7;
const LL inv2=500000004;
void init()
{
phi[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,phi[i]=i-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0){
phi[i*su[j]]=phi[i]*su[j];
break;
}
phi[i*su[j]]=phi[i]*phi[su[j]];
}
}
for(int i=1;i<N;++i)phi[i]=(phi[i]+phi[i-1])%mod;
}
LL sum(LL n)
{
n%=mod;
return n*(n+1)%mod*inv2%mod;
}
LL djs(LL n)
{
if(n<N)return phi[n];
if(mp[n])return mp[n];
LL ans=sum(n);
for(LL L=2,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans-(R-L+1)%mod*djs(n/L)%mod+mod)%mod;
}
return mp[n]=ans;
}
int main()
{
init();
LL n;
scanf("%lld",&n);
printf("%lld\n",djs(n));
return 0;
}
51Nod 1237 - 最大公约数之和 V3
回目录
∑ i = 1 n ∑ j = 1 n g c d ( i , j ) = ∑ i = 1 n ∑ j = 1 n ∑ d ∣ g c d ( i , j ) ϕ ( d ) = ∑ d = 1 n ϕ ( d ) ∑ i = 1 n ∑ j = 1 n [ d ∣ g c d ( i , j ) ] = ∑ d = 1 n ϕ ( d ) ∑ i = 1 n d ∑ j = 1 n d 1 = ∑ d = 1 n ϕ ( d ) ⌊ n i ⌋ 2 \begin{aligned}\sum\limits_{i=1}^n\sum\limits_{j=1}^ngcd(i,j) &=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\sum\limits\limits_{d|gcd(i,j)}\phi(d)\\ &=\sum\limits_{d=1}^n\phi(d)\sum\limits_{i=1}^n\sum\limits_{j=1}^n[d|gcd(i,j)]\\ &=\sum\limits_{d=1}^n\phi(d)\sum\limits_{i=1}^{\frac{n}d}\sum\limits_{j=1}^{\frac{n}d}1\\ &=\sum\limits_{d=1}^n\phi(d){\left \lfloor \frac{n}{i} \right \rfloor}^2\\ \end{aligned} i=1∑nj=1∑ngcd(i,j)=i=1∑nj=1∑nd∣gcd(i,j)∑ϕ(d)=d=1∑nϕ(d)i=1∑nj=1∑n[d∣gcd(i,j)]=d=1∑nϕ(d)i=1∑dnj=1∑dn1=d=1∑nϕ(d)⌊in⌋2
杜教筛加分块即可
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=5e6+9;
int su[N],phi[N],num,vis[N];
unordered_map<LL,LL> mp;
const LL mod=1e9+7;
const LL inv2=500000004;
void init()
{
phi[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,phi[i]=i-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0){
phi[i*su[j]]=phi[i]*su[j];
break;
}
phi[i*su[j]]=phi[i]*phi[su[j]];
}
}
for(int i=1;i<N;++i)phi[i]=(phi[i]+phi[i-1])%mod;
}
LL sum(LL n)
{
n%=mod;
return n*(n+1)%mod*inv2%mod;
}
LL djs(LL n)
{
if(n<N)return phi[n];
if(mp[n])return mp[n];
LL ans=sum(n);
for(LL L=2,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans-(R-L+1)%mod*djs(n/L)%mod+mod)%mod;
}
return mp[n]=ans;
}
LL outside(LL n)
{
LL ans=0;
for(LL L=1,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans+(djs(R)-djs(L-1)+mod)%mod*((n/L)%mod)*((n/L)%mod)+mod)%mod;
}
return ans;
}
int main()
{
init();
LL n;
scanf("%lld",&n);
printf("%lld\n",outside(n));
return 0;
}
51Nod 1238 - 最小公倍数之和 V3
回目录
∑ i = 1 n ∑ j = 1 n l c m ( i , j ) = ∑ d = 1 n ∑ i = 1 n ∑ j = 1 n i j d [ d = = g c d ( i , j ) ] = ∑ d = 1 n d ∑ i = 1 n / d ∑ j = 1 n / d i j [ 1 = = g c d ( i , j ) ] = ∑ d = 1 n d ∑ i = 1 n d i 2 ϕ ( i ) \begin{aligned}\sum\limits_{i=1}^n\sum\limits_{j=1}^nlcm(i,j) &=\sum\limits_{d=1}^n\sum\limits_{i=1}^n\sum\limits_{j=1}^n\frac{ij}{d}[d==gcd(i,j)] \\ &=\sum\limits_{d=1}^nd\sum\limits_{i=1}^{n/d}\sum\limits_{j=1}^{n/d}ij[1==gcd(i,j)]\\ &=\sum\limits_{d=1}^nd\sum\limits_{i=1}^{\frac{n}d}i^2\phi(i)\\ \\ \end{aligned} i=1∑nj=1∑nlcm(i,j)=d=1∑ni=1∑nj=1∑ndij[d==gcd(i,j)]=d=1∑ndi=1∑n/dj=1∑n/dij[1==gcd(i,j)]=d=1∑ndi=1∑dni2ϕ(i)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=5e6+9;
const LL mod=1e9+7;
const LL inv2=500000004;
const LL inv6=166666668;
int su[N],num,vis[N];
LL phi[N];
unordered_map<LL,LL> mp;
void init()
{
phi[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,phi[i]=i-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0){
phi[i*su[j]]=phi[i]*su[j];
break;
}
phi[i*su[j]]=phi[i]*phi[su[j]];
}
}
for(int i=1;i<N;++i)phi[i]=(phi[i]*i%mod*i%mod+phi[i-1])%mod;
}
LL sum(LL n)
{
n%=mod;
return n*(n+1)%mod*inv2%mod;
}
LL get(LL n)
{
n%=mod;
return n*(n+1)%mod*(2*n+1)%mod*inv6%mod;
}
LL djs(LL n)
{
if(n<(LL)N)return phi[n];
if(mp[n])return mp[n];
LL ans=sum(n);
ans=ans*ans%mod;
for(LL L=2,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans-(get(R)-get(L-1)+mod)%mod*djs(n/L)%mod+mod)%mod;
}
return mp[n]=ans;
}
LL outside(LL n)
{
LL ans=0;
for(LL L=1,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans+(sum(R)-sum(L-1)+mod)%mod*djs(n/L)%mod+mod)%mod;
}
return ans;
}
int main()
{
init();
LL n;
scanf("%lld",&n);
printf("%lld\n",outside(n));
return 0;
}
51Nod 1227 - 平均最小公倍数
回目录
∑ i = 1 n 1 i ∑ j = 1 i l c m ( i , j ) = ∑ d = 1 n ∑ i = 1 n ∑ j = 1 i j d [ d = = g c d ( i , j ) ] = ∑ d = 1 n ∑ i = 1 n ∑ j = 1 i [ 1 = = g c d ( i , j ) ] = ∑ d = 1 n ∑ i = 1 n d i ϕ ( i ) + [ i = = 1 ] 2 = ∑ d = 1 n ∑ i = 1 n i ϕ ( i ) + n 2 \begin{aligned}\sum\limits_{i=1}^n\frac{1}{i}\sum\limits_{j=1}^ilcm(i,j) &=\sum\limits_{d=1}^n\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac{j}{d}[d==gcd(i,j)] \\ &=\sum\limits_{d=1}^n\sum\limits_{i=1}^n\sum\limits_{j=1}^i[1==gcd(i,j)]\\ &=\sum\limits_{d=1}^n\sum\limits_{i=1}^{\frac{n}d}\frac{i\phi(i)+[i==1]}{2}\\ &=\frac{\sum\limits_{d=1}^n\sum\limits_{i=1}^ni\phi(i)+n}{2}\\ \\ \end{aligned} i=1∑ni1j=1∑ilcm(i,j)=d=1∑ni=1∑nj=1∑idj[d==gcd(i,j)]=d=1∑ni=1∑nj=1∑i[1==gcd(i,j)]=d=1∑ni=1∑dn2iϕ(i)+[i==1]=2d=1∑ni=1∑niϕ(i)+n
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=5e6+9;
const LL mod=1e9+7;
const LL inv2=500000004;
const LL inv6=166666668;
int su[N],num,vis[N];
LL phi[N];
unordered_map<LL,LL> mp;
void init()
{
phi[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,phi[i]=i-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0){
phi[i*su[j]]=phi[i]*su[j];
break;
}
phi[i*su[j]]=phi[i]*phi[su[j]];
}
}
for(int i=1;i<N;++i)phi[i]=(phi[i]*i%mod+phi[i-1])%mod;
}
LL sum(LL n)
{
n%=mod;
return n*(n+1)%mod*inv2%mod;
}
LL get(LL n)
{
n%=mod;
return n*(n+1)%mod*(2*n+1)%mod*inv6%mod;
}
LL djs(LL n)
{
if(n<(LL)N)return phi[n];
if(mp[n])return mp[n];
LL ans=get(n);
for(LL L=2,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans-(sum(R)-sum(L-1)+mod)%mod*djs(n/L)%mod+mod)%mod;
}
return mp[n]=ans;
}
LL outside(LL n)
{
LL ans=0;
for(LL L=1,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans+(R-L+1+mod)%mod*djs(n/L)%mod+mod)%mod;
}
return (ans+n)%mod*inv2%mod;
}
int main()
{
init();
LL n,m;
scanf("%lld%lld",&n,&m);
printf("%lld\n",(outside(m)-outside(n-1)+mod)%mod);
return 0;
}
51Nod 1220 - 约数之和
回目录
∑ i = 1 n ∑ j = 1 i d ( i , j ) = ∑ i = 1 n ∑ j = 1 n ∑ x ∣ i ∑ y ∣ j x j y [ 1 = = g c d ( x , y ) ] = ∑ d = 1 n μ ( d ) ∑ i = 1 n ∑ j = 1 n ∑ x ∣ i ∑ y ∣ j x j y [ d ∣ g c d ( x , y ) ] = ∑ d = 1 n μ ( d ) d ∑ i = 1 n d ∑ j = 1 n d ∑ x ∣ i ∑ y ∣ j x j y = ∑ d = 1 n μ ( d ) d ∑ i = 1 n d ∑ x ∣ i x ∑ j = 1 n d ∑ y ∣ j j y = ∑ d = 1 n d μ ( d ) ( ∑ i = 1 n d σ ( i ) ) 2 = ∑ d = 1 n d μ ( d ) ( ∑ i = 1 n d i ⌊ n d i ⌋ ) 2 \begin{aligned}\sum\limits_{i=1}^n\sum\limits_{j=1}^id(i,j) &=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\sum\limits_{x|i}\sum\limits_{y|j}\frac{xj}{y}[1==gcd(x,y)] \\ &=\sum\limits_{d=1}^n\mu(d)\sum\limits_{i=1}^n\sum\limits_{j=1}^n\sum\limits_{x|i}\sum\limits_{y|j}\frac{xj}{y}[d|gcd(x,y)] \\ &=\sum\limits_{d=1}^n\mu(d)d\sum\limits_{i=1}^{\frac{n}d}\sum\limits_{j=1}^{\frac{n}d}\sum\limits_{x|i}\sum\limits_{y|j}\frac{xj}{y}\\ &=\sum\limits_{d=1}^n\mu(d)d\sum\limits_{i=1}^{\frac{n}d}\sum\limits_{x|i}x\sum\limits_{j=1}^{\frac{n}d}\sum\limits_{y|j}\frac{j}{y}\\ &=\sum\limits_{d=1}^nd\mu(d)(\sum\limits_{i=1}^{\frac{n}d}\sigma(i))^2\\ &=\sum\limits_{d=1}^nd\mu(d)(\sum\limits_{i=1}^{\frac{n}d}i{\left \lfloor \frac{n}{di} \right \rfloor})^2\\ \\ \end{aligned} i=1∑nj=1∑id(i,j)=i=1∑nj=1∑nx∣i∑y∣j∑yxj[1==gcd(x,y)]=d=1∑nμ(d)i=1∑nj=1∑nx∣i∑y∣j∑yxj[d∣gcd(x,y)]=d=1∑nμ(d)di=1∑dnj=1∑dnx∣i∑y∣j∑yxj=d=1∑nμ(d)di=1∑dnx∣i∑xj=1∑dny∣j∑yj=d=1∑ndμ(d)(i=1∑dnσ(i))2=d=1∑ndμ(d)(i=1∑dni⌊din⌋)2
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=5e6+9;
const LL mod=1e9+7;
const LL inv2=500000004;
const LL inv6=166666668;
int su[N],num,vis[N],mu[N];
unordered_map<int,LL> mp;
void init()
{
mu[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,mu[i]=-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0)break;
mu[i*su[j]]=-mu[i];
}
}
for(int i=1;i<N;++i)mu[i]=(mu[i]*i+mu[i-1]+mod)%mod;
}
LL sum(int n)
{
return (LL)n*(n+1)%mod*inv2%mod;
}
LL djs(int n)
{
if(n<N)return mu[n];
if(mp[n])return mp[n];
LL ans=1;
for(int L=2,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans-(sum(R)-sum(L-1)+mod)%mod*djs(n/L)%mod+mod)%mod;
}
return mp[n]=ans;
}
LL inside(int n)
{
LL ans=0;
for(int L=1,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans+(sum(R)-sum(L-1)+mod)%mod*(n/L)%mod+mod)%mod;
}
return ans*ans%mod;
}
LL outside(int n)
{
LL ans=0;
for(int L=1,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans+(djs(R)-djs(L-1)+mod)%mod*inside(n/L)%mod+mod)%mod;
}
return ans;
}
int main()
{
init();
int n;
scanf("%d",&n);
printf("%lld\n",outside(n));
return 0;
}
HDU 6706 - huntian oy
回目录
∑ i = 1 n ∑ j = 1 i g c d ( i a − j a , i b − j b ) [ g c d ( i , j ) = = 1 ] = ∑ i = 1 n ∑ j = 1 i ( i − j ) [ g c d ( i , j ) = = 1 ] = ∑ i = 1 n ∑ j = 1 i i [ g c d ( i , j ) = = 1 ] − ∑ i = 1 n ∑ j = 1 i j [ g c d ( i , j ) = = 1 ] = ∑ i = 1 n i ϕ ( i ) − ∑ i = 1 n i ϕ ( i ) + [ n = = 1 ] 2 = 1 + ∑ i = 2 n i ϕ ( i ) − ( 1 + ∑ i = 2 n i ϕ ( i ) 2 ) = ∑ i = 2 n i ϕ ( i ) 2 = ∑ i = 1 n i ϕ ( i ) − 1 2 \begin{aligned}\sum\limits_{i=1}^n\sum\limits_{j=1}^igcd(i^a-j^a,i^b-j^b)[gcd(i,j)==1] &=\sum\limits_{i=1}^n\sum\limits_{j=1}^i(i-j)[gcd(i,j)==1] \\ &=\sum\limits_{i=1}^n\sum\limits_{j=1}^ii[gcd(i,j)==1]- \sum\limits_{i=1}^n\sum\limits_{j=1}^ij[gcd(i,j)==1]\\ &=\sum\limits_{i=1}^ni\phi(i)-\frac{\sum\limits_{i=1}^ni\phi(i)+[n==1]}{2}\\ &=1+\sum\limits_{i=2}^ni\phi(i)-(1+\frac{\sum\limits_{i=2}^ni\phi(i)}{2})\\ &=\frac{\sum\limits_{i=2}^ni\phi(i)}{2}\\ &=\frac{\sum\limits_{i=1}^ni\phi(i)-1}{2}\\ \\ \end{aligned} i=1∑nj=1∑igcd(ia−ja,ib−jb)[gcd(i,j)==1]=i=1∑nj=1∑i(i−j)[gcd(i,j)==1]=i=1∑nj=1∑ii[gcd(i,j)==1]−i=1∑nj=1∑ij[gcd(i,j)==1]=i=1∑niϕ(i)−2i=1∑niϕ(i)+[n==1]=1+i=2∑niϕ(i)−(1+2i=2∑niϕ(i))=2i=2∑niϕ(i)=2i=1∑niϕ(i)−1
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=5e6+9;
const LL mod=1e9+7;
const LL inv2=500000004;
const LL inv6=166666668;
int su[N],num,vis[N];
LL phi[N];
unordered_map<int,LL> mp;
void init()
{
phi[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,phi[i]=i-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0){
phi[i*su[j]]=phi[i]*su[j];
break;
}
phi[i*su[j]]=phi[i]*phi[su[j]];
}
}
for(int i=1;i<N;++i)phi[i]=(phi[i]*i%mod+phi[i-1]+mod)%mod;
}
LL sum(int n)
{
return (LL)n*(n+1)%mod*inv2%mod;
}
LL djs(int n)
{
if(n<N)return phi[n];
if(mp[n])return mp[n];
LL ans=(LL)n*(n+1)%mod*(2*n+1)%mod*inv6%mod;
for(int L=2,R;L<=n;L=R+1){
R=n/(n/L);
ans=(ans-(sum(R)-sum(L-1)+mod)%mod*djs(n/L)%mod+mod)%mod;
}
return mp[n]=ans;
}
int main()
{
init();
int n,T,a,b;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&a,&b);
printf("%lld\n",(djs(n)-1+mod)%mod*inv2%mod);
}
return 0;
}
洛谷 P2257 - YY的GCD
回目录
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\begin{aligned}\sum\limits_{p\in prime}\sum\limits_{i=1}^n\sum\limits_{j=1}^n[gcd(i,j)==p] &=\sum\limits_{p\in prime}\sum\limits_{i=1}^{\frac{n}{p}}\sum\limits_{j=1}^{\frac{n}{p}}[gcd(i,j)==1]\\ &=\sum\limits_{p\in prime}\sum\limits_{d=1}^{min(\frac{n}{p},\frac{m}{p})}\mu(d)\sum\limits_{i=1}^{\frac{n}{p}}\sum\limits_{j=1}^{\frac{n}{p}}[d|gcd(i,j)]\\ &=\sum\limits_{p\in prime}\sum\limits_{d=1}^{min(\frac{n}{p},\frac{m}{p})}\mu(d)\sum\limits_{i=1}^{\frac{n}{pd}}\sum\limits_{j=1}^{\frac{n}{pd}}1\\ &=\sum\limits_{p\in prime}\sum\limits_{d=1}^{min(\frac{n}{p},\frac{m}{p})}\mu(d){\left \lfloor \frac{n}{pd} \right \rfloor}{\left \lfloor \frac{m}{pd} \right \rfloor}\\ \\ \end{aligned}
p∈prime∑i=1∑nj=1∑n[gcd(i,j)==p]=p∈prime∑i=1∑pnj=1∑pn[gcd(i,j)==1]=p∈prime∑d=1∑min(pn,pm)μ(d)i=1∑pnj=1∑pn[d∣gcd(i,j)]=p∈prime∑d=1∑min(pn,pm)μ(d)i=1∑pdnj=1∑pdn1=p∈prime∑d=1∑min(pn,pm)μ(d)⌊pdn⌋⌊pdm⌋
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\begin{aligned}\sum\limits_{p\in prime}\sum\limits_{d=1}^{min(\frac{n}{p},\frac{m}{p})}\mu(d){\left \lfloor \frac{n}{pd} \right \rfloor}{\left \lfloor \frac{m}{pd} \right \rfloor} &=\sum\limits_{T=1}^{min(n,m)}{\left \lfloor \frac{n}{T} \right \rfloor}{\left \lfloor \frac{m}{T} \right \rfloor}\sum\limits_{t|T,t\in prime}\mu(\frac{T}{t}) \\ \end{aligned}
p∈prime∑d=1∑min(pn,pm)μ(d)⌊pdn⌋⌊pdm⌋=T=1∑min(n,m)⌊Tn⌋⌊Tm⌋t∣T,t∈prime∑μ(tT)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e7+9;
const LL mod=1e9+7;
const LL inv2=500000004;
const LL inv6=166666668;
int mu[N],mm[N],su[N],num;
bool vis[N];
void init()
{
mu[1]=1;
for(int i=2;i<N;++i){
if(!vis[i])su[++num]=i,mu[i]=-1;
for(int j=1;j<=num&&i*su[j]<N;++j){
vis[i*su[j]]=1;
if(i%su[j]==0)break;
mu[i*su[j]]=-mu[i];
}
}
for(int i=1;i<=num;++i)
for(int j=1;su[i]*j<N;++j)mm[su[i]*j]+=mu[j];
for(int i=1;i<N;++i)mm[i]+=mm[i-1];
}
LL outside(int n,int m)
{
LL ans=0;
for(int L=1,R;L<=n;L=R+1){
R=min(n/(n/L),m/(m/L));
ans+=(LL)(n/L)*(m/L)*(mm[R]-mm[L-1]);
}
return ans;
}
int main()
{
init();
int T,n,m;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
if(n>m)swap(n,m);
printf("%lld\n",outside(n,m));
}
return 0;
}
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