Description

A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8). Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4

Source

問題分析：這是一個最長上升子序列問題，使用DP演算法實現。

  定義f[i]=以a[i]為末尾的最長上升子序列的長度。

  那麼，以a[i]為末尾的最長上升子序列有以下兩種情形：

  1.只包含a[i]的子序列

  2.滿足j<i並且a[j]<a[i]的以a[j]為結尾的上升子序列末尾，追加上a[i]後得到的子序列，通過迴圈找到f[j]+1的最大值。

  即：f[i]=max{1,f[j]+1}

#include <iostream>
using namespace std;
///////////ac
int a[1005],f[1005];
int res(int n)
{
	int s=0;
	for(int i=0;i<n;i++)
	{
		f[i]=1;//為f[i]設定初值，最初只包含a[i] 
		for(int j=0;j<i;j++)
		{
			if(a[j]<a[i])
			{
				f[i]=max(f[i],f[j]+1);
			}
		}
		s=max(s,f[i]);//記錄f[i]的最大值 
	}
	return s;
}
int main()
{
	int n;
	while(cin>>n)
	{
		for(int i=0;i<n;i++)
		{
			cin>>a[i];
		}
		cout<<res(n)<<endl;
	}
	return 0;
}