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CodeforcesRound#191(Div.2)-A.FlippingGame_html/css

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CodeforcesRound#191(Div.2)-A.FlippingGame_html/css

CodeforcesRound#191(Div.2)-A.FlippingGame_html/css_WEB-ITnose: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。..
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导读CodeforcesRound#191(Div.2)-A.FlippingGame_html/css_WEB-ITnose: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。..


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,?...,?an. 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 ak for which their positions are in range [i,?j] (that is i?≤?k?≤?j). Flip the value of xmeans 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,?...,?an. 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

51 0 0 1 0

output

input

41 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.







解题思路:题意是讲有一个0,1序列,现允许你对任意一个子序列取反,问操作后得到的序列,最多能有多少个1。

数据不大,直接暴力。直接两层循环枚举取反序列的起点和终点,统计每种情况下1的个数,保存一个最大值即可。






AC代码:

#include #include using namespace std;int a[105];int main(){// freopen("in.txt","r",stdin); int n; while(cin>>n){ for(int i =0; i>a[i]; int ans = 0; for(int i=0; i=i && k<=j) sum += a[k]^1; //用异或^取反 else sum += a[k]; } if(ans < sum) ans = sum; } } cout< 




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CodeforcesRound#191(Div.2)-A.FlippingGame_html/css

CodeforcesRound#191(Div.2)-A.FlippingGame_html/css_WEB-ITnose: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。..
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