HDU 1003 Max Sum

最新推荐文章于 2022-03-03 08:27:19 发布

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最新推荐文章于 2022-03-03 08:27:19 发布

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本文介绍了一种求解整数序列中具有最大和的连续子序列的问题，并提供了一个具体的Java实现示例。该问题旨在寻找给定序列中能够构成最大和的子序列及其起始与终止位置。

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Max Sum

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

Code

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        for (int z = 0; z < t; z++) {
            int len = s.nextInt();
            int i = 1;
            int temp = s.nextInt();
            int pos1, pos2, x;
            pos1 = pos2 = x = 1;
            int maxSum, sum;
            maxSum = sum = temp;
            while (i < len) {
                temp = s.nextInt();
                i++;
                if (sum + temp < temp) {
                    x = i;
                    sum = temp;
                } else {
                    sum += temp;
                }
                if (sum > maxSum) {
                    maxSum = sum;
                    pos1 = x;
                    pos2 = i;
                }
            }
            System.out.println("Case " + (z + 1) + ":");
            if (z != t - 1) {
                System.out.println(maxSum + " " + pos1 + " " + pos2);
                System.out.println();
            } else
                System.out.println(maxSum + " " + pos1 + " " + pos2);

        }
    }
}