又是一篇题解：Longest Ordered Subsequence

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


题意(翻译)
如果a1 当给定数字序列时，程序必须找到其最长有序子序列的长度。
输入：
输入文件的第一行包含序列N的长度，第二行包含序列N整数的元素，每个元素的范围从0到10000，用空格隔开。1<=N<=1000
输出：
输出文件必须包含一个整数-给定序列的最长有序子序列的长度。

这个就是找子序列的问题，找到最长有序子序列，输出最长有序子序列的长度。

#include
#include
#include

using namespace std;
int main() {
	int n, a, cnt;
	int ans[1005] = {0};
	while(cin >> n) {
		cnt = 0;
		while(n--) {
			cin >> a;
			if(cnt == 0)
				ans[cnt++] = a;
			else {
				if(a > ans[cnt-1]) {
					ans[cnt++] = a;
				} else {
					int pos = lower_bound(ans, ans+cnt-1, a)-ans;
					ans[pos] = a;
				}
			}
		}
		cout << cnt << endl;
	}
}