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
以下是我的代码，不知道是不是我哪里看漏了，麻烦大家帮忙指正下...谢谢

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

新号没什么分，望大家包涵....

希望能指出我的错误所在，不求新的代码...
http://acm.hdu.

我没看你的程序，给你测试了下，用上面的测试数据可以啊！程序编译又没有错误

你的错误所在在于算法压根就不对。这个问题叫最大子段和，是动态规划的教学级案例，不是简单地卡个头尾就能解决的。举个例子 1 -100 10 你觉得结果该是什么。

，算法效率可以优化到o(n).给你个提示先预处理算出所有sj.代表重从a0加到aj的。用sj-si可以算出ai+1到aj的和，懂了没～～～别说还不懂～～

哦哦...多谢，学习了...

吸收大家的意见去试了下，但是还是不行

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

 }

说说我的思路吧方便大家查错
1.用for输入n个数
2.最长字段算法算出max
并找出最后的i
3,重新for找到第一个>=0的数
这就是我的思路

第三布明显错了

那该怎么弄，给思路就好。。。。