Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length N. The i-th term in the sequence is a_i.
In one operation, you can select a term and either increment or decrement it by one.

At least how many operations are necessary to satisfy the following conditions?

For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero.

For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.

Constraints

2 ≤ n ≤ 10^5

|a_i| ≤ 10^9

Each a_i is an integer.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print the minimum necessary count of operations.

Sample Input 1

4
1 -3 1 0

Sample Output 1

4

For example, the given sequence can be transformed into 1, -2, 2, -2 by four operations. The sums of the first one, two, three and four terms are 1, -1, 1 and -1, respectively, which satisfy the conditions.

Sample Input 2

5
3 -6 4 -5 7

Sample Output 2

0

The given sequence already satisfies the conditions.

Sample Input 3

6
-1 4 3 2 -5 4

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)


// **** WHEN THE GOING GETS TOUGH, THE TOUGH GETS GOING. ****

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ll long long int
#define ull unsigned long long int
#define ld long double
#define ordered_set tree < ll ,  null_type ,  less<ll> ,  rb_tree_tag ,  tree_order_statistics_node_update >
//added two more functions to set
//(1)*(set.find_by_order(k))[kth element in the sorted set] 
//(2)set.order_of_key(k)[count of elements strictly less than k]

typedef vector< int > vi;
typedef vector< ll > lvi;
typedef vector< vi > vvi;
typedef vector< lvi > lvvi;
typedef pair< int,int > ii;
typedef pair< ll,ll > lii;
typedef vector<ii> vii;
typedef vector<lii> lvii;

#define min_prq(t) priority_queue<t,vector<t>,greater<t>>
#define max_prq(t) priority_queue<t>
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(auto i = (c).begin(); i != (c).end(); i++)//remember i is an iterator
#define present(c,x) ((c).find(x) != (c).end())//for sets,maps,multimaps
#define cpresent(c,x) (find(all(c),x) != (c).end())//for vectors
#define tc(t) int (t); cin>>(t);while((t)--)
#define ff first
#define ss second
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " is " << a << endl;
	err(++it, args...);
}

//-----------------------------------------------------------------------------------------------------------------------------------------------------------------
#define mod 1000000007
//-----------------------------------------------------------------------------------------------------------------------------------------------------------------

ll mod_sum() { return 0LL; }
template<typename T, typename... Args>
T mod_sum(T a, Args... args) { return ((a + mod_sum(args...))%mod + mod)%mod; }

ll mod_product() { return 1LL; }
template<typename T, typename... Args>
T mod_product(T a, Args... args) { return (a*mod_product(args...))%mod; }
//-----------------------------------------------------------------------------------------------------------------------------------------------------------------


signed main()
{
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);	
	ll n;
	cin>>n;
	ll x;
	cin>>x;
	ll ans=(x==0)?1:0;
	for(int i=1;i<n;i++)
	{
		ll y;
		cin>>y;
		if(x==0)
		{
			if(y==0) 
			{
				ans+=2;
				continue;
			}
			if(y>0) x=-1;
			else x=1;
		}
		ll k=y+x;
		if(x<0)
		{
			ans+=max(0LL,1LL-k);
			x=max(1LL,k);
		}
		else if(x>0)
		{
			k=y+x;
			ans+=max(0LL,k+1LL);
			x=min(-1LL,k);
		}
	}
	cout<<ans<<endl;
	return 0;
}