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)

// Math = ♥️

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

#define ll long long
#define ld long double
typedef pair<ll, ll> ii;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;
typedef vector<vii> vvii;
#define pq priority_queue
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(), (c).end()
#define sz(c) ((ll)c.size())
#define ordered_set                            \
    tree<ll, null_type, less<ll>, rb_tree_tag, \
         tree_order_statistics_node_update>

int main(void) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll n;
    cin >> n;

    vi arr(n);
    for (auto &it : arr) {
        cin >> it;
    }

    vi pre(n);
    pre[0] = arr[0];
    for (int i = 1; i < n; i++) {
        pre[i] = pre[i - 1] + arr[i];
    }

    ll ans = LLONG_MAX;

    ll lazy = 0, curr = 0, curr_ans = 0;
    for (int i = 0; i < n; i++) {
        if (curr == 0) {
            if (pre[i] + lazy <= 0) {
                curr_ans += llabs(pre[i] + lazy) + 1;
                lazy += llabs(pre[i] + lazy) + 1;
            }
        } else {
            if (pre[i] + lazy >= 0) {
                curr_ans += (pre[i] + lazy) + 1;
                lazy -= (pre[i] + lazy) + 1;
            }
        }
        curr ^= 1;
    }

    ans = min(ans, curr_ans);
    lazy = 0, curr = 1, curr_ans = 0;
    for (int i = 0; i < n; i++) {
        if (curr == 0) {
            if (pre[i] + lazy <= 0) {
                curr_ans += llabs(pre[i] + lazy) + 1;
                lazy += llabs(pre[i] + lazy) + 1;
            }
        } else {
            if (pre[i] + lazy >= 0) {
                curr_ans += (pre[i] + lazy) + 1;
                lazy -= (pre[i] + lazy) + 1;
            }
        }
        curr ^= 1;
    }
    ans = min(ans, curr_ans);

    cout << ans << "\n";
}