Maxmum subsequence sum problem
We have a lot of ways to solve the maximum subsequence sum problem, but different ways take different time.
1、Brute-force algorithm
int maxSubSum1(const vector<int> &a)
{
int maxSum=0; for(int i=0;i<a.size();i++)
for(int j=i;j<a.size();j++)
{
int sum=0;
for(int k=i;k<=j;k++)
sum+=a[k]; if(sum>maxSum)
maxSum=sum;
} return maxSum;
}
/*The running time is O(n^3)
It takes too much time.
*/
2、a little imporvement
int maxSubSum2(const vector<int>& a )
{
int maxSum=0; for(int i=0;i<a.size();i++)
{
int sum=0; for(int j=i;j<a.size();j++)
{
sum+=a[j];
if(maxSum<sum)
{
maxSum=sum;
}
}
} return maxSum;
}
3. Divide-conquer algorithm
We can divide this problem into three parts:
(1) First half;
(2) cross the middle parts;
(3) second part;
What we need to do is to find the max sum of the three part.
int max3(int a, int b, int c)
{
if(a>b)
{
if(a>c)return a;
else return c;
}
else
{
if(c>b)return c;
else return b;
}
} int maxSubSum3(cosnt vector<int >& a, int left, int right)
{
if(left==right)
if(a[left]>0) return a[left];
else return 0; int center= (left+right)/2;
int maxLeftSum=maxSumRec(a, left, center);
int maxRightSum=maxSumRec(a, center+1, right); int maxLeftBoderSum=0, leftBoderSum=0;
for(int i=center;i>=left;i--)
{
leftBoderSum+=a[i];
if(leftBoderSum>maxLeftBoderSum)
maxLeftBoderSum=leftBoderSum;
} int maxRightBoderSum=0, leftBoderSum=0;
for(int i=center+1;i<=right;i++)
{
rightBoderSum+=a[i];
if(rightBoderSum>maxRightBoderSum)
maxRightBoderSum=rightBoderSum;
} return max3(maxLeftSum, maxLeftBoderSum+maxRightBoderSum,maxRightSum);
}
4. The best algorithm
If the start is negative, the sum of the subsequence can not be the max. Hence, any negative subsequence cannot possibly be a prefix of the optimal subsequence.
int maxSubSum4(const vector<int> & a)
{
int maxSum=0, sum=0; for(int i=0;i<a.size();i++)
{
sum+=a[i]; if(sum>maxSum)
maxSum=sum;
else if(sum<0)
sum=0;
} return maxSum;
}
Maxmum subsequence sum problem的更多相关文章
- Solutions for the Maximum Subsequence Sum Problem
The maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional ...
- MAXIMUM SUBSEQUENCE SUM PROBLEM
排除不合理的项(负值), 设定一个标杆sum, 往后扫描看是否有比sum好的情况. We should ensure the following conditions: 1. The result m ...
- HD2058The sum problem
The sum problem Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tot ...
- HDU 2058 The sum problem(枚举)
The sum problem Problem Description Given a sequence 1,2,3,......N, your job is to calculate all the ...
- HDU 2058:The sum problem(数学)
The sum problem Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- 【BZOJ-3638&3272&3267&3502】k-Maximum Subsequence Sum 费用流构图 + 线段树手动增广
3638: Cf172 k-Maximum Subsequence Sum Time Limit: 50 Sec Memory Limit: 256 MBSubmit: 174 Solved: 9 ...
- summary of k Sum problem and solutions in leetcode
I found summary of k Sum problem and solutions in leetcode on the Internet. http://www.sigmainfy.com ...
- Subset sum problem
https://en.wikipedia.org/wiki/Subset_sum_problem In computer science, the subset sum problem is an i ...
- HDu 1001 Sum Problem 分类: ACM 2015-06-19 23:38 12人阅读 评论(0) 收藏
Sum Problem Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...
随机推荐
- SQL总结之创建实例表空间监听
[创建数据库实例]cmd------>dbca[创建表空间-sql创建]create tablespace NSTC_WS logging datafile 'D:\app\dell\orada ...
- sphinx cmd command
D:\iso\gaoqiao\app\sphinx\bin\indexer.exe -c D:\iso\gaoqiao\app\sphinx\bin\sphinx.conf --all --rotat ...
- hibernate子查询
对于支持子查询的数据库,Hibernate支持在查询中使用子查询.一个子查询必须被圆括号包围起来(经常是SQL聚集函数的圆括号). 甚至相互关联的子查询(引用到外部查询中的别名的子查询)也是允许的. ...
- python项目练习地址
作者:Wayne Shi链接:http://www.zhihu.com/question/29372574/answer/88744491来源:知乎著作权归作者所有,转载请联系作者获得授权. 目前是3 ...
- ORACLE 使用sqluldr2和sqlldr进行导入导出
oracle数据导出工具sqluldr2可以将数据以csv.txt等格式导出,适用于大批量数据的导出,导出速度非常快.导出后可以使用oracle loader工具将数据导入. 简介: Sqluldr2 ...
- shell中的特殊符号
Shell符号及各种解释对照表: Shell符号 使用方法及说明 # 注释符号(Hashmark[Comments]) 1.在shell文件的行首,作为shebang标记,#!/bin/bash; 2 ...
- 转:KVC/KVO原理详解及编程指南
作者:wangzz 原文地址:http://blog.csdn.net/wzzvictory/article/details/9674431 转载请注明出处 如果觉得文章对你有所帮助,请通过留言或 ...
- Python之软件管理
常用软件包管理工具 一般python软件包管理工具,主要有以下: 图 常用python包管理工具 可以看到distribute是setuptools的替代方案(因为Setuptools包不再维护了), ...
- 淘宝npm镜像使用方法
镜像使用方法(三种办法任意一种都能解决问题,建议使用第三种,将配置写死,下次用的时候配置还在): 通过config命令npm config set registry https://registry. ...
- cursor属性
cursor光标类型 auto default none context-menu help pointer progress wait cell crosshair text vertical-te ...