HDU 1003 Max Sum（dp+维护max）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 204582    Accepted Submission(s): 47829

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

Author

Ignatius.L

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题解：

每输入一个数就对比一下前几项的和，不断维护最大值就OK。

注意：If there are more than one result, output the first one！！！

AC代码：

#include <cstdio>
int main()
{
    int i,j,x,n,m,sum,max;
    int start,end,nowstart;
    scanf("%d",&n);
    {
        for (i=1;i<=n;i++)
        {
            scanf("%d%d",&m,&x);
            sum=max=x;
            start=end=nowstart=1;
            for (j=2;j<=m;j++)
            {
                scanf("%d",&x);
                if (sum+x<x)
                {
                    sum=x;
                    nowstart=j;
                }
                else
                {
                    sum+=x;
                }
                if (sum>max)
                {
                    max=sum;
                    start=nowstart;
                    end=j;
                }
            }
            printf("Case %d:\n%d %d %d\n",i,max,start,end);
            if (i!=n)
            {
                printf("\n");
            }
        }
    }
    return 0;
}