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
解题思路：这个题目的意思是求一组数的最大子段和，最大子序列是要找出由数组成的一维数组中和最大的连续子序列。 找最大子序列的方法很简单，只要前i项的和还没有小于0那么子序列就一直向后扩展，否则丢弃之前的子序列开始新的子序列，同时我们要记下各个子序列的和，最后找到和最大的子序列。如果枚举来求解显然是行不通的，他的时间复杂度约为O(n^3),改用递归也会超时，所以最好利用动态规划和分治法相结合（时间复杂度为O(n)）。每一次求解时还要注意子序列的起始位置和重点位置。
程序代码：
#include<stdio.h>
int a[100005],str[100005],start[100005];
int main()
{
int t,n,i,num=1,end,max,k;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
str[1]=a[1];
start[1]=1;
for(i=2;i<=n;i++)
{
if(str[i-1]>=0)
{
str[i]=str[i-1]+a[i];
start[i]=start[i-1];
}
else
{
str[i]=a[i];
start[i]=i;
}
}
max=str[1];
end=1;
for(k=2;k<=n;k++)
{
if(str[k]>max)
{
max=str[k];
end=k;
}
}
printf("Case %d:\n",num);
num++;
printf("%d %d %d\n",max,start[end],end);
if(t)
printf("\n");
}
return 0;
}