B - Longest Ordered Subsequence（动态规划）

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B - Longest Ordered Subsequence

A numeric sequence of ai is ordered if a1 < a2 < … < aN. Let the subsequence of the given numeric sequence (a1, a2, …, aN) be any sequence (ai1, ai2, …, 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

题意：输入一串数字，求出最大的递增序列的长度。

思路：读入每个数字后遍历一遍，使用a[i]>a[j]作为判定条件，后转换成等式dp[i]=max(dp[i],dp[j]+1),最后输出最长递增序列转换成等式sum=max(sum,dp[i])

AC代码

#include
#include
#include
using namespace std;
int main()
{
    int a[1050],dp[1050],n;
    cin>>n;
     int sum=0;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    for(int i=1;i<=n;i++)
    {   dp[i]=1;
        for(int j=1;j<i;j++)
        {
         if(a[i]>a[j])
        dp[i]=max(dp[i],dp[j]+1);
          }
         sum=max(sum,dp[i]);
          }
          cout<<sum<<endl;return 0;
       }