# CodeForces839D[莫比乌斯反演] Codeforces Round #428 (Div. 2)

/*CodeForces839D[莫比乌斯反演]*/
#include <bits/stdc++.h>
typedef long long LL;
const LL MOD = 1000000007LL;
using namespace std;
int n, maxa = 0, mina = 0x3f3f3f3f;
LL m[200005], F[1000005], f[1000005];
int sum[1000005], a[200005];
int  prime[1000005], vis[1000005], mu[1000005];
void mo_init(int N) {
memset(vis, 0, sizeof(vis));
mu[1] = 1;
int cnt = 0;
for (int i = 2; i < N; i++)
{
if (!vis[i])
{
prime[cnt++] = i;
mu[i] = -1;
}
for (int j = 0; j < cnt && i * prime[j] < N; j++)
{
vis[i * prime[j]] = 1;
if (i % prime[j]) mu[i * prime[j]] = -mu[i];
else
{
mu[i * prime[j]] = 0;
break;
}
}
}
}
void solve() {
for (int i = 0; i < n; i++) {
for (int j = 1; j * j <= a[i]; j++) {
if (a[i] % j == 0) {
if (j * j == a[i]) {
sum[j]++;
}
else {
sum[j]++, sum[a[i] / j]++;
}
}
}
}
LL ans = 0;
for (int i = 2; i <= maxa; i++) {
F[i] = sum[i] * m[sum[i] - 1] % MOD;
}
for (int i = 2; i <= maxa; i++) {
for (int j = i; j <= maxa; j += i) {
f[i] = (f[i] + mu[j / i] * F[j] % MOD) % MOD;
}
ans = (ans + f[i] * i % MOD) % MOD;
}
printf("%lld\n", ans);
}
void init() {
mo_init(1000002);
m[0] = 1LL;
for (int i = 1; i <= n; i++) {
m[i] = m[i - 1] * 2LL % MOD;
}
}
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
maxa = max(maxa, a[i]);
mina = min(mina, a[i]);
} init(), solve();
return 0;
}

posted @ 2017-08-14 14:41 UnderSilence 阅读(...) 评论(...) 编辑 收藏