CodeForces - 327AFlipping Game

A. Flipping Game

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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, …, 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.

Examples

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.

題意：

可以對一段區間裡的01取反一次，n<=100，

1.暴力枚舉區間在計算答案O(n^3)

2.暴力枚舉區間字首和優化算答案O(n^2)

3.觀察每次用字首和計算[L,R]答案(ai記錄1-i中，1的個數,bi記錄1-i中，0的個數）

ans=a[L-1]+a[n]-a[R]+a[R]-b[L-1];

現在考慮以R為右端點的區間，a[n]-a[R]+b[R]這一部分始終是相同的，是以我們隻要找到a[L-1]-b[l-1]這段區間的最大值即可，而這個最大值可以直接開一個變量維護

複雜度O(n)

O(n^2)代碼

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn=+;
int a[maxn],b[maxn],n,x;

int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=;i<=n;++i)
    {
      scanf("%d",&x);
      a[i]=a[i-]+(x==);
      b[i]=b[i-]+(x==);
    }
    int ans=;
    for(int i=;i<=n;++i)
      for(int j=i;j<=n;++j)
      ans=max(ans,a[i-]+a[n]-a[j]+b[j]-b[i-]);

    //for(int i=1;i<=n;++i)printf("a=%d b=%d\n",a[i],b[i]);

    printf("%d\n",ans);
  }
  return ;
}
           

O(n)代碼

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn=+;
int a[maxn],b[maxn],n,x;

int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=;i<=n;++i)
    {
      scanf("%d",&x);
      a[i]=a[i-]+(x==);
      b[i]=b[i-]+(x==);
    }
    int ans=,res=;
    for(int i=;i<=n;++i)
    {
      ans=max(ans,res+a[n]-a[i]+b[i]);
      if(a[i]-b[i]>res)res=a[i]-b[i];
    }


    //for(int i=1;i<=n;++i)printf("a=%d b=%d\n",a[i],b[i]);

    printf("%d\n",ans);
  }
  return ;
}