程序代码：

Max Sum

 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

 Total Submission(s): 98153    Accepted Submission(s): 22614

 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
#include <stdio.h>
#define N 100000 + 10
int main() {
    int T, n, a[N], sum, t, i, k, j, b, sum1, c;
    scanf("%d", &T);
        for(t = 1; t <= T; t++) {
            scanf("%d", &n);
            sum = 0; b = 0;
            for(i = 1; i <= n; i++) {
                scanf("%d", &a[i]);
                if(b > 0) b += a[i];
                else b = a[i];
                if(b > sum) {
                    sum = b;
                    k = i;
                }
            }
            sum1 = 0; c = 0;
            for(i = k; i >= 1; i--) {
                if(c > 0) c += a[i];
                else c = a[i];
                if(c >= sum1) {
                    sum1 = c;
                    j = i;
              }
          }
          printf("Case %d:\n%d %d %d\n", t, sum, j, k);
          if(t < T) printf("\n");
      }
      return 0;

 }
上次听了B版的建议后，昨晚想了很久，经过一番修改但是还是WA，求解释