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: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main(int argc, char *argv[])
{
  const int INF= 1000000000;
  int n;
  long long int a[100000];
  scanf("%d",&n);
  for (int i = 0; i < n; i++) {
    scanf("%ld",&a[i]);
  }
  //1桁目が正
  long long int ans = INF;
  long long int sum =0;
  long long int ansp=0;
  long long int ansm=0;
  for (int i = 0; i < n; i++) {
    if(i%2==0){
      if(sum+a[i]>0){
	sum+=a[i];continue;
      }	
      else{
	ansp+=abs(1-a[i]-sum);
	sum=1;
      }
    }else{
      if(sum+a[i]<0){
	sum+=a[i];continue;
      }	
      else{
	ansp+=abs(-1-a[i]-sum);
	sum=-1;
      }      
    }
  }
  sum=0;
  for (int i = 0; i < n; i++) {
    if(i%2==1){
      if(sum+a[i]>0){
	sum+=a[i];continue;
      }	
      else{
	ansm+=abs(1-a[i]-sum);
	sum=1;
      }
    }else{
      if(sum+a[i]<0){
	sum+=a[i];continue;
      }	
      else{
	ansm+=abs(-1-a[i]-sum);
	sum=-1;
      }      
    }
  }
  printf("%lld\n",min(ansp,ansm));
  return 0;
}