《 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 }