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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {

    static void solve() {
        int n = nextInt();
        long[] a = nextLongArray(n);

        long ans;
        long sum;
        boolean expectPlus;

        // a1 > 0 の場合
        ans = 0;
        sum = 0;
        expectPlus = true;
        for (int i = 0; i < n; i++) {
            sum += a[i];
            if (expectPlus && sum <= 0) {
                ans += -sum + 1;
                sum = 1;
            } else if (!expectPlus && sum >= 0) {
                ans += sum + 1;
                sum = -1;
            }
            expectPlus = !expectPlus;
        }
        long ans1 = ans;

        // a1 < 0 の場合
        ans = 0;
        sum = 0;
        expectPlus = false;
        for (int i = 0; i < n; i++) {
            sum += a[i];
            if (expectPlus && sum <= 0) {
                ans += -sum + 1;
                sum = 1;
            } else if (!expectPlus && sum >= 0) {
                ans += sum + 1;
                sum = -1;
            }
            expectPlus = !expectPlus;
        }
        long ans2 = ans;

        ans = Math.min(ans1, ans2);
        out.println(ans);
    }

    static final int MOD = 1_000_000_007;
    static long[] fac, finv, inv;

    // 階乗(n!)
    static long factorial(long n) {
        long ans = 1;
        for (long i = n; i > 0; i--) {
            ans = ans * i % MOD;
        }
        return ans;
    }

    // nCkの初期化
    static void comInit(int max) {
        fac = new long[max];
        finv = new long[max];
        inv = new long[max];
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;

        for (int i = 2; i < max; i++) {
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    // nCkの計算
    static long com(int n, int k) {
        if (n < k)
            return 0;
        if (n < 0 || k < 0)
            return 0;
        return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
    }

    static PrintWriter out;
    static Scanner sc;

    static int[][] newIntArray(int h, int w, int value) {
        int[][] ret = new int[h][w];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                ret[i][j] = value;
            }
        }
        return ret;
    }

    static int nextInt() {
        return Integer.parseInt(sc.next());
    }

    static long nextLong() {
        return Long.parseLong(sc.next());
    }

    static String nextString() {
        return sc.next();
    }

    static int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    static long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextLong();
        }
        return a;
    }

    static List<Integer> nextIntList(int n) {
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextInt());
        }
        return list;
    }

    static List<Double> nextDoubleList(int n) {
        List<Double> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add((double) nextInt());
        }
        return list;
    }

    static List<Long> nextLongList(int n) {
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextLong());
        }
        return list;
    }

    static char[][] nextCharArray(int h, int w) {
        char[][] c = new char[h][w];
        for (int i = 0; i < h; i++) {
            String str = nextString();
            for (int j = 0; j < w; j++) {
                c[i][j] = str.charAt(j);
            }
        }
        return c;
    }

    static <T extends Comparable<? super T>> void sort(List<T> list) {
        Collections.sort(list);
    }

    // greatest common divisor
    // 最大公約数
    static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    // least common multiple
    // 最小公倍数
    static long lcm(long a, long b) {
        if (a >= b && a % b == 0)
            return a;
        if (b > a && b % a == 0)
            return b;

        if (a > b) {
            return (a / gcd(a, b)) * b;
        } else {
            return (b / gcd(a, b)) * a;
        }
        // return a * b / gcd(a, b);
    }

    // baseのn乗を計算を返す
    static int pow(int base, int n) {
        int ret = 1;
        for (int i = 0; i < n; i++) {
            ret *= base;
        }
        return ret;
    }

    // return n^k mod m
    static long powMod(long n, long k, long m) {
        if (k == 0) {
            return 1;
        } else if (k % 2 == 1) {
            return powMod(n, k - 1, m) * n % m;
        } else {
            long tmp = powMod(n, k / 2, m);
            return tmp * tmp % m;
        }
    }

    // intをlength桁のbit文字列に変換
    static String toBitString(int length, int n) {
        StringBuilder sb = new StringBuilder();
        for (int i = length - 1; i >= 0; i--) {
            if ((n >> i) % 2 == 1) {
                sb.append("1");
            } else {
                sb.append("0");
            }
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        out = new PrintWriter(System.out);
        sc = new Scanner(System.in);

        solve();

        out.flush();
        sc.close();
    }

}