hdu-1003 Max Sum

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

题目：

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 241930 Accepted Submission(s): 57107

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

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

题意概括：

这是一道水DP，是一个可以增加刚学会DP的童鞋们成就感的题，大概题意是，给你一个T，代表T组样例，然后每组样例给你一个数N，然后后面有N个数，问这N个数当中，那个区间的和最大，输出最大和最大和的区间左下标最大和的有下标

注：下标是从1开始的

解题思路：

dp的经典解题思路，利用循环，查找当前位置与之前记录的和是否比当前值还小（之前的和是否为负数），如果是，则将和赋值为当前值，记录当前值的下标，然后对记录的最大值进行比较，如果当前值更大，则更新最大区间左右下标和最大值。

AC代码：

 # include <stdio.h>
 int a[100010];
 int main ()
 {
     int t,n,ret,sum,max,i,sta,end,flag=1;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(i=0;i<n;i++)
             scanf("%d",&a[i]);
         sta=0; end=0; ret=0;
         max=a[0]; sum=a[0];
         for(i=1;i<n;i++)
         {
             if(sum+a[i]<a[i])
             {
                 ret=i;
                 sum=a[i];
            }
            else 
                sum+=a[i];
                
            if(max<sum)
            {
                max=sum;
                sta=ret;
                end=i;
            }
        }
        
        printf("Case %d:
",flag);
        printf("%d %d %d
",max,++sta,++end);
        flag++;
        if(t)
            printf("
");
    }
 }

易出错分析：

第一次提交RE了一次，当时没有看清题，结果数组开小了。然后根据题意，数组开大之后，将数组放入全局变量（数组大概在1000，000左右要放入全局变量，不能放入主函数中），便AC了