hdu 1003(동적 계획 시작) Max Sum

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.
 
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 <iostream>
#include <cstdio>
using namespace std;
const int maxn=100000+10;

int main()
{
	int n,T,a[maxn],max,begin,end,k,i,t,j;
	scanf("%d",&n);
	for (j=1;j<=n;j++)
	{
		t=0;max=-9999;k=1;//max ， 
		scanf("%d",&T);
		for (i=1;i<=T;i++)
		{
			scanf("%d",a+i);
			t+=a[i];
			if (t>max)// ， ， 
			{
				max=t;
				begin=k;
				end=i;
			}
			if (t<0)// 
			{
				t=0;
				k=i+1;// i+1 begin， 
			}
		}
		printf("Case %d:
",j);
		if (j!=n)
		{
			printf("%d %d %d

",max,begin,end);
		}
		else
			printf("%d %d %d
",max,begin,end);
	}
	return 0;
}

어려운 시작 파이팅!!!