Java解题-杭电OJ-1003题

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 222921 Accepted Submission(s): 52449

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

import java.util.Scanner;public class Main{    public static void main(String[] args)    {        Scanner s=new Scanner(System.in);        int cas=s.nextInt();        for (int i = 1; i <= cas; i++)        {            int n=s.nextInt();            int sum = 0, max = -0xfffffff;            int start=1, last=1, temp = 1;            for (int  j= 1; j<= n; j++)            {                int a = s.nextInt();                sum += a;                if (sum > max)                {                    max = sum;                    start = temp;                    last = j;                }                if (sum < 0)                {                    temp = j+ 1;                    sum = 0;                }            }            System.out.println("Case "+i+":");            System.out.println(max+" "+start+" "+last);            if(i!=cas)                System.out.println();        }    }}

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