poj 2533 Longest Ordered Subsequence

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 31787		Accepted: 13903

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_{i_K}), 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

7
1 7 3 5 9 4 8

Sample Output

题意：求最长子序列的长度

#include<stdio.h>
int main ()
{
  int n,i,j;
  int t=0;
  int a[1006],dp[1006];
  scanf("%d",&n);
  for(i=0;i<n;i++)
  {
   scanf("%d",&a[i]);
      dp[i]=1;
  }

  for(i=0;i<n;i++)
  {
    for(j=0;j<i;j++)
	{
	  if(a[i]>a[j] && dp[i]<dp[j]+1)  //  a[j]和a[i]能构成一个上升子序列,然后把当前上升子序列的长度存到dp[i]中

		  dp[i]=dp[j]+1;
	}
  }

  for(i=0;i<n;i++)
  {
    if(t<dp[i])
		t=dp[i];
  }
  printf("%d
",t);
  return 0;
}