UVA 787 Maximum Sub-sequence Product

Bob needs money, and since he knows you are here, he decided to gamble intelligently. The game is rather simple: each player gets a sequence of integers. The players must determine, by using their mega-pocket computers, which is the maximum product value which can be computed with non empty sub-sequences of consecutive numbers from the given sequence. The winner is the one which gets first the right result.
Can you help Bob with a program to compute quickly the wanted product, particularly when the sequence is quite long?
Input

The input file contains sequences of numbers. Each number will have at most 5 digits. There will be at most 100 numbers in each sequence. Each sequence starts on a new line and may continue on several subsequent lines. Each sequence ends with the number -999999 (which is not part of the sequence).
Output
The maximum sub-sequence product for each sequence must be written on the standard output, on a different line. A simple example is illustrated in the sample below.
Sample Input
1 2 3 -999999
-5 -2 2 -30 -999999
-8 -999999
-1 0 -2 -999999
Sample Output
6 120 -8 0
最大子序列乘积和分别记录最小和最大的乘积。类似与最大子序列和的dp思想，由于是乘积用到了java BigInterger类否则会超出数据范围。代码如下。

package com.company;

import java.math.BigInteger;
import java.util.Scanner;

/**
 * Created by wangruoxuan on 16/2/6.
 */
public class MainDp {
    public static BigInteger maxTwo(BigInteger a,BigInteger b){
        if(a.compareTo(b)>0){
            return a;
        }else {
            return b;
        }
    }
    public static BigInteger maxThree(BigInteger a, BigInteger b, BigInteger c){
        return maxTwo(a,maxTwo(b,c));
    }
    public static BigInteger minThree(BigInteger a,BigInteger b,BigInteger c){
        if(a.compareTo(b)>0){
            if(b.compareTo(c)>0){
                return c;
            }else {
                return b;
            }
        }else {
            if(a.compareTo(c)>0){
                return c;
            }else {
                return a;
            }
        }
    }
    public static void main(String[] args){
        final int max_n = 10000;
        int[] a = new int[max_n];
        int x;
        Scanner input = new Scanner(System.in);
        BigInteger[] dp_max = new BigInteger[max_n];
        BigInteger[] dp_min = new BigInteger[max_n];
        while (input.hasNext()){
            int pos = 0;
            while (input.hasNext()&&(x=input.nextInt())!=-999999){
                a[pos++] = x;
            }
            dp_max[0] = BigInteger.valueOf(a[0]);
            dp_min[0] = BigInteger.valueOf(a[0]);
            for (int i = 1;i<pos;i++){
                BigInteger bigElement = BigInteger.valueOf(a[i]);
                dp_max[i] = maxThree(dp_max[i-1].multiply(bigElement),bigElement,dp_min[i-1].multiply(bigElement));
                dp_min[i] = minThree(dp_max[i-1].multiply(bigElement),bigElement,dp_min[i-1].multiply(bigElement));
            }
            BigInteger ans = dp_max[0];
            for(int i = 1;i<pos;i++){
                ans = maxTwo(ans,dp_max[i]);
            }
            System.out.println(ans);


        }
    }
}