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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
    int i, j, N;
    long long a[100010], b[100010], P1, P2, Tc, Tp;
    scanf("%d", &N);
    for (i=0;i<N;i++) scanf("%lld", &a[i]);
    memmove(b, a, N*sizeof(long long));

    P1 = 0LL; Tc = 0LL;
    for (i=0;i<N;i++) {
        Tp = Tc;
        Tc += a[i];
        if (Tc == 0LL) {
            if (Tp > 0LL) { a[i]--; Tc--; }
            else { a[i]++; Tc++; }
            P1++;
        } else if (Tc > 0LL && Tp > 0LL) {
            a[i] -= Tc + 1LL;
            P1 += Tc + 1LL;
            Tc = -1LL;
        } else if (Tc < 0LL && Tp < 0LL) {
            a[i] += -Tc + 1LL;
            P1 += -Tc + 1LL;            
            Tc = 1LL;
        }
    }
    P2 = 0LL; Tc = 0LL;
    if (b[0]>0LL) {
        P2 += b[0] + 1LL;
        b[0] = -1LL;
    } else if (b[0]<0LL) {
        P2 += -b[0] + 1LL;
        b[0] = 1LL;
    }
    Tc = b[0]; Tp = 0LL;
    for (i=1;i<N;i++) {
        Tp = Tc;
        Tc += b[i];
        if (Tc == 0LL) {
            if (Tp > 0LL) { b[i]--; Tc--; }
            else { b[i]++; Tc++; }
            P2++;
        } else if (Tc > 0LL && Tp > 0LL) {
            b[i] -= Tc + 1LL;
            P2 += Tc + 1LL;
            Tc = -1LL;
        } else if (Tc < 0LL && Tp < 0LL) {
            b[i] += -Tc + 1LL;
            P2 += -Tc + 1LL;            
            Tc = 1LL;
        }
    }
    if (P1<P2) printf("%lld\n", P1);
    else printf("%lld\n", P2);
//    for (i=0;i<N;i++) printf("%lld\t", a[i]);  printf("\n");
    return 0;
}