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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
 
const int MAX = 1e5 + 111;

int n;

int a[MAX];

ll calc(int id, ll sum) {
    ll ans = 0;
    for (int i = id; i < n; ++i) {
        ll cur = sum + a[i];
        //cerr << i << " " << sum << " " << cur << "\n";
        if (sum < 0 && cur > 0) {
            sum = cur; 
            continue;
        }
        if (sum > 0 && cur < 0) {
            sum = cur;
            continue;
        }
        if (cur == 0) {
            if (sum < 0)
                cur = 1;
            else
                cur = -1;
            sum = cur;
            ans++;
            continue;
        }
        if (sum < 0) {
            ans += 1 - cur;
            cur = 1;
        } else {
            ans += cur + 1;
            cur = -1;
        }
        sum = cur;
    }
    return ans;
}

int main() {
#ifdef LOCAL
    freopen("xxx.in", "r", stdin);
    freopen("xxx.out", "w", stdout);    
#endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    int id = -1;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        if (a[i] != 0 && id == -1)
            id = i;
    }
    if (id == -1) {
        cout << 1 + (n - 1) * 2;
        return 0;
    }
    if (id != 0) {
        ll ans1 = calc(id, -1);
        ll ans2 = calc(id, 1);
        cout << min(ans1, ans2) + 1 + 2 * id;
    } else {
        ll ans1 = calc(1, a[0]);
        cout << ans1;
    }
    return 0;
}