D探索的时光
//这是个数学问题
//整理f(i)=(x-i)^2*a[i]这个式子
//得到f(i)=(x^2+i^2-2*x*i)*a[i]
//再求和得到sum=x*x*(a1+a2+...+an)+1^2*a1+2^2*a2+...+n^2*an-2*x*(1*a1+2*a2+...+n*an)
//然后利用高中的数学知识(一元二次方程)求解
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
ll a[100005];
void solve(){
ll n;
ll ans=1e18+9,sum=0,sum1=0,sum2=0;
cin>>n;
for(ll i=1;i<=n;i++){
cin>>a[i];
sum+=a[i];
sum1+=i*a[i];
sum2+=i*i*a[i];
}
ll x;
double gg=sum1*1.0/sum;
if(gg>=1&&gg<=n){
ll xx=ceil(sum1*1.0/sum);
ll yy=floor(sum1*1.0/sum);
if(xx>=1&&xx<=n){
x=xx;
ans=min(ans,x*x*sum+sum2-2*x*sum1);
}
if(yy>=1&&yy<=n){x=yy;
ans=min(ans,x*x*sum+sum2-2*x*sum1);
}
}else if(gg>n){
x=n;
ans=x*x*sum+sum2-2*x*sum1;
}else{
x=1;
ans=x*x*sum+sum2-2*x*sum1;
}
cout<<ans;
return ;
}
int main(){
ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
int _;
//cin>>_;
_=1;
while(_--){
solve();
}
return 0;
}