HDU 1003 Max Sum

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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);        }    }}

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