hdu 1003 Max Sum (DP)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158421    Accepted Submission(s): 37055

Problem Description

Given 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.

 

 

Input

The 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).

 

 

Output

For 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

2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

 

 

Sample Output

Case 1:

14 1 4

Case 2:

7 1 6

 

题目大意：给一串数字，然后要求出和最大的连续子序列。并且题目要求这个和最大的连续子序列输出起始位置和终止位置。

注意:初始化问题，还有空格的这个格式问题！

 

详见代码。

 1 #include <iostream>

 2 #include <cstdio>

 3 

 4 using namespace std;

 5 

 6 struct node

 7 {

 8     int s,e,smax;

 9 } sum[100010];

10 

11 int main ()

12 {

13     int t,n,Max,k,j;

14     int num[100010];

15     while (~scanf("%d",&t))

16     {

17         int flag=1;

18         while (t--)

19         {

20 

21             scanf("%d",&n);

22             for (int i=0; i<n; i++)

23             {

24                 scanf("%d",&num[i]);

25             }

26             sum[0].smax=num[0];

27             sum[0].s=sum[0].e=0;

28             k=0,j=0;

29             Max=sum[0].smax;

30             for (int i=1; i<n; i++)

31             {

32                 if (num[i]>sum[i-1].smax+num[i])

33                 {

34                     sum[i].smax=num[i];

35                     sum[i].s=i;

36                     sum[i].e=i;

37                 }

38                 else

39                 {

40                     sum[i].smax=sum[i-1].smax+num[i];

41                     sum[i].s=sum[i-1].s;

42                     sum[i].e=i;

43                 }

44                 if (Max<sum[i].smax)

45                 {

46                     Max=sum[i].smax;

47                     k=sum[i].s;

48                     j=sum[i].e;

49                 }

50             }

51             printf ("Case %d:\n",flag++);

52             printf ("%d %d %d\n",Max,k+1,j+1);

53             if (t)

54                 printf ("\n");

55         }

56     }

57     return 0;

58 }