Maximum continuous subsequence and

Maximum continuous subsequence and
Maximum continuous subsequenceTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 19156 Accepted Submission (s): 8532


The Problem Description is a sequence of {N1, N2,..., NK} Given K integers. Any continuous subsequence can be represented as {Ni, Ni + 1 ,...,
Nj}, where 1 <= I <= j <= K. The maximum continuous subsequence is the element and the largest of all consecutive subsequences,
For example, for a given sequence {-2, 11,-4, 13,-5,-2}, its maximum continuous subsequence is {11,-4, 13}, the largest and
Is 20.
In this year's data structure examination paper, the maximum sum of programming requirements is required. Now, a requirement is added, that is,
The first and last elements of the subsequence.

The Input test Input contains several test cases. Each test case occupies two rows, Row 1 provides a positive integer K (<1st), and row 3 provides K integers, separated by spaces. When K is 0, the input ends and the case is not processed.

For each test case, Output the first and last elements of the largest sum and maximum continuous subsequences in one row.
, Separated by spaces. If the maximum continuous subsequence is not unique, the smallest sequence numbers I and j are output (for example, 2nd and 3 groups in the input sample ). If all K elements are negative, the maximum value is 0, and the first and last elements of the entire sequence are output.

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

20 11 1310 1 410 3 510 10 100 -1 -20 0 0HintHint 
#include 
  
   #include #include 
   
    #define MAXN 10000 + 10using namespace std;int num[MAXN];int main(){    int n;    int MAX,left,right;    int flag;    while(~scanf("%d",&n))    {        memset(num,0,sizeof(num));        if(n == 0)        {            break;        }        int i;        for(i=0; i
    
      MAX)            {                right = num[i];                MAX = count;                left = flag;            }        }        if(MAX >= 0)        {            printf("%d %d %d\n",MAX,left,right);        }        else        {            printf("0 %d %d\n",num[0],num[n-1]);        }    }    return 0;}