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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

大致思路：

用dp[ i ]表示当前到达第 i 项时，以第 i 项作为子数列的结尾能得到的最大和，那么可以发现转移方程：

dp[ i + 1] = max(dp[ i ] + a[ i + 1], a[ i + 1])

而最终的答案就是dp[ 1 ~ n ] 的最大值

在转移时记录当前的状态的来源即可知道起点和终点

#include<stdio.h>

int main()
{
	int t, n, num, sum, max, begin, end, temp, count, i;
	scanf("%d", &t); //输入例子数
	for (count = 1; count <= t; count++) {
		scanf("%d", &n); //输入数组长度
		sum = 0;
		temp = 1; //存储新的起始位置
		end = 1;  
		max = -1001; //存储最大值
		for (i = 1; i <= n; i++) {
			scanf("%d", &num);
			sum += num;
			if (sum < num) { 
				temp = i;
				sum = num;
			}
			if (sum > max) {
				max = sum;
				begin = temp;
				end = i;
			}

		}
		printf("Case %d:\n%d %d %d\n", count, max, begin, end);
		if (count != t)
			printf("\n");
	}
	return 0;

}