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
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 = ;
int a[maxn];
int dp[maxn];
int main()
{
int t;
scanf("%d",&t);
int cas=;
while(t--)
{
int n;
scanf("%d",&n);
for(int i=;i<n;i++) scanf("%d",a+i);
dp[]=a[];
int ans = dp[];
int end_pos=;
for(int i=;i<n;i++)
{
dp[i]=max(dp[i-]+a[i],a[i]);
if(ans<dp[i])
{
ans=dp[i];
end_pos=i;
}
}
int tmp=;
int sta_pos;
for(int i= end_pos;i>-;i--)
{
tmp+=a[i];
if(tmp==ans)
{
sta_pos=i;
}
}
printf("Case %d:\n%d %d %d\n",cas++,ans,sta_pos+,end_pos+);
if(t>) cout<<endl;
}
return ;
}