HDOJ 1003 Max Sum

原创 陈 浩翔  2016-02-06 01:48  阅读 111 次

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

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        int time = 1;
        while(t-->0){
            int m = sc.nextInt();
            int p1,p2=1;
            int n = sc.nextInt();
            int now=n;
            int max=n;
            int x=1;
            p1=1;
            for(int i=1;i<m;i++){
                n=sc.nextInt();
                if(n>n+now){
                    x=i+1;
                    now=n;
                }else{
                    now = n+now;
                }
                if(now>max){
                    max=now;
                    p1=x;
                    p2=i+1;
                }


            }

            System.out.println("Case "+time+":");
            time++;
            System.out.println(max+" "+p1+" "+p2);
            if(t!=0){
                System.out.println();
            }

        }



    }

}

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