HDU 1003 Max Sum

Max SumTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 153057 Accepted Submission (s): 35699


Problem 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 is b etween-1000 and 1000).

Outputfor Each test case, you should output of the lines. The first line was "Case #:", # means the number of the the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end posi tion of the sub-sequence. If there is more than one result, output the first one. Output a blank line between 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

Authorignatius.l
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AC CODE:

#include <stdio.h>main () {    int n, I, a[100002], B, J, S, M, K;    Long Max, sum;    while (scanf ("%d", &n)! = EOF)    {        for (j = 1; J <= N; j + +)        {            scanf ("%d", &m);            for (i = 0; i < m; i++)                scanf ("%d", &a[i]);            sum = 0;            b = 1;            max = -1001;            for (i = 0; i < m; i++)            {                sum + = A[i];                if (Sum > Max)                {                    max = sum;                    K = b;                    s = i + 1;                }                if (Sum < 0)                {                    sum = 0;                    b = i + 2;                }            }            if (j! = N)                printf ("Case%d:\n%ld%d%d\n\n", J, Max, K, s);            else                printf ("Case%d:\n%ld%d%d\n", J, Max, K, s);}}    

Hdu 1003 Max Sum