The meaning of problems
Calculation \ (\ sum_ {i = 1 } ^ {n} \ sum_ {j = 1} ^ {m} lcm (i, j) \)
answer
\[ \begin{aligned}Ans=\sum_{i=1}^{N}\sum_{j=1}^{M}lcm(i,j)&=\sum_{i=1}^{N}\sum_{j=1}^{M}\frac{ij}{gcd(i,j)}=\sum_{d=1}^{N}\sum_{i=1}^{N}\sum_{j=1}^{M}[gcd(i,j)=d]\frac{ij}{d}\\&=\sum_{d=1}^{N}\frac{1}{d}\sum_{i=1}^{N}\sum_{j=1}^{M}[gcd(i,j)=d]ij\\\end{aligned} \]
设：
\[ f(d)=\sum_{i=1}^{N}\sum_{j=1}^{M}[gcd(i,j)=d]ij\\F(n)=\sum_{n|d}f(d)=\sum_{n|i}^{N}\sum_{n|j}^{M}ij=\sum_{i=1}^{\frac{N}{n}}\sum_{j=1}^{\frac{M}{n}}in\cdot jn=n^2\sum_{i=1}^{\frac{N}{n}}\sum_{j=1}^{\frac{M}{n}}ij \]
然后反演 \(f(n)\)：
\[ \begin{aligned}f(n)&=\sum_{n|d}\mu(\frac{d}{n})F(d)=\sum_{n|d}\mu(\frac{d}{n})d^2\sum_{i=1}^{\frac{N}{d}}\sum_{j=1}^{\frac{M}{d}}ij=\sum_{t=1}^{\frac{N}{n}}\mu(t)n^2t^2\sum_{i=1}^{\frac{N}{nt}}\sum_{j=1}^{\frac{M}{nt}}ij\\&=n^2\sum_{t=1}^{\frac{N}{n}}\mu(t)t^2\sum_{i=1}^{\frac{N}{nt}}i\sum_{j=1}^{\frac{M}{nt}}j\end{aligned} \]
然后代回去求 \(Ans\)：
\ [\ Begin {aligned} Ans = \ sum_ {n = 1} ^ {N} \ frac {1} {n} f (n) & = \ sum_ {n = 1} ^ {N} n \ sum_ {t = 1} ^ {\ frac { N} {n}} \ mu (t) t ^ 2 \ sum_ {i = 1} ^ {\ frac {N} {nt}} i \ sum_ {j = 1} ^ { \ frac {M} {nt} } j = \ sum_ {T = 1} ^ {N} T \ sum_ {t | T} \ mu (t) t \ sum_ {i = 1} ^ {\ frac {N} {T}} I \ sum_ {J =. 1} ^ {\ FRAC {M} {T}} J \ End {the aligned} \]
\ (G (T) = \ sum_ {T | T} \ MU (T) t \) I do not really understand how to find, just know that when \ (t, T \) when relatively prime, \ (G (t) \) is a multiplicative function, then \ (g (Tt) = g (T ) G (t) \) , or \ (G (Tt) = G (T) \) .
Code
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
typedef long long LL;
const int N=1e7+10;
const int mod=20101009;
int n,m,vis[N],prime[N],cnt;
LL sum[N],f[N];
void init(int n){
f[1]=1;
for(int i=2;i<=n;i++){
if(!vis[i]) {prime[++cnt]=i;f[i]=(1-i+mod)%mod;}
for(int j=1;1ll*i*prime[j]<=n;j++){
vis[i*prime[j]]=1;
if(i%prime[j]==0) {f[i*prime[j]]=f[i];break;}
f[i*prime[j]]=f[i]*f[prime[j]]%mod;
}
}
for(int i=1;i<=n;i++) sum[i]=(f[i]*i+sum[i-1])%mod;
}
int main(){
init(1e7);
scanf("%d%d",&n,&m);
if(n>m) swap(n,m);
LL ans=0;
for(int l=1,r;l<=n;l=r+1){
r=min(n/(n/l),m/(m/l));
LL temp1=1ll*(n/l+1)*(n/l)/2%mod;
LL temp2=1ll*(m/l+1)*(m/l)/2%mod;
ans=(ans+(sum[r]-sum[l-1]+mod)%mod*temp1%mod*temp2%mod)%mod;
}
printf("%lld\n",ans);
return 0;
}