#枚举和排序 A - Flipping Game

Description

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.

Sample Input
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Hint
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.

主要思路：暴力求解
写代码时思路不够清晰，代码不简洁，落了几种情况。有待改进。

#include <stdio.h>#include <stdlib.h>int main(){    int i,j,m,n;    int a[110];    scanf("%d",&n);    int e=0;    for(i=0; i<n; i++)    {        scanf("%d",&a[i]);        if(a[i]==1)e++;    }    if(e==n-1&&a[n-1]==0)    {        printf("%d\n",n);    }    else if(n==1&&a[0]==1)printf("0\n");    else if(e==n)printf("%d\n",n-1);    else if(n==1&&a[0]==0)printf("1\n");    else    {        int max=0;        for(i=0; i<n; i++)        {            if(a[i]==0)            {                int cnt1=1,cnt2=0;                for(j=i+1; j<n; j++)                {                    if(a[j]==1)cnt2++;                    else cnt1++;                    int q=cnt1-cnt2;                    max=max>q?max:q;                }            }        }        for(i=0; i<n; i++)        {            if(a[i]==0)            {                int cnt1=1,cnt2=0;                for(j=i+1; j<n; j++)                {                    if(a[j]==1)cnt2++;                    else cnt1++;                    int q=cnt1-cnt2;                    if(max==q)                    {                        int l,r;                        for( l=i-1; l>=0; l--)if(a[l]==1)cnt1++;                        for( r=j; r<n; r++)if(a[r]==1)cnt1++;                        printf("%d\n",cnt1);                        return 0;                    }                }            }        }    }    //system("pause");    return 0;}

附简单代码（转）

#include <stdio.h>int main(){    int arr[105];    int n, mx=0, x, y,i, j, m;    scanf("%d", &n);    arr[0]=0;    for(i=1; i<=n; i++){        scanf("%d", &m);        arr[i]=arr[i-1]+m; //区间【1，i】的元素和    }    arr[n+1]=0;    for(i=1; i<=n; i++){        for(j=i; j<=n; j++){            x=arr[j]-arr[i-1];            y=(j-i+1)-x+arr[i-1]+arr[n]-arr[j];  //变化区间和+左右区间和            if(y>mx)mx=y;        }    }    printf("%d\n", mx);    return 0;}