HDOJ 1003 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

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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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/50640219