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>
#define rep(i, a, b) for(int i = a; i < b; i++)
#define Rep(i, a, b) for(int i = a; i <= b; i++)
#define repr(i, a, b) for(int i = a; i >= b; i--)
#define _GLIBCXX_DEBUG
#define Vl vector<ll>
#define Vs vector<string>
#define Vp vector<pair<ll, ll>>
#define ld long double
using ll = long long;
#define ALL(v) (v).begin(),(v).end()
#define endl "\n"
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define co(x) cout << x << endl
#define coel cout << endl
#define pb push_back
#define sz(v) ((ll)(v).size())
using namespace std;  
const double pi = acos(-1.0);
const ll MOD = 1e9 + 7;
const ll INF = 1LL << 60;
#define pp pair<ll, pair<ll, ll>> 
#define fi first
#define se second

void print(Vl vec){
    rep(i, 0, sz(vec)){
        if(i) cout << " ";
        cout << vec[i];
    }
    coel;
}

////////////////////////////////////////////////////////

ll dp[100005];

int main() {
    ll n; cin >> n;
    Vl a(n);
    rep(i, 0, n) cin >> a[i];

    dp[0] = a[0];
    ll ans = 0;
    rep(i, 1, n){
        if(dp[i-1] < 0){
            if(dp[i-1]+a[i] > 0){
                dp[i] = dp[i-1]+a[i];
            }else{
                ll ai = 1 - dp[i-1];
                ans += abs(ai - a[i]);
                dp[i] = 1;
            }
        }
        if(dp[i-1] > 0){
            if(dp[i-1]+a[i] < 0){
                dp[i] = dp[i-1]+a[i];
            }else{
                ll ai = -1 - dp[i-1];
                ans += abs(ai - a[i]);
                dp[i] = -1;
            }
        }
    }

    ll cnt = 0;
    if(a[0] > 0) {
        dp[0] = -1;
        cnt += abs(a[0]+1);
    } else {
        dp[0] = 1;
        cnt += abs(a[0]-1);
    }
    rep(i, 1, n){
        if(dp[i-1] < 0){
            if(dp[i-1]+a[i] > 0){
                dp[i] = dp[i-1]+a[i];
            }else{
                ll ai = 1 - dp[i-1];
                cnt += abs(ai - a[i]);
                dp[i] = 1;
            }
        }
        if(dp[i-1] > 0){
            if(dp[i-1]+a[i] < 0){
                dp[i] = dp[i-1]+a[i];
            }else{
                ll ai = -1 - dp[i-1];
                cnt += abs(ai - a[i]);
                dp[i] = -1;
            }
        }
    }

    chmin(ans, cnt);

    co(ans);

    return 0;
}