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 100000int 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\n",sum);            if( sum > max)            {                max = sum;                start = temp;                end = i;            }            if( sum < 0)            {                sum = 0;                temp = i + 1;            }        }        printf("Case %d:\n",x-t);        printf("%d %d %d\n",max,start,end);        if( x-t < x)            printf("\n");    }    return 0; }