Longest Ordered Subsequence 最长上升子序列+DP

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

题目大意是让你求给的数字序列的最大上升子序列，直接上DP，每个位置的dp数组值就是当前位置之前的所有比自己小的上升子序列的长度，两重循环就可以解决。

AC代码

#include
#include
#include
using namespace std;
int num[1005];
int dp[1005];
int main()
{
	int N;
	while(scanf("%d",&N)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		int mmax=-1;
		for(int i=1;i<=N;i++)
			cin>>num[i];
		for(int i=1;i<=N;i++)
		{
			for(int j=1;jnum[j])
					dp[i]=max(dp[i],dp[j]+1);
			if(dp[i]==0) 
				dp[i]=1; 
			mmax=max(dp[i],mmax);
		}
		cout<