2486: Stock Wave

最新推荐文章于 2024-09-30 09:30:00 发布

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最新推荐文章于 2024-09-30 09:30:00 发布

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分类专栏： joj acm_动态规划文章标签： output integer less input each up

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本文介绍了一种算法，用于从一系列历史价格数据中找出最长的波动序列（即“波浪”），这是一种先升后降再重复的模式。输入是一系列正数价格，输出是最长波动序列的长度。

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2486: Stock Wave

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	32768K	1208	188	Standard

As a stock analyst, Tom can forecast the trend according to a series of historical prices. To find the "Wave" is the first thing that he needs to do.

A "Wave" is a series of prices, the value in even place is greater than the former one, and the value in odd place is less than the former one. If you draw these values in one curve, you will see "up, down, up, down ...." Yes, it is a "Wave".

Can you help Tom to find out the longest "Wave"?

Input

The first line of each case is a positive integer n (n < 1000). The next n line(s) is a price with two decimal places.

Output

Print the length of the longest "Wave" in the input, if the length is less than 3, output '0'.

Sample Input

Sample Output

5
0

Hint

The longest "Wave" is a subsequence of initial price series, the neighboring values of "Wave" may be NOT neighboring in the original series, but they must keep order of input.

Problem Source: provided by skywind

#include<iostream>
using namespace std;
const int maxn=1000;
int main()
{
int n;
double a[maxn+1];
while(cin>>n)
{
  int i,j;
   for(i=0;i<n;i++)
   cin>>a[i];
  int sum[maxn+1];//表示前n个数的最大波长
  int lab[maxn+1];//标记状态 0表示位于波峰 1表示位于波谷
  lab[0]=sum[0]=1;
  int t=-1;
  for(int k=1;k<n;k++)
  {
   if(a[k]>a[k-1])
   {
    t=k;
    break;
   }
  }
  if(t==-1) {cout<<"0"<<endl;continue;}
  for(i=0;i<t;i++)
   sum[i]=lab[i]=1;
  for(i=t;i<n;i++)
  {
   int max=0;
   int f=2;
   for(j=i-1;j>=0;j--)
   {
    if(lab[j]==0&&a[i]<a[j])//表示可形成波谷
    {
     if(sum[j]>max)
     {
      max=sum[j];
      f=1;//及当前状态为波峰
     }
    }
    else if(lab[j]==1&&a[i]>a[j]) //表示可形成波峰
    {
     if(sum[j]>max)
     {
      max=sum[j];
      f=0;//记当前状态为波谷
     }
    }

   }

   sum[i]=max+1;
    lab[i]=f;

  }
  if(sum[n-1]<3) sum[n-1]=0;
  cout<<sum[n-1]<<endl;
}
return 0;
}