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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)


import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Main {

	private static final int mod =(int)1e9+7;
	public static void main(String[] args) throws Exception  {
	      Scanner  sc=new Scanner(System.in);
	      PrintWriter out=new PrintWriter(System.out);
	  int n=sc.nextInt();
	  int a[]=new int[n];
	  for(int i=0;i<n;i++) {
	    a[i]=sc.nextInt();
	  }
	 long sum=a[0];
	 long operations=0;
	 if(a.length==1) {
	   if(a[0]!=0) {
	     System.out.println(0);
	   }else {
	     System.out.println(1);
	   }
	 }else {
	 if(sum==0) {
	   if(a.length>=2&&sum+a[1]>0)
	   sum--;
	   else 
	     sum++;
	     
	   operations++;
	 }
	  for(int i=1;i<n;i++) {
	    if(sum>0) {
	      if(sum+a[i]<0) {
	        sum+=a[i];
	      }else {
	        if(sum+a[i]==0) {
	          sum+=a[i]-1;
	          operations++;
	        }else {
	          long req=(long)-1-1l*sum;
	          sum=-1;
	          operations+=(-1l*req+a[i]);     
	        }
	      }
	    }else {
	      if(sum+a[i]>0) {
	        sum+=a[i];
	      }else {
	        if(sum+a[i]==0) {
	          sum+=a[i]+1;
	          operations++;
	        }else {
	          long req=(long)1+-1l*sum;
	          sum=1;
	          operations+=(req-a[i]);
	        }
	      }
	    }

	  }
	  System.out.println(operations);
	    }
	}
	
	static boolean vis[]=new boolean[10001];
	      static int gcd(int a, int b) 
	      { 
	          if (a == 0) 
	              return b; 
	          return gcd(b % a, a); 
	      } 
	    
	      // Function to find gcd of array of 
	      // numbers 
	      static int f(int arr[], int n) 
	      { 
	          int result = n; 
	          int max=-1;
	          int ans=0;
	          for (int element: arr){ 
	                if(vis[element]==false)
	                     
	              result = gcd(n, element); 
	              if(result>max) {      
	                    max=result;
	                    ans=element;
	           
	              }
	           
	            
	          } 
	     
	               
	          return ans; 
	      } 
	     
	      
}