POJ 2533 Longest Ordered Subsequence

Longest Ordered Subsequence

Description

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

71 7 3 5 9 4 8

Sample Output

4

题意：求出最长的连续递增序列长度。

简单的动态规划题，定义一个数组记录每个数之前的比它小的并且权值最大的一个数。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[1001],s[1001];int main(){    int t,i;    while(~scanf("%d",&t))    {        memset(s,0,sizeof(s));        int maxn=0;        i=0;        while(t--)        {            scanf("%d",&a[i]);            int j=i+1;            s[i]=1;            while(--j)            {                if(a[j-1]<a[i]&&s[j-1]+1>=s[i])                s[i]=s[j-1]+1;            }            if(s[i]>=maxn)                maxn=s[i];            i++;        }        printf("%d\n",maxn);    }    return 0;}