\[\begin{split} A&=\sum_{i=1}^{n}\sum_{j=1}^n{ijgcd(i,j)} \\ &=\sum_{d=1}^n\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ijd^3[gcd(i,j)=1]\\ &=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij[gcd(i,j)=1] \end{split} \]
设\(f(n)=\sum_{i=1}^{n}\sum_{j=1}^nij[gcd(i,j)=1]\),\(s(n)=\sum_{i=1}^{n}\sum_{j=1}^nij\)
则
\[\begin{split}A&=\sum_{d=1}^nd^3f(\lfloor\frac{n}{d}\rfloor)\end{split}\]
and
\[\sum_{i=1}^{n}i^3=(\sum_{i=1}^ni)^2=[\frac{n(n+1)}{2}]^2\]
Have
\[\begin{split}s(n)&=\sum_{i=1}^{n}\sum_{j=1}^nij=[\frac{n(n+1)}{2}]^2\\ &=\sum_{i=1}^{n}\sum_{j=1}^nij \\ &=\sum_{t=1}^{n}\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{t}\rfloor}t^2ij[gcd(i,j)=1] \\ &=\sum_{t=1}^{n}t^2f(\lfloor\frac{n}{t}\rfloor)=f(n)+\sum_{t>1}t^2f(\lfloor\frac{n}{t}\rfloor) \end{split}\]
\[f(n)=s(n)-\sum_{t>1}t^2f(\lfloor\frac{n}{t}\rfloor)=[\frac{n(n+1)}{2}]^2-\sum_{t>1}t^2f(\lfloor\frac{n}{t}\rfloor)\]
On the block number to
\ [\ sum_ {i = 1 } ^ {n} i ^ 2 = \ frac {n (n + 1) (2n + 1)} {6} \]
预处理 \(f(n)\)
\[f(n)=\sum_{i=1}^{n}\sum_{j=1}^nij[gcd(i,j)=1]=f(n-1)+2\sum_{i=1}^{n}in[gcd(i,n)=1]\]
而
\ [\ sum_ {i = 1} ^ or [GCD (i, n) = 1] n = \ frac {\ varphi (n)} {2} \]
故
\, [f, (n), = f, (n-the 1), the 2 +, n ^ \ varphi, (n), \]
Mentally disabled athletes accidentally n also die out so tune more than half an hour
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 5e6+5;
int i2,i4,i6,mod,n;
bool isNotPrime[N + 5];
int mu[N + 5], phi[N + 5], primes[N + 5], cnt;
inline void euler() {
isNotPrime[0] = isNotPrime[1] = true;
mu[1] = 1;
phi[1] = 1;
for (int i = 2; i <= N; i++) {
if (!isNotPrime[i]) {
primes[++cnt] = i;
mu[i] = -1;
phi[i] = i - 1;
}
for (int j = 1; j <= cnt; j++) {
int t = i * primes[j];
if (t > N) break;
isNotPrime[t] = true;
if (i % primes[j] == 0) {
mu[t] = 0;
phi[t] = phi[i] * primes[j];
break;
} else {
mu[t] = -mu[i];
phi[t] = phi[i] * (primes[j] - 1);
}
}
}
}
inline void exgcd(int a,int b,int &x,int &y) {
if(!b) {
x=1,y=0;
return;
}
exgcd(b,a%b,x,y);
int t=x;
x=y,y=t-(a/b)*y;
}
inline int inv(int a,int b) {
int x,y;
return exgcd(a,b,x,y),(x%b+b)%b;
}
int sum[N+1];
int h(int n) { n%=mod;
return n*n%mod*(n+1)%mod*(n+1)%mod*i4%mod;
}
int u(int n) { n%=mod;
return n*(n+1)%mod*(2ll*n+1)%mod*i6%mod;
}
map<int,int> mp;
int S(int n) {
if(n<N) return sum[n];
if(mp[n]) return mp[n];
int l=2, r, ans=h(n);
while(l<=n) {
r=n/(n/l);
ans -= (((u(r)-u(l-1))%mod+mod)%mod*S(n/l))%mod;
ans %= mod;
ans += mod;
ans %= mod;
l=r+1;
}
mp[n]=ans;
return ans;
}
int read() {
long long t;
cin>>t;
return t;
}
signed main() {
ios::sync_with_stdio(false);
mod=read();
n=read();
i2=inv(2,mod);
i4=inv(4,mod);
i6=inv(6,mod);
euler();
sum[0]=phi[0];
for(int i=1;i<=N;i++) sum[i]=(sum[i-1]+phi[i]%mod*i%mod*i%mod)%mod;
int ans = 0, l=1, r;
while(l<=n) {
//cout<<l<<endl;
r=(n/(n/l));
ans += h(n/l) * ((S(r)-S(l-1)+mod)%mod) % mod;
ans %= mod;
l=r+1;
}
cout<<(long long)ans;
}