A. Flipping Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub got bored, so he invented a game to be played on paper.

He writes n integers a₁, a₂, ..., a_n. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values a_k for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.

The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a₁, a₂, ..., a_n. It is guaranteed that each of those n values is either 0 or 1.

Output

Print an integer — the maximal number of 1s that can be obtained after exactly one move.

Sample test(s)

Input

5 1 0 0 1 0

Output

4

Input

4 1 0 0 1

Output

4

Note

In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].

In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.

思路：设数列为array，one[i]表示从array[1]到array[i]（包括上下界）1的个数。故当对[i,j]范围内的数执行flip操作后，数列1的个数为：

one[n] - (one[j] - one[i-1]) + (j - i + 1 - (one[j] - one[i-1]));

式子中(one[j] - one[i-1])为[i,j]范围内1的个数，(j - i + 1 - (one[j] - one[i-1]))自然就是[i,j]中0的个数。

AC Code：

1 #include 
        
          2 #include 
         
           3  4 using namespace std; 5  6 const int maxn = 105; 7 int one[maxn], n; 8  9 int main()10 {11     while(scanf("%d", &n) != EOF)12     {13         int b;14         one[0] = 0;15         for(int i = 1; i <= n; i++)16         {17             one[i] = one[i-1];18             scanf("%d", &b);19             one[i] += b;20         }21         int cnt = -1;22         for(int i = 1; i <= n; i++)23         {24             for(int j = i; j <= n; j++)25             {26                 int tmp = one[n] - (one[j] - one[i-1]) + (j - i + 1 - (one[j] - one[i-1]));27                 if(cnt < tmp) cnt = tmp;28             }29         }30         printf("%d\n", cnt);31     }32     return 0;33 }