HDU 1231: maximum continuous subsequence (DP)

Maximum continuous subsequence Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 18461 accepted submission (s): 8202


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 Huge input, scanf is recommended.

The same is true for hdu1003 Max sum .. Water ..

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<stdlib.h>#include<vector>#include<queue>#include<cmath>using namespace std;const int maxn = 100000 + 50;int n;int a[maxn];int start;int last;int temp;int ans;int sum;int t;int main(){    while(scanf("%d", &n)==1&&n)    {        t = 0;        memset(a, 0, sizeof(a));        for(int i=1; i<=n; i++)        {            scanf("%d", &a[i]);            if(a[i]<0)            t++;        }        if( t==n ){printf("0 %d %d\n", a[1], a[n]);continue;}        start = 1;        last = 1;        temp = 1;        sum = a[1];        ans = a[1];        for(int i=2; i<=n; i++)        {            if( sum<0 )            {                temp = i;                sum = 0;            }            sum += a[i];            if( sum>ans )            {                ans = sum;                start = temp;                last = i;            }        }        printf("%d %d %d\n", ans, a[start], a[last]);    }    return 0;}