hdu 1003 MAX SUM 简单的dp，测试样例之间输出空行

测试样例之间输出空行，if(t>0) cout<<endl;

这样出最后一组测试样例之外，其它么每组测试样例之后都会输出一个空行。

dp[i]表示以a[i]结尾的最大值，则：dp[i]=max(dp[i]+a[i],a[i])

解释：

 以a[i]结尾的最大值，要么是以a[i-1]为结尾的最大值+a[i],要么是a[i]自己本身，就是说，要么是连同之前的

构成一个多项的字串，要么自己单独作为一个字串，不会有其他的可能了。

状态规划的对状态的要求是：当前状态只与之前的状态有关，而且不影响下一个状态。

Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1003

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

#include<iostream>
#include<stdio.h>
using namespace std;
const int maxn = 100000;
int a[maxn];
int dp[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    int cas=1;
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d",a+i);
        dp[0]=a[0];
        int ans = dp[0];
        int end_pos=0;
        for(int i=1;i<n;i++)
        {
            dp[i]=max(dp[i-1]+a[i],a[i]);
            if(ans<dp[i])
            {
                ans=dp[i];
                end_pos=i;
            }
        }
        int tmp=0;
        int sta_pos;
        for(int i= end_pos;i>-1;i--)
        {
            tmp+=a[i];
            if(tmp==ans)
            {
                sta_pos=i;
            }
        }
        printf("Case %d:
%d %d %d
",cas++,ans,sta_pos+1,end_pos+1);
        if(t>0) cout<<endl;
    }
    return 0;
}