*Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 236675 Accepted Submission(s): 55857*
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
这道题是用动态规划求得最大和子序列，子序列指的是连续的一组数列，不是断断续续的数列。
array[i]的值表示 以第i个数字为结尾的子序列的最大和
array[1]; array[2] = max{array[1]+array[2], array[1]}; array[3] = max{array[1]+array[2]+array[3], array[2]+array[3], array[3]}; array[4] = max{array[1]+array[2]+array[3]+array[4], array[2]+array[3]+array[4], array[3]+array[4], array[4]}; ……
最后的最大和的值就是这里面的最大值。
简化一下，用一个变量maxSum来记录array计算运行当中的最大值，sum表示连续子序列的和,x和y表示起始点和终止点。 如果sum > maxSum，则maxSum=sum,并更新x和y的值； 如果sum < maxSum，表示当前的array值是负值，但是之后的数字可能会让sum值再次大于maxSum的值（相等同理）； 如果sum<0，则舍弃这一段子序列，因为这一段子序列只会使整个序列的和减小，更新temp标志。 （注意 只有当sum>maxSum时才能更新x和y的值，也就是说temp标记更改之后可能并没有出现比maxSum还大的值）
