Longest Ordered Subsequence（dp）

A numeric sequence of a_i is ordered ifa₁ < 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：

求给定序列的最长上升子序列的长度，如：（1，7），（1，3，5，9）（4，8）（1，3，4，8）都是原序列的子序列，但最长的长度为4

思路：

设 F（i）为以a（i）为最大值的上升子序列的长度，a（i）为原序列中第i个数字

F（i）=1                                 i=1;

F（i）=F（j）+1                  1<j<i      F（j）为小于a（i）的数字a（j）中以a（i）为最大值的上升子序列的长度

代码：

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int dp[1005];int a[1005];int main(){    int n;    scanf("%d",&n);    memset(dp,0,sizeof(dp));    int i;    for(i=1;i<=n;i++)        scanf("%d",&a[i]);    int j;    int temp;    dp[1]=1;//第一个为1    for(i=2;i<=n;i++){ //从第二个开始找        temp=0;        for(j=1;j<i;j++){            if(a[i]>a[j]&&temp<dp[j])//在比a【i】小的这些数中找到上升子序列最大的                temp=dp[j];        }        dp[i]=temp+1;    }    int maxx=0;    for(i=1;i<=n;i++)        maxx=max(dp[i],maxx);    printf("%d\n",maxx);return 0;}