Max Sum

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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 141904 Accepted Submission(s): 33027

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
思路：用到了动态规划，理解清这一步：当m+now<m时，就只需要从下一步开始计数。

注意：如果一般方法例如用两个for会超时。
#include<stdio.h>
int main()

{
int now,max,x,m,n,i,j,test,pos1,pos2;
scanf("%d",&test);
for(i=1; i<=test; i++)

{
   scanf("%d %d",&n,&m);
   max=now=m;
   pos1=pos2=x=1;
   for(j=2; j<=n; j++)

   {
    scanf("%d",&m);
      if(m+now<m)

      {
         now=m;
           x=j;
       }
           else now+=m;
           if(now>max)

           {
             max=now;
             pos1=x;
              pos2=j;
          }
      }
printf("Case %d:\n%d %d %d\n",i,max,pos1,pos2);
if(i!=test)
printf("\n");
}
// while(1);
return 0;
}