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"

using namespace std;
typedef long long int ll;
typedef pair<ll, ll > pi;  
typedef pair<pair<ll, ll >, ll > pii;  
vector<ll > vec;
vector<vector<ll > > vec2;
ll MOD = 1000000007;

int main(){

    ll N;
    cin >> N;

    vector<ll > vec(N, 0);

    for(ll i = 0; i < N; i++){
        cin >> vec[i];
    }

    ll ans = 0;
    ll g_ans = 0;
    ll k_ans = 0;
    ll sum = 0;

    //奇数番目を正に揃える
    bool flg = true;
    for(ll i = 0; i < N; i++){
        sum += vec[i];
        if(flg == true){
            if(sum <= 0){
                //cout << i << " " << sum << " " << abs(1 - sum) << endl;
                k_ans += abs(sum) + 1;
                sum += 1;
            }
            flg = false;
        }else{
            if(sum >= 0){
                //cout << i << " " << sum << " " << abs(-1 - sum) << endl;
                k_ans += abs(sum) + 1;
                sum += -1;
            }
            flg = true;
        }
    }

    //cout << k_ans << endl;
    //奇数番目を負に揃える
    flg = true;
    sum = 0;
    for(ll i = 0; i < N; i++){
        sum += vec[i];
        //cout << sum << endl;
        if(flg == false){
            if(sum <= 0){
                //cout << i << " " << sum << " " << abs(sum - vec[i]) << endl;
                g_ans += abs(sum) + 1;
                sum += 1;
            }
            flg = true;
        }else{
            if(sum >= 0){
                //cout << i << " " << sum << " " << abs(sum - vec[i]) << endl;
                g_ans += abs(sum) + 1;
                sum += -1;
            }
            flg = false;
        }
    }
    
    ans = min(k_ans, g_ans);

    cout << ans << endl;


}