杜教筛求 \(\phi(n)\),
\[ S(n)=n(n+1)/2-\sum_{d=2}^n S(\frac{n}{d}) \]
答案为
\[ \sum_{d=1}^n \phi(d) h(\frac{n}{d}) \]
其中 \(h(n)=\sum_{i=1}^n i^2\)
顺便学习了一波 unordered_map
#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define int long long
const int N = 2e6+5;
int i2,i4,i6,mod,n;
bool isNotPrime[N + 5];
int mu[N + 5], phi[N + 5], primes[N + 5], cnt, sum[N+5];
unordered_map<int,int> mp;
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 S(int n) {
if(n<=N) return sum[n];
if(mp[n]) return mp[n];
int ans=n*(n+1)%mod*i2%mod;
int l=2,r;
while(l<=n) {
r=n/(n/l);
ans-=(r-l+1)*S(n/l)%mod;
ans%=mod;
ans+=mod;
ans%=mod;
l=r+1;
}
mp[n]=ans;
return ans;
}
int h(int n) {
return (n*(n+1)%mod*(2*n+1)%mod*i6%mod);
}
signed main() {
cin>>n>>mod;
i2=inv(2,mod); i4=inv(4,mod); i6=inv(6,mod);
euler();
for(int i=1;i<=N;i++) sum[i]=sum[i-1]+phi[i], sum[i]%=mod;
int l=1,r,ans=0;
while(l<=n) {
r=n/(n/l);
ans+=(S(r)-S(l-1))*h(n/l)%mod;
ans%=mod;
ans+=mod;
ans%=mod;
l=r+1;
}
cout<<ans;
}