预处理约数就可以了。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 40005
#define eps 1e-10
#define mod 998244353
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
//#pragma comment (linker,"/STACK:102400000,102400000")
typedef long long LL;
//typedef int LL;
using namespace std;
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
// head
vector<int> g[maxn];
int a[maxn], next[maxn];
int b[maxn], c[maxn];
int n;
void init(void)
{
int tmp;
for(int i = 1; i <= 100000; i++) {
tmp = sqrt(i);
for(int j = 1; j <= tmp; j++)
if(i%j == 0) {
g[i].push_back(j);
if(j != i/j)
g[i].push_back(i/j);
}
}
}
void work(void)
{
LL ans = 0;
memset(next, -1, sizeof next);
for(int i = 1; i <= n; i++) {
if(next[a[i]] == -1) b[i] = a[i];
else b[i] = a[next[a[i]]];
int d = g[a[i]].size();
for(int j = 0; j < d; j++) next[g[a[i]][j]] = i;
}
memset(next, -1, sizeof next);
for(int i = n; i >= 1; i--) {
if(next[a[i]] == -1) c[i] = a[i];
else c[i] = a[next[a[i]]];
int d = g[a[i]].size();
for(int j = 0; j < d; j++) next[g[a[i]][j]] = i;
}
for(int i = 1; i <= n; i++) ans += (LL)b[i]*c[i];
printf("%I64d\n", ans);
}
int main(void)
{
init()
while(scanf("%d", &n), n != 0) {
for(int i = 1; i <= n; i++) scanf(a[i]);//scanf("%d", &a[i]);
work();
}
return 0;
}
