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: Accepted)

#include <bits/stdc++.h>

#define int long long
#define fs first
#define sc second
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define mp make_pair
#define len(v) ((int)v.size())
#define vc vector
#define pr pair
#define endl "\n"
#define endd "\n"; return 0;
#define all(v) v.begin(), v.end()

using namespace std;

const int MAXN = 1e5;
const int INF = 1e10;
const int MOD = 1e9 + 7;

main() {
    ios_base::sync_with_stdio(0);
    int n;
    cin >> n;
    int a[n];
    int ans1 = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    {
        int prefsum[n];
        prefsum[0] = a[0];
        int ans = 0;
        if (a[0] <= 0){
            ans += 1 - a[0];
            prefsum[0] = 1;
        }
        for (int i = 1; i < n; i++) {
            prefsum[i] = prefsum[i - 1] + a[i];
            if (prefsum[i] == 0) {
                if (i % 2 == 0){
                    ans++;
                    prefsum[i] = 1;
                }
                else{
                    ans++;
                    prefsum[i] = -1;
                }
            }
            else{
                if (i % 2 == 0){
                    if (prefsum[i] < 0){
                        ans += 1 - prefsum[i];
                        prefsum[i] = 1;
                    }
                }
                else{
                    if (prefsum[i] > 0){
                        ans += prefsum[i] + 1;
                        prefsum[i] = -1;
                    }
                }
            }
        }
        ans1 = ans;
    }
    {
        for (int i = 0; i < n; i++){
            a[i] *= -1;
        }
        int prefsum[n];
        prefsum[0] = a[0];
        int ans = 0;
        if (a[0] <= 0){
            ans += 1 - a[0];
            prefsum[0] = 1;
        }
        for (int i = 1; i < n; i++) {
            prefsum[i] = prefsum[i - 1] + a[i];
            if (prefsum[i] == 0) {
                if (i % 2 == 0){
                    ans++;
                    prefsum[i] = 1;
                }
                else{
                    ans++;
                    prefsum[i] = -1;
                }
            }
            else{
                if (i % 2 == 0){
                    if (prefsum[i] < 0){
                        ans += 1 - prefsum[i];
                        prefsum[i] = 1;
                    }
                }
                else{
                    if (prefsum[i] > 0){
                        ans += prefsum[i] + 1;
                        prefsum[i] = -1;
                    }
                }
            }
        }
        ans1 = min(ans, ans1);
    }
    cout << ans1;

}