Question

杭电ACM1003帮看下哪里错了，老是WA，谢谢

代码在这里
#include <stdio.h>
int main()
{
int i,j,m,n,k,l,h,max,a[20],s[20];
scanf("%d",&m);
h=0;
while(h<m)
{
h++;
scanf("%d",&n);
if(n>=1&&n<=100000)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
max=a[0];
for(i=0;i<n;i++)
{
for(j=i;j<n;j++)
{
if(j==0)
s[j]=a[j];
else
s[j]=s[j-1]+a[j];
if(s[j]>max)
{
k=i;
l=j;
max=s[j];
}
}
s[i]=0;
}
printf("Case %d:\n",h);
printf("%d %d %d\n\n",max,k+1,l+1);
}
}
}
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

不好意思，第一次提问...

Answer 1

楼主注意读题啊
那个100000是数据范围
你数组只开20怎么装得下？- -
还有就是
这道题你用的是O(n2)的复杂度
就算不错也一定会TLE的
应该是要用到DP的思想（动态规划）
希望楼主能先自己想想，因为这题不难
如果实在想不出再问吧

Answer 2

连个题目说明都不给

我们怎么知道你写的是什么题目的程序啊？

至少给个链接吧 要不然怎么帮你