Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 265910 Accepted Submission(s): 63244

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

Solution

贪心问题，计算最大子序列，计算以每个节点为终点的最大子序列，若前一个节点最大子序列和为负则当前节点最大子序列和为其本身，否则当前节点加上前一节点的最大子序列和

Code

/**
 * date:2017.11.18
 * author:孟小德
 * function:杭电acm1003
 *  Max Sum     贪心算法，计算每一个点做为终点的最大子序列和
 */
 
 
 
import java.util.*;
 
public class acm1003
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
 
        int T = input.nextInt();
 
        for (int i=0;i<T;i++)
        {
            int N = input.nextInt();
 
            int[] num_N = new int[N];
            // 输入数列
            for (int j=0;j<N;j++)
            {
                num_N[j] = input.nextInt();
            }
 
            int[][] result = new int[N+1][3];
            result[0][0] = num_N[0];
            result[0][1] = 1;
            result[0][2] = 1;
            result[N][0] = num_N[0];
            result[N][1] = 1;
            result[N][2] = 1;
            for (int j=1;j<N;j++)
            {
                if (result[j-1][0] < 0)
                {
                    result[j][0] = num_N[j];
                    result[j][1] = j+1;
                    result[j][2] = j+1;
 
                }
                else
                {
                    result[j][0] = result[j-1][0] + num_N[j];
                    result[j][1] = result[j-1][1];
                    result[j][2] = j+1;
                }
                if (result[N][0] <= result[j][0])
                {
                    result[N][0] = result[j][0];
                    result[N][1] = result[j][1];
                    result[N][2] = result[j][2];
                }
            }
            System.out.println("Case " + (i+1) + ":");
            System.out.println(result[N][0] + " " +result[N][1]
                + " " + result[N][2]);
            if (i<T-1)
            {
                System.out.println();
            }
        }
 
    }
 
}