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## 1003 Max Sum

(2012-09-20 21:59:48)

### 谢先斌

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

Author

Ignatius.L

 6802937 2012-09-20 12:25:29 Accepted 1003 62MS 304K 501 B C++ hixxb

这是一个求最大子数列的题目，参考代码如下：

#include

using namespace std;

int main()

{

int start, end, n, t, k, max, temp, i, j, a;

cin>>t;

k=t;

while(k--)

{

max=-1002;

start=end=0;

temp=0;

cin>>n;

for(i=0, j=0; i

{

cin>>a;

temp+=a;

if(temp>max)

{

max=temp;

start=j;

end=i;

}

if(temp<0)

{

temp=0;

j=i+1;

}

}

cout<<"Case "<<t-k<<":"<<endl;

cout<<max<<" "<<start+1<<" "<<end+1<<endl;

if(k!=0)

cout<<endl;

}

return 0;

}

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