标签：first   space   open   closed   opera   des   ttl   ted   ret   

for(int i=1;i<=n;i++)
    {
        nmax+=a[i];
        maxx=max(maxx,nmax);
        nmax=max(nmax,0);
    }

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 akfor 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

Examples

正确解法：

一直在想O（n）怎么写

题目是说 把一个子序列翻转，求最大1的个数。

我们把1的收益为 -1,0的收益为 1

求一个序列的最大收益，也就是求这个序列的最大子序列和

最大收益加上原本1的个数就是答案了。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 3000000+10;
 5 const ll inf = 1e17;
 6 int n;
 7 int a[150],yi,maxx=-999999,nmax;
 8 int main()
 9 {
10     scanf("%d",&n);
11     for(int i=1;i<=n;i++)
12     {
13         scanf("%d",&a[i]);
14         if(a[i]==1)
15         {
16             yi++;
17             a[i]=-1;
18         }
19         else
20             a[i]=1;
21     }
22     for(int i=1;i<=n;i++)
23     {
24         nmax+=a[i];
25         maxx=max(maxx,nmax);
26         nmax=max(nmax,0);
27     }
28     printf("%d\n",maxx+yi);
29 
30     return 0;
31 }

View Code

Kadane Algorithm

标签：first   space   open   closed   opera   des   ttl   ted   ret   

原文地址：https://www.cnblogs.com/Kaike/p/11270240.html