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
题意：
给你n个数a[1],a[2],a[3]...a[n],要你求出最大连续子段和，而且输出这个和，和这个子段的起始位置和终止位置，假设这个子段有多种可能，输出第一种结果就可以。
思路：
前段时间做个类似的题目，用的是暴力求解，用的两重循环。可是这个题目要求是n<100000，用两重循环肯定会超时，因此我们必须换一种思路，必须降低循环次数，当输入的时候，累计求和，用pos记录位置，star记录起点。end记录终点，假设sum小于0.就不必往以下累加了。
代码:
#include<cstdio>
using namespace std;
#define INF 0x7fffffff
int const maxn=100000+10;
int a[maxn];
int main()
{
int kcase=1;
int T,n,max,star,end,pos,sum;//max为最大字段和。star为枚举起始位置，end为枚举终止位置。sum为字段和。pos用来更新枚举起始位置
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
max=-INF;
pos=sum=0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
if(sum>max)
{
max=sum;
star=pos;
end=i;
}
if(sum<0)
{
sum=0;
pos=i+1;
}
}
printf("Case %d:
",kcase++);
printf("%d %d %d
",max,star+1,end+1);//数组下标是从0開始的，而数的位置是从1開始的
if(T)
printf("
");
}
return 0;
}