HDU 1003----Max Sum（最大连续子序列和）

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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 

#include<stdio.h>
int main(void)
{
	int T,n,i,k,s,e,st,max,num;
	int a[100005];
	scanf("%d",&T);
	for(k=1;k<=T;k++)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++) scanf("%d",&a[i]);
		printf("Case %d:\n",k);
		max=num=a[0];
		s=st=e=0;
		for(i=1;i<n;i++)
		{
			if(num<0) 
			{
			   num=a[i];
			   st=i;
			}
			else num+=a[i];
			if(max<num)
			{
			    max=num;
			    s=st;
				e=i;
			}
		}
	   	printf("%d %d %d\n",max,s+1,e+1);
		if(k!=T) printf("\n");
	}
	return 0;
}