hdu杭电1003 dp 连续子序列最大值

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.
题意：
输出连续子序列的最大值，且输出起点和终点
思路
我是参考了大佬的思路，自己纯暴力写超时，后来才知道是dp题。
先确定状态，用数组存储每一项的子序列最大值。状态转移方程，如果第i-1项大于0，则第i项的子序列最大值可以由第i-1项的值相加得到否则加0.
代码

#include<stdio.h>
#include<string.h>

#define N 100000
int num[N+10];

int main()
{
    int t,i,n,x,sum,temp,end,start,max;
    scanf("%d",&t);
    x = t ;
    while( t --)
    {
        scanf("%d",&n);
        sum = 0;
        temp = 1;
        max = -1001;
        memset(num,0,sizeof(num));
        for( i = 1; i <= n; i ++)
        {
            scanf("%d",&num[i]);
            sum += num[i];//printf("sum=%d
",sum);
            if( sum > max)
            {
                max = sum;
                start = temp;
                end = i;
            }

            if( sum < 0)
            {
                sum = 0;
                temp = i + 1;
            }

        }
        printf("Case %d:
",x-t);
        printf("%d %d %d
",max,start,end);
        if( x-t < x)
            printf("
");
    }
    return 0;
 }