《 动态规划_ 入门_最大连续子序列_HDU_1003 》

Posted 2021-10-27 kangxinxin

题目描述:

 

 

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 321851    Accepted Submission(s): 76533

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

 

 

 

Java 代码实现，其中有个小坑，就是dp数组要开的大些，不然一直WA，不知道错误在哪里

 

 

 1 import java.util.Scanner;
 2 
 3 public class Main {
 4 
 5     public static void main(String[] args) {
 6         Scanner cin = new Scanner(System.in);
 7         int[] array = new int[100010];
 8         int[] dp = new int[100010];
 9         int t = cin.nextInt();
10         for (int count = 0; count < t; count++) {
11             
12             int n = cin.nextInt();
13             for(int i = 0;i<n;i++){
14                 array[i] = cin.nextInt();
15             }
16             
17             dp[0] = array[0];
18             
19             for(int i =1;i<n;i++){
20                 
21                 dp[i] = Math.max(dp[i-1]+array[i], array[i]);
22             }
23             
24             int max = dp[0];
25             
26             int endIndex = 0;
27             
28             for(int i = 1;i<n;i++){
29                 if(dp[i]>max){
30                     max = dp[i];
31                     endIndex = i;
32                 }
33             }
34             
35             
36             int temp =0,l = endIndex;
37             
38             for(int i = endIndex;i>=0;i--){
39                 temp+=array[i];
40                 if(temp==max){
41                     l = i;
42                 }
43             }
44             
45             if(count!=0){
46                 System.out.println();
47             }
48             System.out.println("Case "+(count+1)+":");
49             System.out.println(max+" "+(l+1)+" "+(endIndex+1));
50         }
51     }
52 
53 }