经典DP水题A

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：给一串数字，输出从左往右由小到大的最大长度，即 1 7 3 5 9 4 8,最长为1 3 5 8 或者1 3 4 8等等，长度为4.

思路：DP，从左往右遍历一遍，更新比它小的数，统计此刻长度。

AC java 代码：

import java.util.Scanner;
//3812 KB 1563 ms
public class sdupractice0724DPA {


public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[][] arr = new int[n][2];
for (int i = 0; i < n; i++) {
arr[i][0] = scan.nextInt();
arr[i][1] = 1;
}
for (int i = 1; i < n; i++) {
int count = 0;
for (int j = 0; j < i; j++) {
if (arr[i][0] > arr[j][0] && arr[j][1] > count) {
count = arr[j][1];
}
}
arr[i][1] += count;
}
int max = 0;
for (int i = 0; i < n; i++) {
if (arr[i][1] > max)
max = arr[i][1];
}
System.out.println(max);
}


}