Longest Ordered Subsequence

A numeric sequence of  a_i is ordered if  a₁ <  a₂ < ... <  a_N. Let the subsequence of the given numeric sequence (  a₁,  a₂, ...,  a_N) be any sequence (  a_i1,  a_i2, ...,  a_iK), where 1 <=  i₁ <  i₂ < ... <  i_K <=  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
思路：
   DP；最长上升子序列；建立dp[i];当就ja[j],则可以将以a[j]为末尾的子序列追加a[i]后得到的得到的子序列，然后取max；

代码：

#include
#include
#include
using namespace std;
int a[1002]={0};
int dp[1002];
int main() {
 int t;
 scanf("%d",&t);
 for(int i=0;i  scanf("%d",&a[i]);
 }
    for(int i=0;i<1002;i++){
         dp[i]=1;
 }

    int maxx=0;
    for(int i=0; i        for(int j=0; j            if(a[i]>a[j]) dp[i]=max(dp[i],dp[j]+1);
            }
            maxx=max(maxx,dp[i]);
        }
        printf("%d\n",maxx);
    return 0;
}