题目链接:

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 158421    Accepted Submission(s): 37055

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

 

题目大意：给一串数字，然后要求出和最大的连续子序列。并且题目要求这个和最大的连续子序列输出起始位置和终止位置。

注意:初始化问题，还有空格的这个格式问题！

 

详见代码。

1 #include 
       
         2 #include 
        
          3  4 using namespace std; 5  6 struct node 7 { 8     int s,e,smax; 9 } sum[100010];10 11 int main ()12 {13     int t,n,Max,k,j;14     int num[100010];15     while (~scanf("%d",&t))16     {17         int flag=1;18         while (t--)19         {20 21             scanf("%d",&n);22             for (int i=0; i
         
          sum[i-1].smax+num[i])33                 {34                     sum[i].smax=num[i];35                     sum[i].s=i;36                     sum[i].e=i;37                 }38                 else39                 {40                     sum[i].smax=sum[i-1].smax+num[i];41                     sum[i].s=sum[i-1].s;42                     sum[i].e=i;43                 }44                 if (Max