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: Accepted)

#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_ARRAY_MAX		100000									// 最大配列数

// 内部変数
static int si1Array[D_ARRAY_MAX];								// 配列
static int siACnt;												// 配列数
static long long llMin;											// 最小回数

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpT, *szpFpA;
#endif

// 最小回数の更新
int
fUpMin(
	int piSign					// <I> 符号 [1]＋ [-1]－
)
{
	int i;

	long long llCnt = 0;
	long long llSum = 0;

	for (i = 0; i < siACnt; i++) {

		// 符号の反転
		piSign *= -1;

		// 値の加算
		llSum += si1Array[i];

		// 符号のチェック
		if (llSum * (long long)piSign > 0) {	// OK
			continue;
		}

		// 回数加算
		llCnt += llabs(llSum) + (long long)1;

		// 合計値更新
		llSum = piSign;
	}

	// 最小回数更新
	if (llMin > llCnt) {
		llMin = llCnt;
	}

	return 0;
}

// 実行メイン
int
fMain(
	int piTNo					// <I> テスト番号 1～
)
{
	int i;
	char lc1Buf[1024], lc1Out[1024];

	// データ初期化
	llMin = LLONG_MAX;											// 最小回数

	// テストファイルオープン
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", piTNo);
	szpFpT = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", piTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#endif

	// 配列数取得
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
	fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif
	sscanf(lc1Buf, "%d", &siACnt);

	// 配列取得
	for (i = 0; i < siACnt; i++) {
#ifdef D_TEST
		fscanf(szpFpT, "%d", &si1Array[i]);
#else
		fscanf(stdin, "%d", &si1Array[i]);
#endif
	}
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
	fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif

	// 最小回数の更新
	fUpMin(1);
	fUpMin(-1);

	// 結果セット
	sprintf(lc1Out, "%lld\n", llMin);

	// 結果表示
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, lc1Out)) {
		siRes = -1;
	}
#else
	printf("%s", lc1Out);
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpT);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", piTNo);
	}
	else {
		printf("NG %d\n", piTNo);
	}
#endif

	return 0;
}

int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		fMain(i);
	}
#else
	fMain(0);
#endif

	return 0;
}