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 <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <unordered_map>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

#define dbg(x1) cout<<#x1<<" = "<<(x1)<<endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<(x1)<<" "<<#x2<<" = "<<(x2)<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<(x1)<<" "<<#x2<<" = "<<(x2)<<" "<<#x3<<" = "<<(x3)<<endl

ll a[maxn];

int main() {
    //freopen(".in", "r", stdin);
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    ll prefix = a[1];
    ll ans = 0;
    if (prefix == 0) {
        ll ans1 = 0, ans2 = 0;
        ll prefix1 = 1;
        ans1 = 1;
        for (int i = 2; i <= n; i++) {
            if (prefix1 > 0) {
                prefix1 += a[i];
                if (prefix1 >= 0) {
                    ans1 += (prefix1 - (-1));
                    prefix1 = -1;
                }
            } else if (prefix1 < 0) {
                prefix1 += a[i];
                if (prefix1 <= 0) {
                    ans1 += (1 - prefix1);
                    prefix1 = 1;
                }
            }
        }
        prefix1 = -1;
        ans2 = 1;
        for (int i = 2; i <= n; i++) {
            if (prefix1 > 0) {
                prefix1 += a[i];
                if (prefix1 >= 0) {
                    ans2 += (prefix1 - (-1));
                    prefix1 = -1;
                }
            } else if (prefix1 < 0) {
                prefix1 += a[i];
                if (prefix1 <= 0) {
                    ans2 += (1 - prefix1);
                    prefix1 = 1;
                }
            }
        }
        ans = min(ans1, ans2);
    } else {
        for (int i = 2; i <= n; i++) {
            if (prefix > 0) {
                prefix += a[i];
                if (prefix >= 0) {
                    ans += (prefix - (-1));
                    prefix = -1;
                }
            } else if (prefix < 0) {
                prefix += a[i];
                if (prefix <= 0) {
                    ans += (1 - prefix);
                    prefix = 1;
                }
            }
        }
    }

    printf("%lld\n", ans);
    return 0;
}