题目链接：D. Armchairs
思路：我们将0的位置放在一个数组中，将1位置放在一个数组中，我们规定每一个1位置都是有序的，即顺序不可以被改变，\(f(i,j)\)表示处理完前i个人，且第i个人坐在第j个板凳上的最小花费，显然\(f(i,j) = min_{p=i-1}^{p=j-1}f(i-1,p) + dis(a[i],a[j])\)然后可以写出\(\Theta(n^3)\)的动态规划，发现\(min_{p=i-1}^{p=j-1}f(i-1,p)\)可以通过处理前缀取\(min\)。\(\Theta(1)\)求解，所以降为\(\Theta(n^2)\)
\(Code:\)

/* -*- encoding: utf-8 -*-
'''
@File    :   255.cpp
@Time    :   2021/05/31 21:05:02
@Author  :   puddle_jumper
@Version :   1.0
@Contact :   1194446133@qq.com
'''

# here put the import lib*/
#include<set>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
#define ch() getchar()
#define pc(x) putchar(x)
#include<stack>
#include<unordered_map>
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define ll long long
#define ull unsigned long long
#define pb emplace_back
#define mp make_pair
#define PI acos(-1)
using namespace std;
template<typename T>void read(T&x){
    static char c;
    static int f;
    for(c=ch(),f=1; c<'0'||c>'9'; c=ch())if(c=='-')f=-f;
    for(x=0; c>='0'&&c<='9'; c=ch())x=x*10+(c&15);
    x*=f;
}
template<typename T>void write(T x){
    static char q[65];
    int cnt=0;
    if(x<0)pc('-'),x=-x;
    q[++cnt]=x%10,x/=10;
    while(x)
        q[++cnt]=x%10,x/=10;
    while(cnt)pc(q[cnt--]+'0');
}


const int N = 5e3+10;
int n,a[N];
int f[N][N];
int b[N],c[N],totb ,totc;
int d[N][N];
void solve(){
    read(n);
    memset(f,0x3f,sizeof f);
    rep(i,1,n)read(a[i]);
    rep(i,1,n){
        if(a[i] == 1)b[++totb] = i;
        else c[++totc] = i;
    }
    f[0][0] = 0;
    rep(i,1,totb){
        rep(j,i,totc-totb+i){
            f[i][j] = d[i-1][j-1] + abs(b[i]-c[j]);
            if(j == i)d[i][j] = f[i][j];
            else d[i][j] = min(f[i][j],d[i][j-1]);
           // rep(p,0,j-1)f[i][j] = min(f[i][j],f[i-1][p]+abs(b[i]-c[j]));
        }
    }
    int ans = 9999999;
    if(totb == 0)ans = 0;
    rep(i,totb,totc)ans = min(ans,f[totb][i]);
    write(ans);
}


signed main(){solve();return 0; }