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最新推荐文章于 2023-04-26 00:51:38 发布

转载最新推荐文章于 2023-04-26 00:51:38 发布 · 78 阅读

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原文链接：http://www.cnblogs.com/graph/archive/2013/03/26/2983161.html

本文探讨了最大子序列和问题的求解方法，通过动态规划实现算法，并提供了完整的C语言代码示例。该问题旨在从给定的整数序列中找出连续子序列的最大和及其边界元素。

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Maximum Subsequence Sum

Given a sequence of K integers { N₁, N₂, ..., N_K }. A continuous subsequence is defined to be { N_i, N_i+1, ..., N_j } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

NP 问题一直是我的短板，需要多加练习。
状态转移方程： sum[i] = max (sum[i-1]+a[i], a[i])
其中sum[i]表示的是以a[i]元素结尾的连续子序列的最大和

View Code

 1 #include<stdio.h>
 2 int a[10001];
 3 int main()
 4 {
 5     int K;;
 6     int first,last ,i,maxA, sumA;
 7     int maxFirst, maxLast;
 8     scanf("%d", &K);
 9     
10     i = 0;
11     while (i < K)
12     {
13         scanf("%d",&a[i]);
14         i++;
15     }
16     maxA = a[0]; sumA = a[0];
17     first = last = 0;
18     maxFirst = maxLast = 0;
19 
20     for(i = 1; i< K; i++){
21 
22         if(sumA + a[i]> a[i]){
23 
24             sumA = sumA + a[i];
25             last ++;
26         }
27         else if(sumA + a[i] <= a[i]){
28 
29             first = last = i;
30             sumA = a[i];
31         }
32 
33         if(sumA > maxA){
34 
35             maxA = sumA;
36             maxFirst = first;
37             maxLast = last;
38         }
39     }
40     if(maxA < 0)
41         printf("0 %d %d",a[0], a[K-1]) ;
42     else
43      printf("%d %d %d", maxA, a[maxFirst], a[maxLast]) ;
44 
45     return 0;
46 }

转载于:https://www.cnblogs.com/graph/archive/2013/03/26/2983161.html