杭电1003（Max Sum） 首次dp

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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

代码实现：

import java.util.Scanner;
class Main{
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        int k=t;
        while(t-->0){
            int n=sc.nextInt();
            int[] a=new int[n];
            int[] dp=new int[n];
            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
                dp[i]=a[i];
            }
            int start=1;
            int end=1;
            for(int i=1;i<n;i++){
                dp[i]=Math.max(dp[i],dp[i-1]+a[i]);
            }
            int max=dp[0],sum=0,j=0;
            for(int i=0;i<n;i++){
                if(max<dp[i]){
                    max=dp[i];
                    end=i+1;
                    start=j+1;
                }
                if(dp[i]<0){//当dp[i]<0时起始位置移至i+1
                    j=i+1;
                }
            }
            System.out.println("Case "+(k-t)+":");
            System.out.println(max+" "+start+" "+end);
            if(t!=0){
                System.out.println();
            }
        }
    }
}