Gym-100989L(搜索)

Gym-100989L

AbdelKader enjoys math. He feels very frustrated whenever he sees an incorrect equation and so he tries to make it correct as quickly as possible!

Given an equation of the form: A₁ o A₂ o A₃ o ... o A_n  =  0, where o is either + or -. Your task is to help AbdelKader find the minimum number of changes to the operators + and -, such that the equation becomes correct.

You are allowed to replace any number of pluses with minuses, and any number of minuses with pluses.

Input

The first line of input contains an integer N (2 ≤ N ≤ 20), the number of terms in the equation.

The second line contains N integers separated by a plus + or a minus -, each value is between 1 and 10⁸.

Values and operators are separated by a single space.

Output

If it is impossible to make the equation correct by replacing operators, print  - 1, otherwise print the minimum number of needed changes.

Examples

Input

7
1 + 1 - 4 - 4 - 4 - 2 - 2

Output

3

Input

3
5 + 3 - 7

Output

-1

题目大意：n个数字之间有n-1个运算符+或-，通过改变运算符使得值为0。

解题思路：可以看到n范围很小，所以可以采用暴力搜索。

代码如下：

#include<stdio.h>
#include<algorithm>
using namespace std;

int n;
int a[50];
int ans=-1;

void dfs(int c,int cnt,int f)
{
    if(c==n+1&&cnt==0)
    {
        if(ans==-1) ans=f;
        else ans=min(ans,f);
    }
    if(c==n+1) return ;
    dfs(c+1,cnt+a[c],f);
    dfs(c+1,cnt-a[c],f+1);
}

void solve()
{
    dfs(2,a[1],0);
    printf("%d
",ans);
}

int main()
{
    char ch[5];
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        //printf("...
");
        if(i==1)
        {
            scanf("%d",&a[i]);
        }
        else
        {
            scanf("%s%d",&ch,&a[i]);
            if(ch[0]=='-')
            {
                a[i]=-a[i];
            }
        }
    }
    solve();
    return 0;
}