PAT(甲级)1007
2015-09-23 16:15
260 查看
1007. Maximum Subsequence Sum (25)
时间限制400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1,
..., Nj } 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
代码稍后补上,之前的没留下
![](http://static.blog.csdn.net/xheditor/xheditor_emot/default/tongue.gif)
相关文章推荐
- win7 下启动mysql
- 树套树
- 测试用例
- np_xtcxyczjh-III 整理I[小公码(宏 错误包装) Makefile]
- 工厂模式 Factory Pattern
- vs项目中使用c++调用lua5.1
- 聊聊编程范式
- 已经为类型参数“Chart”指定了 constraint 子句。必须在单个 where 子句中指定类型参数的所有约束
- test5.22
- Spring中的数据库操作--使用JdbcTemplate
- 在Android中查看和管理sqlite数据库
- mysql中int、bigint、smallint 和 tinyint的区别
- 二叉树的基本操作
- 腾讯校招面试经验
- 使用C++的类模板实现Stack类
- java中注解的使用与实例(一)
- LA2995(迭代更新)
- Hexo 是一个快速、简洁且高效的博客框架
- php中session_id()函数详细介绍,会话id生成过程及session id长度
- jquery.prop() -Additional Notes