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;


int main(){
    int n;
    cin >> n;
    vector<int> a(n+1);
    for(int i = 0; i < n; i++){
        cin >> a.at(i);
    }
    vector<int> b(n+1);
    b[0] = a[0];

    int ansa = 0,ansb = 0,sum = 0;
    //a[0]をマイナス(偶数がマイナス)
    for(int i = 0; i < n; i++){
        if(i % 2 == 0){
            if(b[i] >= 0){
                ansa += 1 + b[i];
                b[i] = -1;
                b[i + 1] = b[i] + a[i + 1];
            }
            else b[i + 1] = b[i] + a[i + 1];
        }
        else{
            if(b[i] <= 0){
                ansa += 1 - b[i];
                b[i] = 1;
                b[i + 1] = b[i] + a[i + 1];
            }
            else b[i + 1] = b[i] + a[i + 1];
        }
    }

    b[0] = a[0];
    //a[0]をプラス(偶数がプラス)
    for(int i = 0; i < n; i++){
        if(i % 2 == 0){
            if(b[i] <= 0){
                ansb += 1 - b[i];
                b[i] = 1;
                b[i + 1] = b[i] + a[i + 1];
            }
            else b[i + 1] = b[i] + a[i + 1];
        }
        else{
            if(b[i] >= 0){
                ansb += 1 + b[i];
                b[i] = -1;
                b[i + 1] = b[i] + a[i + 1];
            }
            else b[i + 1] = b[i] + a[i + 1];
        }
    }
    

    int ans = min(ansa,ansb);
    cout << ans << endl;
}