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;
}