HDU 1003 Max sum (DP, maximum continuous subsequence and)

Max sumproblem descriptiongiven a sequence a [1], a [2], a [3] ...... A [n], your job is to calculate the max sum of a sub-sequence. for example, given (6,-1, 5, 4,-7), the Max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
 
Inputthe first line of the input contains an integer T (1 <= T <= 20) which means the number of test cases. then T lines follow, each line starts with a number N (1 <= n <= 100000), then n integers followed (all the integers are between-1000 and 1000 ).
Outputfor each test case, you should output two lines. the first line is "case #:", # means the number of the test case. the second line contains three integers, the max sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. if there are more than one result, output the first one. output a blank line between two cases.
Sample Input

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample output

Case 1:14 1 4Case 2:7 1 6

Evaluate the maximum continuous sum of N numbers

D [I] indicates the maximum continuous subsequence with number I a as the right end and it is easy to obtain the transfer equation d [I] = max (d [I-1] + a, a)

Obviously, when the number of I is greater than the maximum number of I-1 on the right and the number of I on the right, the maximum number of I on the right and the number of I on the same time update left end for yourself

#include<cstdio>#include<cstring>using namespace std;const int N = 100005;int main(){    int a, cas, ans, l, le, ri, n, d[N];    scanf ("%d", &cas);    for (int k = 1; k <= cas; ++k)    {        memset (d, 0x8f, sizeof (d));        ans = d[0];        scanf ("%d", &n);        for (int i = 1; i <= n; ++i)        {            scanf ("%d", &a);            if (d[i - 1] + a < a)                d[i] = a, l = i;            else                d[i] = d[i - 1] + a;            if (d[i] > ans)                ans = d[i], le = l, ri = i;        }        if (k > 1) printf ("\n");        printf ("Case %d:\n%d %d %d\n", k, ans, le, ri);    }    return 0;}