求大神指教，这程序是怎么个思想？
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
代码：
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i,t;
scanf("%d",&t);
for (i=1; i<=t; i++)
{
int j,n,max=-1001,sum=0,tmp=1,*a;
int first=0,last=0;
scanf("%d",&n);
a = (int *)malloc(n*sizeof(int));
for (j=0; j<n; j++)
{
scanf("%d",a+j);
sum += a[j];
if (sum>max)
{
max = sum;
first = tmp;
last = j+1;
}
if (sum<0)
{
sum = 0;
tmp = j+2;
}
}
printf("Case %d:\n%d %d %d\n",i,max,first,last);
if (i!=t)
printf("\n");
free(a); a = NULL;
}
return 0;
}
[解决办法]
假设a(k)到a(j)的和最大,那么a(1)到a(k-1)一定是负数,否则a(i)到a(j)的和就大于a(k)到a(j)的和
因此可以把和是负数的地方看成断点然后重新开始累加,
最后记录整个累加过程的最大值就好
