HDU 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 211310    Accepted Submission(s):
49611

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

 

解题心得：

　　这个题是一个简单地动态规划题，题意：输入n个数，求它的最大的子序列和。首先写出状态转移方程： sum[i]=max{sum[i-1]+a[i],a[i]}。

　　s数组是记录开始位置，每当sum加上一个a[i]，值小于0的时候，s取用新的值。ans是记录结束位置，每当sum加上一个a[i]，值大于等于0的时候，更新一下。

　   我又在格式问题上出错了，以后要更加注意格式问题，最后一行不需要换行！

　　也可以参考这个人的思路，http://blog.csdn.net/code_pang/article/details/7772200，通过枚举发现的规律。

　

最后是代码：

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int t;
    int n;
    int a[];//存储序列
    int sum[];//存储以每个数为结尾的子序列和
    int s[];//存储开始位置
    int ans;//结束位置
    scanf("%d",&t);
    for(int i=;i<=t;i++){
        scanf("%d",&n);
        for(int i1=;i1<n;i1++){
            scanf("%d",&a[i1]);
        }
        ans=;
        sum[]=a[];
        s[]=;
        for(int j=;j<n;j++){
            if(sum[j-]>=){
                sum[j]=sum[j-]+a[j];
                s[j]=s[j-];
            }else{
                sum[j]=a[j];
                s[j]=j;
            }
            if(sum[ans]<sum[j])
                ans=j;
        }
        if(i<t){
            printf("Case %d:\n%d %d %d\n",i,sum[ans],s[ans]+,ans+);
            printf("\n");
        }else{
            printf("Case %d:\n%d %d %d\n",i,sum[ans],s[ans]+,ans+);
        }
    }
    return ;
}