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

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

Flipping Game
Time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

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

He writesNIntegersA1, bytes,A2, middle..., middle ,...,AN. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indicesIAndJ(1 digit ≤ DigitILimit ≤ limitJLimit ≤ limitN) And flips all valuesAKFor which their positions are in range [I, Bytes,J] (That isILimit ≤ limitKLimit ≤ limitJ). Flip the valueXMeans to apply operationXBytes = bytes 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 integerN(1 digit ≤ DigitNLimit ≤ limit 100). In the second line of the input there areNIntegers:A1, bytes,A2, middle..., middle ,...,AN. It is guaranteed that each of thoseNValues is either 0 or 1.

Output

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

Sample test (s) input

51 0 0 1 0

Output

4

Input

41 0 0 1

Output

4

Note

In the first case, flip the segment from 2 to 5 (ILimit = Limit 2, limit,JRequired = required 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].

In the second case, flipping only the second and the third element (ILimit = Limit 2, limit,JNumbers = numbers 3) will turn all numbers into 1.

Solution: the question is that there is a sequence 0, 1, which allows you to reverse any subsequence and ask how many subsequences can be obtained after the operation.

The data is not big and direct violence. Perform Two-layer cyclic enumeration to retrieve the start and end points of the reverse sequence, count the number of 1 in each case, and save a maximum value.

AC code:

# Include <iostream> # include <cstdio> using namespace std; int a [105]; int main () {// freopen ("in.txt", "r", stdin ); int n; while (cin> n) {for (int I = 0; I <n; I ++) cin> a [I]; int ans = 0; for (int I = 0; I <n; I ++) {for (int j = I; j <n; j ++) {int sum = 0; for (int k = 0; k <n; k ++) {if (k> = I & k <= j) sum + = a [k] ^ 1; // returns the inverse else sum + = a [k];} if (ans <sum) ans = sum ;}} cout <ans <endl ;} return 0 ;}