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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.Scanner;

public class Main {

    private static Scanner scan = new Scanner(System.in);

    public static void main(String[] args) {
	int N = Integer.parseInt(scan.next());
	long a[] = new long[N];

	for (int i = 0; i < N; i++) {
	    a[i] = Long.parseLong(scan.next());
	}

	long s1 = 0;
	long s2 = 0;
	long count1 = 0;
	long count2 = 0;
	for (int i = 0; i < N; i++) {
	    s1 += a[i];
	    s2 += a[i];
	    if(i%2 == 0) {
		if(s1 <= 0) {
		    count1 += (1-s1);
		    s1 = 1;
		}
		if(s2 >= 0) {
		    count2 += (s2+1);
		    s2 = -1;
		}
	    } else {
		if(s1 >= 0) {
		    count1 += (s1+1);
		    s1 = -1;
		}
		if(s2 <= 0) {
		    count2 += (1-s2);
		    s2 = 1;
		}
	    }
	}

	System.out.println(count1 < count2 ? count1 : count2);
	scan.close();
    }
}