hdu 1003 Max Sum(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
1 import java.util.Scanner; 2 3 public class Test4 { 4 5 public static void main(String[] args) { 6 7 int[] dp=new int[100005]; 8 int[] arrs=new int[100005]; 9 10 Scanner input=new Scanner(System.in); 11 12 int T=input.nextInt();//测试用例个数 13 for(int i=0;i<T;i++) { 14 15 int start,end,left,submax; 16 17 int n=input.nextInt(); 18 for(int j=0;j<n;j++) { 19 20 arrs[j]=input.nextInt(); 21 22 } 23 24 dp[0]=submax=arrs[0]; 25 start=end=left=1; 26 27 for(int j=1;j<n;j++) { 28 29 if(arrs[j]>arrs[j]+submax) { 30 31 submax=arrs[j]; 32 left=j+1; 33 34 }else { 35 36 submax=submax+arrs[j]; 37 38 } 39 40 if(dp[j-1]>=submax) { 41 42 dp[j]=dp[j-1]; 43 44 }else { 45 46 dp[j]=submax; 47 start=left; 48 end=j+1; 49 50 } 51 52 } 53 54 System.out.println("case"+(i+1)+":"+dp[n-1]+" "+left+" "+end); 55 56 } 57 58 } 59 60 }
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