hdu-1231-maximal continuous sub-sequence (JAVA+DP dynamic programming)

Maximum continuous subsequenceTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 21101 Accepted Submission (s): 9361


Problem description A sequence of k integers {N1, N2, ..., NK}, any contiguous subsequence can be expressed as {Ni, ni+1, ...,
Nj}, where 1 <= i <= J <= K. The maximal contiguous subsequence is the element and the largest of all successive subsequence sequences,
For example, given sequence {-2, 11,-4, 13,-5,-2}, its maximum contiguous subsequence is {11,-4, 13}, Max
to 20.
In this year's data structure exam, it is required to write the program to get maximum and, now add a requirement, that also need to output the
The first and last elements of a subsequence.

The input test inputs contain several test cases, each with 2 rows, a positive integer k (< 10000) in line 1th, and a K integer in line 2nd, separated by a space. When k is 0 o'clock, the input ends and the use case is not processed.

Output for each test case, the first and last elements of the largest and longest contiguous subsequence are exported in 1 rows
The middle is separated by a space. If the maximum consecutive subsequence is not unique, the output sequence number I and J is the smallest one (as in the 2nd, 3 groups of the input sample). If all k elements are negative, define their maximum and 0, outputting the entire sequence of the first and the end elements.

Sample Input

6-2 11-4 13-5-210-10 1 2 3 4-5-23 3 7-2165-8 3 2 5 01103-1 -5-23-1 0-20

Sample Output

1310 1 410 3 510 100-1 -20 0 0HintHintHuge input, scanf is recommended.

SOURCE Zhejiang University Computer Postgraduate exam on the machine-2005
recommendjgshining | We have carefully selected several similar problems for you:1003 1087 1176 1203 1257

Undoubtedly, the most common sub-series of the problem, the test is dynamic programming, that is, DP. Although not as simple as the 01 backpack, the idea I chose is very simple and understandable. (See the simplest of ideas in parsing ...)

Import Java.io.*;import java.util.*;p ublic class main{public static void Main (string[] args) {Scanner input = new Scanner (S ystem.in); while (Input.hasnext ()) {int n=input.nextint (); if (n==0) break;int a[] = new Int[n];            Open an array to store the given sequence for (int i=0;i<n;i++) {a[i]=input.nextint ();} int sum,max,start,end,cnt=0;    Sum is current and, Max is present Max and, start is the starting element point, end is the end element point int temp=0;                     Used to temporarily save the starting point, that is, the next start,cnt record negative number. sum=start=end=0;max=a[0];for (int i=0;i<n;i++) {sum+=a[i];                  if (Sum>max)                 //If sum is greater than the maximum value, I is the last marker point end and the maximum value is sum.{ end=i;max=sum;start=temp;}                           else if (sum<0)//If current sum<0 and Sum<=max, that is, the current sum is certainly not the maximum value {temp=i+1;               The starting position should be the next number subscript sum=0;cnt++;                  Record negative number}}if (cnt==n)                      //cnt==n All the numbers are negative {System.out.println ("0" + "" +a[0]+ "" +a[n-1]);} else {System.out.println (max+ "" "+a[start]+" "+a[end]");}}}

hdu-1231-maximal continuous sub-sequence (JAVA+DP dynamic programming)