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: Wrong Answer)

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

int main(void){
	/*
	サイズnの入力
	*/
	int n;
	if(scanf("%d\n", &n) < 1){
		exit(1);
	}

	/*
	a_1,a_2, ... a_nの入力
	*/
	int *a;
	a = (int*) malloc ( n * sizeof(int));
	int i;
	for(i=0; i<n-1; i++){
		if(scanf("%d ", &a[i]) < 1){ //scanfの戻り値は、成功した変数の数、もしくは-1である
			//printf(stderr, "error: ill-formed fraction.\n");
			exit(1);
		}
	}
	if(scanf("%d", &a[n-1]) < 1){
		//printf(stderr, "error: ill-formed fraction.\n");
		exit(1);
	}

	/*sum[i]はa[0]〜a[i]までの和*/
	int *sum;
	sum = (int*)malloc( n * sizeof(int));
	sum[0] = a[0];

	int evenchangeCount = 0; //何回変更するのかのカウント
	int oddPositive = 1; //iが奇数のsumが正のときに1
	int evenPositive = 0;//iが偶数のsumが正のときに1

	for(i=1; i<n; i++){
		sum[i] = sum[i-1] + a[i];
		if(i%2 == 1){
			/*iが奇数の時*/
			if(oddPositive == 1){
			//正でなければいけないとき
				if(sum[i] <= 0){ //正でないなら
					evenchangeCount += -sum[i] + 1;
					sum[i] = 1;
				}
			}else{		
			//負でなければいけないとき
				if(sum[i] >= 0){ //負でないなら
					evenchangeCount += sum[i] + 1;
					sum[i] = -1;
				}
			}
		}else{
			/*iが偶数の時*/
			if(evenPositive == 1){
			//正でなければいけないとき
				if(sum[i] <= 0){
				//正でないなら
					evenchangeCount += -sum[i] + 1;
					sum[i] = 1;
				}
			}else{
			//負でなければいけないとき
				if(sum[i] >= 0){
				//負でないなら
					evenchangeCount += sum[i] + 1;
					sum[i] = -1;
				}
			}
		}
	}

	int oddchangeCount = 0; //何回変更するのかのカウント
	oddPositive = 0; //iが奇数のsumが正のときに1
	evenPositive = 1;//iが偶数のsumが正のときに1

	sum[0] = a[0];
	if(a[0] <= 0){
		oddchangeCount += -sum[0] + 1;
		sum[0] = 1;
	}

	for(i=1; i<n; i++){
		sum[i] = sum[i-1] + a[i];
		if(i%2 == 1){
			/*iが奇数の時*/
			if(oddPositive == 1){
			//正でなければいけないとき
				if(sum[i] <= 0){ //正でないなら
					oddchangeCount += -sum[i] + 1;
					sum[i] = 1;
				}
			}else{		
			//負でなければいけないとき
				if(sum[i] >= 0){ //負でないなら
					oddchangeCount += sum[i] + 1;
					sum[i] = -1;
				}
			}
		}else{
			/*iが偶数の時*/
			if(evenPositive == 1){
			//正でなければいけないとき
				if(sum[i] <= 0){
				//正でないなら
					oddchangeCount += -sum[i] + 1;
					sum[i] = 1;
				}
			}else{
			//負でなければいけないとき
				if(sum[i] >= 0){
				//負でないなら
					oddchangeCount += sum[i] + 1;
					sum[i] = -1;
				}
			}
		}
	}
	
	if(evenchangeCount <= oddchangeCount){
		printf("%d", evenchangeCount);
	}else{
		printf("%d", oddchangeCount);
	}	
	return 0;
	

}