Description

Problem Description
Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.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.
 
Input
Input contains multiple test cases. Each test case is described in a line as follow: 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.
 
Output
For each case, print the maximum according to rules, and one line one case.
 
Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
 
Sample Output
4
10
3
 
dp
求上升子序列的和的最大值
当增加第i个数啊a[i]时,只需在j=1~(i-1)中寻找在a[j]<a[i]的情况下的dp[j]的最大值max然后更新dp[i]=max+a[i]  ,记录最大值输出。
 
代码实现:

 #include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int n,a[],dp[],i,j,k,ans,max;
while(cin>>n&&n)
{
memset(dp,,sizeof(dp));
for(i=;i<n;i++)
cin>>a[i];
for(i=;i<n;i++)
{
max=;
for(j=;j<i;j++)
{
if(a[j]<a[i])
{
max=max>dp[j]?max:dp[j];
}
}
dp[i]=max+a[i];
}
ans=;
for(i=;i<n;i++)
ans=ans>dp[i]?ans:dp[i];
cout<<ans<<endl;
}
return ;
}

Super Jumping! Jumping! Jumping!杭电1087的更多相关文章

  1. 杭电1087 Super Jumping! Jumping! Jumping!(初见DP)

    Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 ...

  2. 杭电ACM分类

    杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze ...

  3. 杭电dp题集,附链接还有解题报告!!!!!

    Robberies 点击打开链接 背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和- 把状态转移方程写成了f ...

  4. 杭电acm阶段之理工大版

    想參加全国软件设计大赛C/C++语言组的同学,假设前一篇<C和指针课后练习题总结>没看完的,请先看完而且依照上面的训练做完,然后做以下的训练. 传送门:http://blog.csdn.n ...

  5. 杭电ACM题单

    杭电acm题目分类版本1 1002 简单的大数 1003 DP经典问题,最大连续子段和 1004 简单题 1005 找规律(循环点) 1006 感觉有点BT的题,我到现在还没过 1007 经典问题,最 ...

  6. 杭电acm习题分类

    专注于C语言编程 C Programming Practice Problems (Programming Challenges) 杭电ACM题目分类 基础题:1000.1001.1004.1005. ...

  7. acm入门 杭电1001题 有关溢出的考虑

    最近在尝试做acm试题,刚刚是1001题就把我困住了,这是题目: Problem Description In this problem, your task is to calculate SUM( ...

  8. 杭电acm 1002 大数模板(一)

    从杭电第一题开始A,发现做到1002就不会了,经过几天时间终于A出来了,顺便整理了一下关于大数的东西 其实这是刘汝佳老师在<算法竞赛 经典入门 第二版> 中所讲的模板,代码原封不动写上的, ...

  9. 杭电OJ——1198 Farm Irrigation (并查集)

    畅通工程 Problem Description 某省调查城镇交通状况,得到现有城镇道路统计表,表中列出了每条道路直接连通的城镇.省政府"畅通工程"的目标是使全省任何两个城镇间都可 ...

随机推荐

  1. easyui datagrid自定义操作列

    通过formatter方法给Jquery easyui 的datagrid 每行增加操作链接 我们都知道Jquery的EasyUI的datagrid可以添加并且自定义Toolbar, 这样我们选择一行 ...

  2. #define命令的一些高级用法

    =========================================================== define中的三个特殊符号:#,##,#@ ================= ...

  3. linux学习之linux的hostname修改详解《转》

    linux的hostname是一个kernel变量,可以通过hostname命令来查看本机的hostname.也可以直接cat /proc/sys/kernel/hostname查看. #hostna ...

  4. 半同步半异步模式的实现 - MSMQ实现

    半同步半异步模式的实现 - MSMQ实现 所谓半同步半异步是指,在某个方法调用中,有些代码行是同步执行方式,有些代码行是异步执行方式,下面我们来举个例子,还是以经典的PlaceOrder来说,哈哈. ...

  5. 原生javascript-图片按钮切换

    原生javascript-图片按钮切换 即上次被分配写原生JS相册弹窗后,这次又接到了写原生JS,图片按钮切换,真激情: 个人在线实例:http://www.lgyweb.com/picSwitch/ ...

  6. enode框架step by step之框架的物理部署思路

    enode框架step by step之框架的物理部署思路   enode框架系列step by step文章系列索引: enode框架step by step之开篇 enode框架step by s ...

  7. Hadoop YARN介绍

    YARN产生背景 MRv1的局限 YARN是在MRv1基础上演化而来的,它克服了MRv1中的各种局限性.在正式介绍YARN之前,先了解下MRv1的一些局限性,主要有以下几个方面: 扩展性差.在MRv1 ...

  8. C# 关于委托的小例子

    本例子是一个关于委托的小例子[猫叫,狗跳,人喊]. 委托是C#开发中一个非常重要的概念,理解起来也和常规的方法不同,但一旦理解清楚,就可以信手拈来,随处可用. 委托是对方法的抽象.它存储的就是一系列具 ...

  9. linux基础命令大全

    编辑器 ed vi/vim (交互式) sed (非交互) vi/vim 的使用 1.命令模式 移动光标 方向键 hjkl H L M G 1G nG 复制行 yy nyy 粘贴 p 删除行 dd n ...

  10. openui5的资料比较少

    openui5的资料比较少,稳定优秀的开源框架,国内了解的人了了,都在追AngularJS.ExtJS.React. React比较新,非死不可出品而且裹挟Native的噱头.Mobile Nativ ...