# HDU 1003 Max Sum（最大子列和）

Problem Description

Given a sequence a,a,a......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 <cstdio>
#define INF 0x3f3f3f3f
int main()
{
int t;
int n, tt;
scanf("%d",&t);
int k = 0;
while(t--)
{
scanf("%d",&n);
int sum = 0, maxx = -INF;
int start = 1, endd = 1, pos = 1;
for(int i = 1; i <= n; i++)
{
scanf("%d",&tt);
sum+=tt;
if(sum > maxx)
{
start = pos;
endd = i;
maxx = sum;
}
if(sum < 0)
{
sum = 0;
pos = i+1;
}
}
printf("Case %d:\n",++k);
printf("%d %d %d\n",maxx,start,endd);
if(t)
printf("\n");
}
return 0;
}
```

## HDU 1003 Max Sum 最大连续子序列的和

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