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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

#define REP(i,n) for(int i=0;i<n;++i)
#define SORT(name) sort(name.begin(), name.end())
#define ZERO(p) memset(p, 0, sizeof(p))
#define MINUS(p) memset(p, -1, sizeof(p))
#if 0
#  define DBG(fmt, ...) printf(fmt, __VA_ARGS__)
#else
#  define DBG(fmt, ...)
#endif

#define MOD 1000000007
#define INF 1000000000
#define MAX_N 100010

// 300, 60min + 解けず

ll N;

// [1, N]
ll bit1[MAX_N + 1];
ll bit2[MAX_N + 1];
// [1, i] の合計値を返す O(logn)
ll sum1(int i) {
    ll s = 0;
    while(i > 0) {
        s += bit1[i];
        i -= i & -i;
    }
    return s;
}
ll sum2(int i) {
    ll s = 0;
    while(i > 0) {
        s += bit2[i];
        i -= i & -i;
    }
    return s;
}
// ai に x を足して更新　O(logn)
void add1(int i, ll x) {
    while(i <= N) {
        bit1[i] += x;
        i += i & -i;
    }
}
void add2(int i, ll x) {
    while(i <= N) {
        bit2[i] += x;
        i += i & -i;
    }
}

int main()
{
    cin >> N;
    REP(i, N) {
        ll a;
        cin >> a;
        add1(i+1, a);
        add2(i+1, a);
    }
    ll ans, prev, cand;
    auto Check1 = [&]() {
        for(int i = 2; i <= N; ++i) {
            if(sum1(i) == 0) {
                cand++;
                if(sum1(i-1) > 0) { add1(i, -1); }
                else { add1(i, 1); }
            }
            else if(sum1(i) > 0 && sum1(i-1) > 0) {
                ll diff = sum1(i) + 1;
                cand += diff;
                add1(i, -diff);
                assert(sum1(i) < 0);
            }
            else if(sum1(i) < 0 && sum1(i-1) < 0) {
                ll diff = 1 - sum1(i);
                cand+= diff;
                add1(i, diff);
                assert(sum1(i) > 0);
            }
            DBG("i: %d cand: %lld\n", i, cand);
        }
    };
    auto Check2 = [&]() {
        for(int i = 2; i <= N; ++i) {
            if(sum2(i) == 0) {
                cand++;
                if(sum2(i-1) > 0) { add2(i, -1); }
                else { add2(i, 1); }
            }
            else if(sum2(i) > 0 && sum2(i-1) > 0) {
                ll diff = sum2(i) + 1;
                cand += diff;
                add2(i, -diff);
                assert(sum2(i) < 0);
            }
            else if(sum2(i) < 0 && sum2(i-1) < 0) {
                ll diff = 1 - sum2(i);
                cand+= diff;
                add2(i, diff);
                assert(sum2(i) > 0);
            }
            DBG("i: %d cand: %lld\n", i, cand);
        }
    };
    if(sum1(1) == 0) {
        ans = INF;

        add1(1, 1);
        cand = 1;
        prev = sum1(1);
        Check1();
        ans = min(ans, cand);

        add2(1, -1);
        cand = 1;
        prev = sum2(1);
        Check2();
        ans = min(ans, cand);
    } else {
        ans = INF;

        // そのまま
        cand = 0;
        prev = sum1(1);
        Check1();
        ans = min(ans, cand);

        // 反対の符号
        cand = abs(sum2(1)) + 1;
        if(sum2(1) > 0) {
            add2(1, -cand);
        } else {
            add2(1, cand);
        }
        prev = sum2(1);
        Check2();
        ans = min(ans, cand);
    }
    printf("%lld\n", ans);
    return 0;
}