HDU 1087 最大上升子序列的和
Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32134 Accepted Submission(s): 14467

The game can be played by two or more than two players. It consists of a chessboard(棋盘)and some chessmen(棋子), and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
Your task is to output the maximum value according to the given chessmen list.
N value_1 value_2 …value_N
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.
0
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n;
int a[];
int ans[];
int exm;
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n==)
break;
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
ans[i]=a[i];
}
int maxlen=-;
for(int i=;i<=n;i++)
{
int maxn=;
for(int j=;j<i;j++)
{
if(a[j]<a[i]&&maxn<ans[j])
{
maxn=ans[j];
}
}
ans[i]=maxn+a[i];
if(ans[i]>maxlen)
{
maxlen=ans[i];
exm=i;
}
}
cout<<maxlen<<endl;
}
return ;
}
HDU 1087 最大上升子序列的和的更多相关文章
- hdu 1087 最大上升子序列的和(dp或线段树)
Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 ...
- HDU 1087 最大递增子序列
Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 ...
- hdu 1087 最大递增子序列和
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #defin ...
- HDU 1087 简单dp,求递增子序列使和最大
Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 ...
- (最大上升子序列) Super Jumping! Jumping! Jumping! -- hdu -- 1087
http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit:1000MS ...
- HDU 1087 Super Jumping! Jumping! Jumping! 最长递增子序列(求可能的递增序列的和的最大值) *
Super Jumping! Jumping! Jumping! Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64 ...
- HDU 1069&&HDU 1087 (DP 最长序列之和)
H - Super Jumping! Jumping! Jumping! Time Limit:1000MS Memory Limit:32768KB 64bit IO Format: ...
- HDU 1087 Super Jumping! Jumping! Jumping
HDU 1087 题目大意:给定一个序列,只能走比当前位置大的位置,不可回头,求能得到的和的最大值.(其实就是求最大上升(可不连续)子序列和) 解题思路:可以定义状态dp[i]表示以a[i]为结尾的上 ...
- HDU 1231 最大连续子序列 --- 入门DP
HDU 1231 题目大意以及解题思路见: HDU 1003题解,此题和HDU 1003只是记录的信息不同,处理完全相同. /* HDU 1231 最大连续子序列 --- 入门DP */ #inclu ...
随机推荐
- IDEA的使用方法(一)(IDEA基本快捷键)
一个软件的快捷键显得尤为重要,接下来来讲讲快捷键 CTR+N 搜索类 CTR+SHIT+N 搜索文件 CTR+ALT+空格 代码提示(类似于 ALT+/) ALT+F7 查询在某处使用 CTR+Q 查 ...
- Shell学习——变量
1.在Bash中,每一个变量的值都是字符串 2.查看某个进程的环境变量 cat /proc/$PID/environ | tr '\0' '\n' 3.变量赋值 3.1.var=value,注意var ...
- webmin纯web界面管理linux系统
关键字: 摘要:从Windows环境的管理转到Linux环境的管理时所面临的挑战之一是,您需要去学习利用新的工具.作为一个管理员,您希望理解操作系统的细节以发挥它的最大功效.但是,当您还处在学习阶段时 ...
- anaconda+jupyter notebook 安装配置
安装Anaconda 从清华大学开源软件镜像站选择合适自己的版本 wget https://mirrors.tuna.tsinghua.edu.cn/anaconda/archive/Anaconda ...
- python开发基础之字符编码、文件处理和函数基础
字符编码 为什么要有字符编码? 字符编码是为了让计算机能识别我们人写的字符,因为计算机只认识高低电平,也就是二进制数"0","1". 一个文件用什么编码方式存储 ...
- hive-pom.xml
4.0.0 <groupId>com.cenzhongman</groupId> <artifactId>hive</artifactId> <v ...
- Eclipse Java 构建路径 ---Eclipse教程第13课
Eclipse Java 构建路径 设置 Java 构建路径 Java构建路径用于在编译Java项目时找到依赖的类,包括以下几项: 源码包 项目相关的 jar 包及类文件 项目引用的的类库 我们可以通 ...
- 快速登录机器&数据库
本文来自网易云社区. 作者:盛国存 背景 我们日常在使用ApiDoc维护管理api文档,提高了api文档的整体维护性.但在老旧接口中,补充接口注解无疑是一次繁重的体力劳动.仔细查看,大多数接口的格式 ...
- 使用系统的某些block api(如UIView的block版本写动画时),是否也考虑循环引用问题?
系统的某些block api中,UIView的block版本写动画时不需要考虑,但也有一些api 需要考虑 以下这些使用方式不会引起循环引用的问题 [UIView animateWithDuratio ...
- Caliburn micro 学习笔记...
页面跳转 LLS 结合 CM 使用方法 事件处理