HDU ACM 1003

（免费宾馆住宿 + 实习工资）阿里巴巴实习生招聘信息（仅限2013届本、硕毕业生）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66796    Accepted Submission(s): 15297

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

 

 

 

 

import java.util.Scanner;
public class Main {

	public static void main(String[] args) {

		int t, i;
		Scanner in = new Scanner(System.in);
		t = in.nextInt();
		for (i = 0; i < t; i++) {
			anster(in, i + 1);

			if (i != t - 1)
				System.out.println();
		}
	}

	private static void anster(Scanner in, int k) {

		int l = 0;
		int r = 0;
		int sum = 0;
		int count = in.nextInt();
		int num;
		int max = -1001; 
		int tempStep = 1;

		for (int i = 0; i < count; i++) {
			num = in.nextInt();
			sum = sum + num;

			if (sum > max) {
				l = tempStep;
				r = i + 1;
				max = sum;
			}

			if (sum < 0) {

				sum = 0;
				tempStep = i + 2;

			}
		}

		System.out.println("Case " + k + ":");
		System.out.println(max + " " + l + " " + r);
	}

}