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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <cstring>
using namespace std;
typedef long long ll;
int main()
{
    //freopen("input.in", "r", stdin);
    int n;
    while(scanf("%d", &n)!=EOF){

        vector<ll> nums(n, 0);

        for(int i=0;i<n;i++){
            scanf("%lld", &nums[i]);
        }

        ll sum1=0, sum2=0, res1=0, res2=0;
        if(nums[0]>0){
            sum1+=nums[0];
            res1=0;
            sum2=-1;
            res2+=(1+nums[0]);
        }
        else if(nums[0]<0){
            sum1=1;
            res1+=(1-nums[0]);
            sum2+=nums[0];
            res2=0;
        }
        else if(nums[0]==0){
            sum1=1;
            res1=1;
            sum2=-1;
            res1=1;
        }
        //first number is positive
        for(int i=1;i<n;i++){
            if(sum1>0){
                res1+=(sum1+nums[i] < 0? 0 : (sum1+nums[i]+1));
                if(sum1+nums[i] < 0){
                    sum1+=nums[i];
                }
                else {
                    sum1=-1;
                }
            }
            else {
                res1+=(sum1+nums[i] > 0? 0 : (1+abs(sum1+nums[i])));
                if(sum1+nums[i] > 0){
                    sum1+=nums[i];
                }
                else {
                    sum1=1;
                }
            }
        }

        //first number is negative
        for(int i=1;i<n;i++){
            if(sum2>0){
                res2+=(sum2+nums[i] < 0? 0 : (sum2+nums[i]+1));
                if(sum2+nums[i] < 0){
                    sum2+=nums[i];
                }
                else {
                    sum2=-1;
                }
            }
            else{
                res2+=(sum2+nums[i] > 0? 0 : (1+abs(sum2+nums[i])));
                if(sum2+nums[i] > 0){
                    sum2+=nums[i];
                }
                else {
                    sum2=1;
                }
            }
        }
        printf("%lld\n", min(res1, res2));
    }
    return 0;
}