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 5.4.1) (judge verdict: Wrong Answer)


#include <stdio.h>
int n,a[100000];
int min(int a,int b){
    if(a<b)return a;
    else return b;
}
int count(int sf){
    int sum=0,count=0,i;
    for(i=0;i<n;i++){
        sum+=a[i];
        if(!sf){
            if(sum>=0){
                count+=sum+1;
                sum=-1;
            }
        }
        else{
            if(sum<=0){
                count+=-sum+1;
                sum=1;
            }
        }
        sf=!sf;
    }
    return count;
}
int main(void){
    int i,j;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    //偶数番目と奇数番目を比較して小さい方を選択する
    printf("%d\n",min(count(0),count(1)));
    return 0;
}