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[100010];

int main ()
{
    int t,n,Max,k,j;
    int num[100010];
    while (~scanf("%d",&t))
    {
        int flag=1;
        while (t--)
        {

            scanf("%d",&n);
            for (int i=0; i<n; i++)
            {
                scanf("%d",&num[i]);
            }
            sum[0].smax=num[0];
            sum[0].s=sum[0].e=0;
            k=0,j=0;
            Max=sum[0].smax;
            for (int i=1; i<n; i++)
            {
                if (num[i]>sum[i-1].smax+num[i])
                {
                    sum[i].smax=num[i];
                    sum[i].s=i;
                    sum[i].e=i;
                }
                else
                {
                    sum[i].smax=sum[i-1].smax+num[i];
                    sum[i].s=sum[i-1].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+1,j+1);
            if (t)
                printf ("\n");
        }
    }
    return 0;
}