Arithmetic Sequence（dp）

Arithmetic Sequence

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 51  Solved: 19
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Description

    Giving a number sequence A with length n, you should choosingm numbers from A(ignore the order) which can form an arithmetic sequence and make m as large as possible.

Input

   There are multiple test cases. In each test case, the first line contains a positive integer n. The second line contains nintegers separated by spaces, indicating the number sequenceA. All the integers are positive and not more than 2000. The input will end by EOF.

Output

   For each test case, output the maximum  as the answer in one line.

Sample Input

5
1 3 5 7 10
8
4 2 7 11 3 1 9 5

Sample Output

4
6

HINT

   In the first test case, you should choose 1,3,5,7 to form the arithmetic sequence and its length is 4.

   In the second test case, you should choose 1,3,5,7,9,11 and the length is 6.

题解：等差数列最长长度；开个二维dp，记录下就好了；一维dp也能过。。。

java超时；c过；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[2010][2010];
int a[2010];
int main(){
    int N;
        while(~scanf("%d", &N)){
            for(int i = 0; i < N; i++){
                scanf("%d", a + i);
            }
            sort(a, a + N);
            int ans = 1; 
            for(int i = 0; i < N; i++){
                for(int k = 0; k <= 2000; k++){
                        dp[i][k] = 0;
                }
            }
            for(int i = 0; i < N; i++){
                for(int j = 0; j < i; j++){
                    int d = a[i] - a[j];
                    dp[i][d] = dp[j][d] + 1;
                    ans = max(ans, dp[i][d] + 1);
                }
            }
            printf("%d
", ans);
            }
    return 0;
}

java：

import java.util.Arrays;
import java.util.Scanner;

public class  ArithmeticSequence{
    static int[][] dp = new int[2010][2010];
    public static void main(String[] args){
        Scanner cin = new Scanner(System.in);
        int N;
        while(cin.hasNext()){
            N = cin.nextInt();
            int[] a = new int[N];
            for(int i = 0; i < a.length; i++){
                a[i] = cin.nextInt();
            }
            Arrays.sort(a);
            int ans = 1; 
            for(int i = 0; i < N; i++){
                for(int k = 0; k <= 2000; k++){
                        dp[i][k] = 0;
                }
            }
            for(int i = 0; i < a.length; i++){
                for(int j = 0; j < i; j++){
                    int d = a[i] - a[j];
                    dp[i][d] = dp[j][d] + 1;
                    ans = Math.max(ans, dp[i][d] + 1);
                }
            }
            System.out.println(ans);
        }
    }
}

 dp;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[2010];
int a[2010];
int main(){
    int N;
        while(~scanf("%d", &N)){
            for(int i = 0; i < N; i++){
                scanf("%d", a + i);
            }
            sort(a, a + N);
            int ans = 1; 
                for(int i = 0; i <= 2000; i++){
                    for(int k = 0; k <= 2000; k++){
                        dp[k] = 0;
                }
                for(int j = 0; j < N; j++){
                    if(a[j] >= i)dp[a[j]] = max(dp[a[j] - i] + 1, dp[a[j]]);
                    else dp[a[j]] = 1;
                        ans = max(ans, dp[a[j]]);
                }
            }
            printf("%d
", ans);
            }
    return 0;
}