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：给你一个序列，求最大连续序列的和；

解题思路：这个题上学期就写过了，Max Sum，上学期最后期末复习，不爱学高数了，就把杭电的题挨着写，第一页写了一半；但是没用DP，用一个maxn记录当前序列最大的和，从头枚举连续数列，最后输出maxn；

感悟：总结隔着一天才写的，因为时间太长了，早忘了当初写的什么；

代码：

#include

int main()

{

   int t,i,max=-1001,start=0,end=0,temp=0,sum=0;

    int a;

    long n;

    scanf("%d",&t);

    for(int i=1;i<=t;i++)

    {

   if(i!=1)

        printf("\n");

        scanf("%d",&n);

        max=-1001,start=0,end=0,temp=1,sum=0;

        for(int j=1;j<=n;j++)

        {

   scanf("%d",&a);

            sum+=a;

            if(sum>max)

            {

   max=sum;

                end=j;

                start=temp;

            }

            if(sum<0)

            {

  sum=0;

               temp=j+1;

            }

        }

        printf("Case %d:\n",i);

        printf("%d %d %d\n",max,start,end);

    }

    return 0;

}