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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int n;
int a[100000];

long long int min(long long int a, long long int b) {
	return a < b ? a : b; 
}

long long int count(long long int sum, long long int count, int sf) {
	int i;
	for (i = 1; i < n; i++) {
		if (!sf) {
			if (sum + a[i] < 0) {
				sum += a[i];
			}
			else {
				count += sum + a[i] + 1;
				sum = -1;
			}
		}
		else {
			if (sum + a[i] > 0) {
				sum += a[i];
			}
			else {
				count -= sum + a[i] - 1;
				sum = 1;
			}
		}
		sf = !sf;
	}
	return count;
}

int main(void) {
	int i;
	long long int ans;

	scanf("%d", &n);
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	if (a[0] > 0)
		ans = min(count(a[0], 0, 0), count(-1, a[0] + 1, 1));
	else
		ans = min(count(a[0], 0, 1), count(1, -a[0] + 1, 0));

	printf("%lld\n", ans);

	return 0;
}