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。

 #include <iostream>
 #include <cstdio>

 using namespace std;

 struct node
 {
     int s,e,smax;
 } sum[];

 int main ()
 {
     int t,n,Max,k,j;
     int num[];
     while (~scanf("%d",&t))
     {
         int flag=;
         while (t--)
         {

             scanf("%d",&n);
             for (int i=; i<n; i++)
             {
                 scanf("%d",&num[i]);
             }
             sum[].smax=num[];
             sum[].s=sum[].e=;
             k=,j=;
             Max=sum[].smax;
             for (int i=; i<n; i++)
             {
                 if (num[i]>sum[i-].smax+num[i])
                 {
                     sum[i].smax=num[i];
                     sum[i].s=i;
                     sum[i].e=i;
                 }
                 else
                 {
                     sum[i].smax=sum[i-].smax+num[i];
                     sum[i].s=sum[i-].s;
                     sum[i].e=i;
                 }
                 if (Max<sum[i].smax)
                 {
                     Max=sum[i].smax;
                     k=sum[i].s;
                     j=sum[i].e;
                 }
             }
             printf ("Case %d:\n",flag++);
             printf ("%d %d %d\n",Max,k+,j+);
             if (t)
                 printf ("\n");
         }
     }
     return ;
 }