#include<stdio.h>
int main()
{
    int Sum(int a[100001],int m,int n);
    int a[20][100001];
    int bg,ed;
    int times;
    int i,j,k;
    int max[20]={0};
    scanf("%d",&times);
    for(k=0;k<times;k++)
    {
        scanf("%d",&a[k][0]);
        for(i=1;i<=a[k][0];i++)
            scanf("%d",&a[k][i]);
    }
    for(k=0;k<times;k++)
    {
        for(i=1;i<=a[k][0];i++)
            for(j=i;j<=a[k][0];j++)
                if(Sum(a[k],i,j)>max[k])
                {
                    max[k]=Sum(a[k],i,j);
                    bg=i;
                    ed=j;
                }
         printf("%d %d %d\n",max[k],bg,ed);
    }
    return 0;
}
int Sum(int a[100001],int m,int n)
{
    int sum=0;
    int i;
    for(i=m;i<=n;i++)
        sum=sum+a[i];
    return sum;
}

不懂啊，至少也要运行scanf("%d",&times);
这个吧

Max Sum
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 109135    Accepted Submission(s): 25151

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
是在做这道题目，题目输入要求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).
想了种拙劣的方法，就是计算出每种可能的值

那是不是应该改成指针？

翻译一下题目，题目大意是让你输入一些数字，然后算出连续相加的最大值

并输出相加的开始和结束的位置

题目要求测试的数据有点多

输入部分
    scanf("%d",&times);
    for(k=0;k<times;k++)
    {
        scanf("%d",&a[k][0]);
        for(i=1;i<=a[k][0];i++)
            scanf("%d",&a[k][i]);
    }
输入要测试的次数
输入每次测试的数据的长度，输入要测试的数据

我也很无奈呀，每次输入的数据长度要求是1-100000

对不起对不起，我该把注释写好的，害大家烦恼了