Maximum Subsequence Sum

最新推荐文章于 2021-08-11 20:47:17 发布

gzcoder1

最新推荐文章于 2021-08-11 20:47:17 发布

阅读量206

点赞数

CC 4.0 BY-SA版权

分类专栏：前缀和

本文链接：https://blog.youkuaiyun.com/PH_Dream/article/details/79738938

前缀和专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种求解最大连续子序列和及其起始和结束位置的算法，通过使用前缀和与贪心策略实现O(n)的时间复杂度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

传送门

Given a sequence of $K$ integers { $N^{ 1 }$ , $N^{ 2 }$ , ..., $N^{ K }$ }. A continuous subsequence is defined to be { $N^{ i }$ , $N^{ i + 1 }$ , ..., $N^{ j }$ } where $1$ . 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 $K$ ( $\leq$ ). 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.

题意：给出一组序列，求最大连续子序列最大和，并记录起点和终点（若有多个，优先选择第一个得到最大和的起点和终点）

前缀和+贪心思想 = O（n）

解析：对于前缀和s1为负数的连续子序列，直接舍去，更新临时起点（临时！！），当前缀和s2比之前记录的前缀和si要大时,把起点和终点一起更新

  ans = a[0];

 for(int i = 0 ; i < k ; i ++){
          sum +=a[i];
          if(sum < 0 ) {  sum = 0 ;fst = i+1 ; continue;}
          if(sum > ans){ ans = sum; st = fst; ed = i;}
   }

前缀和+暴力枚举（枚举起点和终点）

ans = a[0];

  for(int i = 0 ; i < k ; i ++){
        sum = 0;
     for(int j = i ; j < k ; j++){
          sum +=a[j];
          if(sum > ans){ ans = sum; st = i; ed = j;}
     }
   }