hdu 1003 dp

　　　　　　　　　　　　　　Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 128098    Accepted Submission(s): 29672

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

 

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又来水题了

2bb的错误，想不出的状态方程。。。

import java.util.Scanner;


public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for(int i=0; i<T; i++){
            int n = sc.nextInt();
            int[] dp = new int[n+1], a = new int[n+1];
            for(int j=0; j<n; j++) a[j]=sc.nextInt();
            int maxSum=a[0], right=0, left=0;
            for(int j=0; j<n; j++){ 
                dp[j]=Math.max((j==0?0:dp[j-1]), 0)+a[j];
                if(dp[j]>maxSum){
                    maxSum=dp[j];    right=j;
                }
            }
            for(int j=right; j>-1; j--){
                dp[j]=Math.max((j==right?0:dp[j+1]), 0)+a[j];
                if(maxSum==dp[j])
                    left=j;
            }
            if(i>0) System.out.println();
            System.out.println("Case "+(i+1)+":");
            System.out.println(maxSum+" "+(++left)+" "+(++right));
        }
    }
}