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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
	int n;
	cin>>n;
	int a[n];
	long long int s=0;
	long long int ans=INT_MAX;
	int i;
	for(i=0;i<n;i++)
	cin>>a[i];
	s=a[0];
	long long int p=0;
	if(s>0)
	{
		for(i=1;i<n;i++)
		{
			if(i%2)
			{
				if(s+a[i]<0)
				{
					 s+=a[i];
				}
				else 
				{
					p+=1+s+a[i];
					s=-1;
				}		
			}
			else
			{
				if(s+a[i]>0)
				s+=a[i];
				else
				{
					p+=1-s-a[i];
					s=1;
				}	
			}
		}
		s=-1;
		ans=min(ans,p);
	/*	p=a[0]+1;
		for(i=1;i<n;i++)
		{
			if(i%2==0)
			{
				if(s+a[i]<0)
				{
					 s+=a[i];
				}
				else 
				{
					p+=1+s+a[i];
					s=-1;
				}		
			}
			else
			{
				if(s+a[i]>0)
				s+=a[i];
				else
				{
					p+=1-s-a[i];
					s=1;
				}	
			}
		}
		ans=min(ans,p);     */
		cout<<ans<<endl;
	}
	else if(s<0)
	{
		for(i=1;i<n;i++)
		{
			if(i%2==0)
			{
				if(s+a[i]<0)
				{
					 s+=a[i];
				}
				else 
				{
					p+=1+s+a[i];
					s=-1;
				}		
			}
			else
			{
				if(s+a[i]>0)
				s+=a[i];
				else
				{
					p+=1-s-a[i];
					s=1;
				}	
			}
		}
		s=1;
		ans=min(ans,p);
		/*p=(-1)*a[0]+1;
		for(i=1;i<n;i++)
		{
			if(i%2==1)
			{
				if(s+a[i]<0)
				{
					 s+=a[i];
				}
				else 
				{
					p+=1+s+a[i];
					s=-1;
				}		
			}
			else
			{
				if(s+a[i]>0)
				s+=a[i];
				else
				{
					p+=1-s-a[i];
					s=1;
				}	
			}
		}
		ans=min(ans,p);
		//cout<<p<<endl; */
		cout<<ans<<endl;
	}
	else
	{
		p=1;
		s=1;
		for(i=1;i<n;i++)
		{
			if(i%2)
			{
				if(s+a[i]<0)
				{
					 s+=a[i];
				}
				else 
				{
					p+=1+s+a[i];
					s=-1;
				}		
			}
			else
			{
				if(s+a[i]>0)
				s+=a[i];
				else
				{
					p+=1-s-a[i];
					s=1;
				}	
			}
		}
		s=-1;
		ans=min(ans,p);
		p=1;
		for(i=1;i<n;i++)
		{
			if(i%2==0)
			{
				if(s+a[i]<0)
				{
					 s+=a[i];
				}
				else 
				{
					p+=1+s+a[i];
					s=-1;
				}		
			}
			else
			{
				if(s+a[i]>0)
				s+=a[i];
				else
				{
					p+=1-s-a[i];
					s=1;
				}	
			}
		}
		ans=min(ans,p);
		cout<<ans<<endl;		
	}
	return 0;
}