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最经典也是最简单的动态规划题目 

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

7
1 7 3 5 9 4 8

Sample Output

4

动态转移方程如下（这是O(n^2)级别的）：b[i] = Max{b[j] + 1} (1 <= j < i 且 a[j] < a[i])

程序代码如下

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#include<iostream>
using namespace std;
int a[5001], b[5001];
int main() {
    int i,j,n,summ=-1;
    cin>>n;
    for (i=1;i<=n;i++) {
        cin>>a[i];b[i]=1;
    }
    for (i=1; i <= n; i++) {
        for (j=1;j<i;j++)
            if (a[j]<a[i]&&b[j]+1>b[i]) {
                b[i]=b[j]+1; //方程：b[i] = Max{b[j] + 1} (1 <= j < i 且 a[j] < a[i])
            }
        summ=max(b[i],summ); //求最大的b[i] 
    }
    cout<<summ<<endl;
    //while (1);
    return 0;
}

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