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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

using ll = long long int;
using ull = unsigned long long int;
using P = pair<ll, ll>;
using P3 = pair<P,int>;
using PP = pair<P, P>;
constexpr ll INF = 1LL << 60;
constexpr ll MOD = ll(1e9)+7;
constexpr int di[] = {0, 1, 0, -1};
constexpr int dj[] = {1, 0, -1, 0};
constexpr int di8[] = {0, 1, 1, 1, 0, -1, -1, -1};
constexpr int dj8[] = {1, 1, 0, -1, -1, -1, 0, 1};
constexpr double EPS = 1e-9;


int main(){
    int n;
    cin >> n;
    vector<ll> a(n);
    for(int i=0;i<n;i++) cin >> a[i];
    for(int i=1;i<n;i++) a[i] += a[i-1];
    ll s = 0, ans1 = 0, ans2 = 0;
    for(int i=0;i<n;i++){
        if(i%2){
            if(a[i]+s <= 0){
                ll d = abs(a[i]+s)+1;
                ans1 += d;
                s += d;
            }
        }
        else{
            if(a[i]+s >= 0){
                ll d = abs(a[i]+s)+1;
                ans1 += d;
                s -= d;
            }
        }
        // cout << a[i]+s << endl;
    }
    s = 0;
    for(int i=0;i<n;i++){
        if(i%2){
            if(a[i]+s >= 0){
                ll d = abs(a[i]+s)+1;
                ans2 += d;
                s -= (abs(a[i])+1);
            }
        }
        else{
            if(a[i]+s <= 0){
                ll d = abs(a[i]+s)+1;
                ans2 += d;
                s += d;
            }
        }
        // cout << a[i]+s << endl;
    }
    cout << min(ans1,ans2) << endl;
    return 0;
}