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  • 《 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 }
     
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  • 原文地址:https://www.cnblogs.com/kangxinxin/p/10776324.html
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