Gym - 100989M 0-1背包

Time limit

1000 ms

Memory limit

262144 kB

George met AbdelKader in the corridor of the CS department busy trying to fix a group of incorrect equations. Seeing how fast he is, George decided to challenge AbdelKader with a very large incorrect equation. AbdelKader happily accepted the challenge!

Input

The first line of input contains an integer N (2 ≤ N ≤ 300), 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 300.

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

视作01背包，由于题目数据限制，我们可以得知容量为-90000到90000，由于负数的特殊性，我们对其加上修正值90000便于操作。
i为第几个物品，j为背包容量，Dp[i][j]表示在第i个物品时，使剩余容量为j时的操作次数（改变符号的次数）
ac代码
int dp[305][180005];
int main()
{
    int n;
    cin>>n;
    char b;
    int al[305];
    cin>>al[1];
    int sum=al[1];
    for(int n1=2;n1<=n;n1++)
    {
        getchar();
        cin>>b;
        getchar();
        cin>>al[n1];
        sum+=al[n1];
        if(b=='-')
        {
            al[n1]=-al[n1];
        }
    }
    if(sum%2==1)
    {
        cout<<-1<<endl;
    }
    else
    {
        for(int n1=1;n1<=n;n1++)
        {
            for(int j=0;j<=2*sum;j++)
            {
                dp[n1][j]=99999;
            }
        }
        //for(int j=0;j<=sum;j++)
            dp[1][sum/2+al[1]]=0;
        for(int n1=2;n1<=n;n1++)
        {
            for(int j=0;j<=sum;j++)
            {
                if(j>=al[n1])
                {
                    dp[n1][j]=min(dp[n1-1][j-al[n1]],dp[n1][j]);
                }
                if(j>=-al[n1])
                {
                    dp[n1][j]=min(dp[n1-1][j+al[n1]]+1,dp[n1][j]);
                }
            }
        }
        if(dp[n][sum/2]>=n)
        {
            cout<<-1<<endl;
        }
        else
        {
            cout<<dp[n][sum/2]<<endl;
        }
    }
}