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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>

int main(void){
    long long n,ans=0,tmp=0;
    scanf("%lld",&n);
    long long a[n],sum[n];
    for (long long i=0; i<n; i++){
        scanf("%lld",&a[i]);
    }
    
    for (long long i=0; i<n; i++){
        if (i==0){
            sum[i]=a[i];
        } else {
            sum[i]=sum[i-1]+a[i];
        }
        if (i%2==0){
            if (sum[i]<=0){
                tmp+=llabs(sum[i])+1;
                sum[i]=1;
            }
        } else {
            if (sum[i]>=0){
                tmp+=llabs(sum[i])+1;
                sum[i]=-1;
            }
        }
    }
    
    for (long long i=0; i<n; i++){
        if (i==0){
            sum[i]=a[i];
        } else {
            sum[i]=sum[i-1]+a[i];
        }
        if (i%2==0){
            if (sum[i]>=0){
                ans+=llabs(sum[i])+1;
                sum[i]=-1;
            }
        } else {
            if (sum[i]<=0){
                ans+=llabs(sum[i])+1;
                sum[i]=1;
            }
        }
    }    
    if (tmp<ans){
        ans=tmp;
    }
    printf ("%lld\n",ans);
	return 0 ;
}