A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1a2, ..., aN) be any sequence ( ai1ai2, ..., 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
#include <iostream>
#define M 1005
using namespace std; int arr[M],dp[M],maxn;
int main()
{
int n;
while(cin>>n){
maxn = 0;
for(int i = 0; i < n; i++)
cin>>arr[i];
for(int i = 0; i < n; i++)
{
dp[i] = 1;
for(int j = 0; j < i; j++)
{
if(arr[j] < arr[i])
dp[i] = max(dp[i],dp[j]+1);
}
maxn = max(maxn,dp[i]);
} cout<<maxn<<endl;
} return 0;
}

POJ 2533 裸的LIS的更多相关文章

  1. POJ 2533 动态规划入门 (LIS)

    Longest Ordered Subsequence Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 42914 Accepte ...

  2. POJ 2533 Longest Ordered Subsequence(LIS模版题)

    Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 47465   Acc ...

  3. poj 2533 Longest Ordered Subsequence 最长递增子序列

    作者:jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4098562.html 题目链接:poj 2533 Longest Ordered Subse ...

  4. nyoj 17-单调递增最长子序列 && poj 2533(动态规划,演算法)

    17-单调递增最长子序列 内存限制:64MB 时间限制:3000ms Special Judge: No accepted:21 submit:49 题目描述: 求一个字符串的最长递增子序列的长度 如 ...

  5. POJ 2533 最小上升子序列

    D - POJ 2533 经典DP-最长上升子序列 A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let th ...

  6. POJ 2533 Longest Ordered Subsequence(裸LIS)

    传送门: http://poj.org/problem?id=2533 Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 6 ...

  7. POJ 2533 - Longest Ordered Subsequence - [最长递增子序列长度][LIS问题]

    题目链接:http://poj.org/problem?id=2533 Time Limit: 2000MS Memory Limit: 65536K Description A numeric se ...

  8. LIS(n^2) POJ 2533 Longest Ordered Subsequence

    题目传送门 题意:LIS(Longest Increasing Subsequence)裸题 分析:状态转移方程:dp[i] = max (dp[j]) + 1   (a[j] < a[i],1 ...

  9. POJ 2533 Longest Ordered Subsequence (LIS DP)

    最长公共自序列LIS 三种模板,但是邝斌写的好像这题过不了 N*N #include <iostream> #include <cstdio> #include <cst ...

随机推荐

  1. Bayesian RL and PGMRL

    简介: PGMRL: PGMRL就是把RL问题建模成一个概率图模型,如下图所示: 然后通过variational inference的方法进行学习: PGMRL给RL问题的表示给了一个范例,对解决很多 ...

  2. Swagger UI 传入对象类型参数

    Swagger要传送对象作为参数,只需添加@ModelAttribute或@RequestBody @RestController @RequestMapping("/api/json/re ...

  3. 对List遍历过程中添加和删除的思考

    对List遍历过程中添加和删除的思考 平时开发过程中,不少开发者都遇到过一个问题:在遍历集合的的过程中,进行add或者remove操作的时候,会出现2类错误,包括:java.util.Concurre ...

  4. 抖音圈圈乐 系统搭建H5微信小游戏圈圈乐系统介绍

    网红线下游戏抖音圈圈乐改造而来 一.搭建此系统需要准备如下资料: 1. 认证微信服务号 2. 微信支付商户号 3. 备案域名及云服务器 二.系统功能简介: 1. 游戏闯关 2. 每个商品闯关难度后台自 ...

  5. Android学习Scroller(五)——具体解释Scroller调用过程以及View的重绘

    PS: 该篇博客已经deprecated,不再维护.详情请參见  站在源代码的肩膀上全解Scroller工作机制  http://blog.csdn.net/lfdfhl/article/detail ...

  6. c# 正则

    Regex reg = new Regex("^do(es)(xy)?$"); var result = reg.Match("doesxy"); foreac ...

  7. Oracle同义词(synonym)

    oracle的同义词总结   从字面上理解就是别名的意思,和视图的功能类似.就是一种映射关系.   同义词拥有如下好处:   节省大量的数据库空间,对不同用户的操作同一张表没有多少差别;   扩展的数 ...

  8. 3D Slicer Reconstruct CT/MRI

    3D Slicer Reconstruct CT/MRI 1. Load DCM file of your CT/MRI 2. Go to Volume Rendering, click the ey ...

  9. gym 101858

    我这个傻逼被治了一下午. 大好的橘势,两个小时6T,去看L,哇傻逼题.然后我跑的最短路T到自闭 最后十几分钟去想了下A,一直在想如何表示状态..就是想不到二进制搞一下... 然后游戏结束了..以后我就 ...

  10. Java9之HashMap与ConcurrentHashMap

    HashMap在Java8之后就不再用link data bins了,而是转为用Treeify的bins,和之前相比,最大的不同就是利用了红黑树,所以其由 数组+链表+红黑树 组成.: * This ...