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：
给出一个序列，求出和最大的子序列，并输出最大和 和 和最大的子序列开始下标和结束下标。
思路：
先判断序列第一个元素是否大于0，若大于0，令此时最大和等于第一个元素的值，若小于0，则舍去第一个元素，此时最大和为0，要注意下标的变化，然后依次相加，同理，若和为负数，则归为0，更新下标，若此时和加上后面的一个数变小了，则此时说明和最大的子序列已经到了结束的时候了，记录坐标，注意输出格式！
代码：
#include<stdio.h>
int main()
{
int t,k=0;
scanf("%d",&t);
while(t--)
{
k++;
int n,a,s;
scanf("%d",&n);
scanf("%d",&a);
s=a;
int i,b,c=1,head=1,tail=1;
for(i=2; i<=n; i++)
{
scanf("%d",&b);
if(s<0)
{
s=0;
c=i;
}
s=s+b;
if(s>a)
{
a=s;
head=c;
tail=i;
}
}
if(k==1)
{
printf("Case %d:\n",k);
printf("%d %d %d\n",a,head,tail);
}
else
{
printf("\nCase %d:\n",k);
printf("%d %d %d\n",a,head,tail);
}
}
return 0;
}