题目：
Max Sum
**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 230422 Accepted Submission(s): 54269
**
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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.\
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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).\
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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.\
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Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
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Sample Output
Case 1:
14 1 4
Case 2:
7 1 6
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Author
Ignatius.L
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思路：
题目给了n个数，让你从中截取一段数列，使他的和最大，并且需要最大和的数列的起始和结束位置，思路就是逐渐递推，以第一个样例为准，我们的递推思路是，以当前的为最后一个数，使当前的和最大的数列，则分别为，6,6,10,14,14，所以最大的就是14，在这个问题中，先把最大值初始化为-1001，因为题目的范围最小是它，然后用两个变量记录首尾，用sum逐渐累加当前的和然后用sum和当前最大的进行比较，更新最大值，再更新首尾的位置，因为题目的为置是从1开始的，所以当sum位负值的时候，temp=i+2,其中i+1是当前位置，则i+2表示下一个位置。
代码：
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int main()
{
int t;
scanf("%d",&t);
for(int q=1;q<=t;q++)
{
int a,n;
scanf("%d",&n);
int sum=0,maxsum=-1001,first=0,last=0,temp=1;//最大的数初始化为最大值
for(int i=0; i<n; i++)
{
scanf("%d",&a);
sum+=a;//计算当前的和
if(sum>maxsum)
{
maxsum=sum;//递推最大值
first=temp;//首项的位置
last=i+1;//更新尾项
}
if(sum<0)
{
sum=0;
temp=i+2;//当最大子序列和小于0时，将temp=i+2其中i+1表示当前位置,i+2就表示当前位置的下一个位置了。既此最大子序列和为负值，那么下一个的最大子序列和应该是它本身，而不再累加前边的。
}
}
printf("Case %d:\n%d %d %d\n",q,maxsum,first,last);
if(q!=t)printf("\n");
}
return 0;
}
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