帮忙看看，谢谢了。
Problem DescriptionGiven 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
帮忙看看错在哪里了？
#include<stdio.h>
main()
{
long i,j,t,n,str[100001],k[100001],f[100001],h,c;
while(scanf("%d",&t)!=EOF)
{
for(i=1;i<=t;i++)
{
scanf("%d",&n);
for(j=1;j<=n;j++)
{
scanf("%d",&str[j]);
if(j==1)
{k[j]=str[j];}
else
{k[j]=k[j-1]+str[j];}
}
h=k[1];
for(j=1;j<=n;j++)
{
f[j]=k[j];
if(f[j]>h)
{
h=f[j];
}
}
printf("Case %d:\n",i);
printf("%d ",h);
for(c=1;c<=n;c++)
{
if(k[c]==h)
{printf("%d %d\n",n-c,c);
break;}
}
if(i!=t)
{printf("\n");}
}
}
}
