Codeforces Round #191 (Div. 2) A. Flipping Game(简单)

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

Input

4
1 0 0 1

Output

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 <iostream>
 2 #include <cstdio>
 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 }

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