POJ 2533 Longest Ordered Subsequence

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今天手感不好,来几道裸题。


Longest Ordered Subsequence
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 51872 Accepted: 23089

Description

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

71 7 3 5 9 4 8

Sample Output

4

题目道出天机:最长递增子序列。

f[i]表示到i为止且选i的情况下,最长的子序列。附代码

#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iostream>#include<cmath>#include<cstdlib>#include<ctime>using namespace std;const int MAXN=1010;int f[MAXN],a[MAXN];int main(){int n;scanf("%d",&n);for (int i=1;i<=n;i++){scanf("%d",&a[i]);f[i]=1;}for (int i=2;i<=n;i++){for (int j=1;j<i;j++){if (a[i]>a[j]){f[i]=max(f[i],f[j]+1);}}}int ans=0;for (int i=1;i<=n;i++){ans=max(ans,f[i]);}cout<<ans;    return 0;}