《 动态规划_ 入门_最大连续子序列_HDU_1003 》
《 动态规划_ 入门_最大连续子序列_HDU_1003 》
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 }