算法_hdoj_1003

question：

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 324604    Accepted Submission(s): 77195

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

 

idea：

add up the sum each input

memory the max,and update it when the sum more than the max

set the sum as zero when the sum is less than zero,and meanwhile,set the temp as current order of the number and add two

Thinking process:if the sum is still more than zero ,next addition must bigger than it,and it must be less then the last when the sum is negative number,so initialize the sum only when it is negative

 

source code ：

 

import java.util.Scanner;

public class Main {
    static Scanner sc = new Scanner(System.in);
    public static void main(String[] args) {
//        int[] arr = {1,1,3,-2,-4,5,6};
//        System.out.println(rec_opt(arr,6));
//        System.out.println(rec_opt_circle(arr));
//        System.out.println(new int[2][3].length);
//        System.out.println(sum_extract(arr,arr.length-1,9));
        int group_amount = sc.nextInt();
        int counter = 1;
        while((group_amount--)>0){
            int start = 0;
            int max = -9999;
            int end = 0;
            int sum = 0;
            int temp = 1;

            int len = sc.nextInt();

            for (int i = 0;i<len;i++){
                sum+= sc.nextInt();
                if(sum > max){
             max = sum;
             start = temp;
             end = i+1;
             }

             if(sum<0){
             sum = 0;//重新初始化，然后跳转到当前位置重新开始累加
             temp = i + 2;
             }
             }
             /**
             * Case 1:
             * 14 1 4
             */
            System.out.println("Case "+(counter++)+":");
            System.out.println(max + " " + start + " " + end);
            if(group_amount!=0)System.out.println();
        }
    }
}

hope that I can help you guys XD

that's all