Description
对于\(A_1,A_2,\cdots,A_N\),求 \(\sum_{i=1}^N\sum_{j=1}^N lcm(A_i,A_j)\)
Solution
经过一波推导得到
\[\sum_{i=1}^N\sum_{j=1}^N lcm(A_i,A_j) = \sum_{d=1}^N d (\sum_{i=1}^{N/d} i C_{id})^2 \sum_{k|d}k\mu(k) \]
暴力计算即可
#include
using namespace std;
#define int long long
const int N = 100005;
const int MAXN = 100005;
bool isNotPrime[MAXN + 1];
int mu[MAXN + 1], phi[MAXN + 1], primes[MAXN + 1], cnt;
inline void euler() {
isNotPrime[0] = isNotPrime[1] = true;
mu[1] = 1;
phi[1] = 1;
for (int i = 2; i <= MAXN; 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 > MAXN) 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);
}
}
}
}
int n,c[N];
signed main() {
euler();
ios::sync_with_stdio(false);
cin>>n;
for(int i=1;i<=n;i++) {
int tmp;
cin>>tmp;
c[tmp]++;
}
int ans=0;
for(int d=1;d<=N;d++) {
int a1=0,a2=0;
for(int i=1;i<=N/d;i++) {
a1+=i*c[i*d];
}
vector fac;
for(int i=1;i*i<=d;i++) {
if(d%i==0) {
fac.push_back(i);
if(i*i!=d) fac.push_back(d/i);
}
}
for(int k:fac) {
a2+=k*mu[k];
}
ans+=d*a1*a1*a2;
}
cout<