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 Rust (1.15.1) (judge verdict: Wrong Answer)


use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
  let stdin = stdin();
  let stdin = stdin.lock();
  let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed to read char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
  token.parse().ok().expect("failed to parse token")
}

use std::cmp::{max, min};
use std::collections::BTreeMap;

fn main() {
  let n = read::<i64>();
  let mut vec_a = vec![];

  for i in 0..n {
    vec_a.push(read::<i64>());
  }

  let mut prev_sum = 0 as i64;
  let mut ans_plus = 0;
  let mut prev_sum2 = 0 as i64;
  let mut ans_minus = 0;

  // 最初を正にする
  if vec_a[0] <= 0 {
    prev_sum = 1;
    ans_plus = (1 - prev_sum).abs();
  } else {
    prev_sum = vec_a[0];
  }

  // 2こ目以降は流れで
  for i in 1..vec_a.len() {
    let b = vec_a[i as usize];
    if 0 < prev_sum {
      if 0 <= prev_sum + b {
        ans_plus += (1 + prev_sum).abs() + b;
        prev_sum = -1;
      } else {
        prev_sum += b;
      }
    } else if prev_sum < 0 {
      if prev_sum + b <= 0 {
        ans_plus += (1 - prev_sum).abs() - b;
        prev_sum = 1;
      } else {
        prev_sum += b;
      }
    }
  }

  // 最初を負にする
  if 0 <= vec_a[0] {
    prev_sum2 = -1;
    ans_minus = (1 + prev_sum2).abs();
  } else {
    prev_sum2 = vec_a[0];
  }
  // 2こ目以降は流れで
  for i in 1..vec_a.len() {
    let b = vec_a[i as usize];
    if 0 < prev_sum2 {
      if 0 <= prev_sum2 + b {
        ans_minus += (1 + prev_sum2).abs() + b;
        prev_sum2 = -1;
      } else {
        prev_sum2 += b;
      }
    } else if prev_sum2 < 0 {
      if prev_sum2 + b <= 0 {
        ans_minus += (1 - prev_sum2).abs() - b;
        prev_sum2 = 1;
      } else {
        prev_sum2 += b;
      }
    }
  }

  // println!("plus: {}, minus: {}", ans_minus, ans_plus);
  println!("{}", min(ans_minus, ans_plus));

}