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
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代码：
1 #include<cstdio>
2 int a[100000+10];
3 int main()
4 {
5 int t,n,k=1;
6 scanf("%d",&t);
7 while(t--)
8 {
9 scanf("%d",&n);
10 for(int i=1;i<=n;i++)
11 scanf("%d",&a[i]);
12 int max=-1001,st,endd,sum=0,st1=1; //注意整数的范围
13 for(int i=1;i<=n;i++)
14 {
15 sum+=a[i];
16 if(sum>max)
17 {
18 max=sum;
19 st=st1;
20 endd=i;
21 }
22 if(sum<0)
23 {
24 sum=0;
25 st1=i+1; //st1是临时的开始点，如果后面的sum<0，这个开始点也就不会计入
26 }
27 }
28 printf("Case %d:\n%d %d %d\n",k++,max,st,endd);
29 if(t!=0)
30 {
31 printf("\n");
32 }
33
34 }
35 return 0;
36 }