hdu----1003 Max Sum-优快云博客

本文介绍了一个关于计算给定整数序列中具有最大和的子序列的问题。通过动态规划的方法，逐步迭代并记录最大子序列和及其起始与结束位置，最终输出结果。文章包含完整的C语言代码实现。

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水题，主要是最后输出比较特殊，开始是输出一行空一行，最后只换行输出，不空一行！

对于dp我也没学好，总之思想是从前往后叠加，前面的和与下一个数比较，并更新最大值maxx！

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158606 Accepted Submission(s): 37098

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>
int a[100010];
int main()
{
    int t,n,step=1;
    scanf("%d",&t);
    int sum;
    int start,endl,pos;
    int maxx;
    while(t--)
    {
        start=endl=pos=1;
        sum=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        maxx=sum=a[1];
        for(int i=2;i<=n;i++)
        {
            if(sum+a[i]<a[i])
            {
                sum=a[i];
                pos=i;
            }
            else
            {
                sum+=a[i];
            }
            if(sum>maxx)
            {
                maxx=sum;
                start=pos;
                endl=i;
            }
        }
        printf("Case %d:\n",step);
        printf("%d %d %d\n",maxx,start,endl);
        if(t)
            printf("\n");
        step++;
    }
    return 0;
}