CF:A. Flipping Game

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, ..., 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

5
1 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 (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.

解题报告：

大体意思是给出一串数，a1~an;进行一次操作:找到一个范围i~j;在[ai~aj]范围内进行a[i]=1-a[i];的操作，找出最大值。

这题需要注意的是必须进行一次操作，例如n=1;a[1]=1;最后输出的结果是0而不是1;

我的方法是暴力枚举，O（n*n);

参考代码：

#include <iostream>

#include <string.h>

#include<cstdio>

#include<cstdio>

using namespace std;

int main()

{

//freopen("in.txt","r",stdin);

  int m,n,i,j,s,i2,j2,a[200],b[200],max,t,count;

  while(scanf("%d",&n)!=-1)

  {

      memset(b,0,sizeof(b));

      s=0;

      count=0;

      for(i=1;i<=n;i++)

      {

            scanf("%d",&a[i]);

            s+=a[i];

            if(a[i]==1)

            count++;

      }

      if(n==1)

      {

          if(a[1]==1)

        printf("0\n");

        else

        printf("1\n");

      }

else

      {

      b[1]=a[1];

      b[0]=0;

      b[n+1]=b
;

      max=-1;

      for(i=2;i<=n;i++)

      b[i]=a[i]+b[i-1];

     /* for(i=1;i<=n;i++)

      cout<<b[i]<<" ";

      cout<<endl;*/

      for(i=2;i<=n;i++)

       for(j=1;j<i;j++)

      {

          t=b[j-1]+(i-j+1-(b[i]-b[j]))+b
-b[i];

  // cout<<"j= "<<j<<" i="<<i<<" "<<b[j-1]<<" "<<i-j+1-(b[i]-b[j-1])<<" "<<b
-b[i]<<"  t="<<t<<endl;

          if(t<s)

          t=s;

          if(t>max)

          max=t;

      }

      if(count!=max)

      printf("%d\n",max);

      else

       printf("%d\n",max-1);

      }

  }

    return 0;

}