Flipping Game
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.
Example
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：
給定一串01序列，選擇其中一段翻轉，也就是0變1，1變0.求一次翻轉後的最多1的數目。
思路：
暴力枚舉：
用個數組儲存字首和，表示i之前1的數目，在枚舉每個區間，資料比較小，也能過。
區間DP：
預處理差不多，DP[i][j]表示I到j之間一次翻轉後一最多的數量。
對于每個區間而言，在中間找個斷點k，一次翻轉可能發生在前面I到K，也可能發生在K到j之間，
那我們就得到了狀态轉移方程
dp[I][j]=max(dp[i][k]+sum[j]-sum[k],dp[k+1][j]+sum[k]-sum[i-1]));
(i<=K < j)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<math.h>
#include<string.h>
#define mem(a,b) memset((a),(b),sizeof(b));
using namespace std;
int main()
{
ios::sync_with_stdio(false);
// freopen("text.txt", "r", stdin);
int a[];
int sum[];
int dp[][];
int n;
cin>>n;
sum[]=;
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++)
{
cin>>a[i];
sum[i]=a[i]+sum[i-];
dp[i][i]=-a[i];
}
for(int l=;l<n;l++)
for(int i=;i+l<=n;i++)
{
int j=i+l;
dp[i][j]=j-i+-sum[j]+sum[i-];
for(int k=i;k<j;k++)
dp[i][j]=max(dp[i][j],
max(dp[i][k]+sum[j]-sum[k],
dp[k+][j]+sum[k]-sum[i-]));
//cout<<i<<' '<<j<<' '<<dp[i][j]<<endl;
}
cout<<dp[][n]<<endl;
return ;
}
所謂水題就是拿來練習，隻有對某種方法足夠熟練，有一定了解才能用起來得心應手，雖然我現在菜的不行，但堅持下去，我想會有收獲的。