思路:
质因子分解可以顺着分解,也可以逆着分解
即找到每一个数字的倍数,再找到每一个数字的因数
cpp
const int N = 5e5+10;
vector<int> ff[N];
vector<int> f[N];
vector<int> g[N];
void solve(){
int n;
cin>>n;
vector<int> nums(n+1,0);
ll ans = 0;
for(int i = 1;i<=n;i++){
cin>>nums[i];
}
for(int i = 1;i<=n;i++){
int w = gcd(i,nums[i]);
int x = i/w,y = nums[i]/w;
//cout<<x<<" "<<y<<" eeeeee"<<endl;
if(nums[i]%i == 0) ans--;
f[x].emplace_back(y);
g[y].emplace_back(x);
}
vector<int>cnt(n+1,0);
for(int i = 1;i<=n;i++){
int m = f[i].size();
if(m == 0) continue;
sort(f[i].begin(),f[i].end());
for(int j = i;j<=n;j+=i){
for(auto x : g[j]){
cnt[x]++;
}
}
for(int j = 0;j<m;j++){
int k = 1;
while(j<m-1 && f[i][j] == f[i][j+1]){
j++;
k++;
}
int y = f[i][j];
ll sum = 0;
for(auto son : ff[y]){
sum += cnt[son];
}
sum *= k;
ans += sum;
}
for(int j = i;j<=n;j+=i){
for(auto x : g[j]){
cnt[x]--;
}
}
}
for(int i = 1;i<=n;i++){
f[i].clear();
g[i].clear();
}
cout<<ans/2<<endl;
}