1003：max sum

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

代码：

 

#include<cstdio>
int a[1000];

int  main()
{
 int cas;
 scanf("%d",&cas);
 for(int i=1;i<=cas;i++)
 {
  printf("Case %d:/n",i);
  int n,beg=1,end=1,temp=1;
  scanf("%d",&n);
  int sum=0,max=-9999;
  for(int j=1;j<=n;j++)
  {
   int num;
   scanf("%d",&num);
   sum+=num;
   if(sum>max)
   {
    max=sum;
    end=j;
    beg=temp;
   }
   if(sum<0)
   {
    sum=0;
    temp=j+1;
   }
  }
  printf("%d %d %d/n",max,beg,end);
  if(i!=cas)
   printf("/n");
 }
 return 0;
}