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

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=;
int num[maxn],sum[maxn];
int main()
{
int n,i,j,ans;
while(scanf("%d",&n)!=EOF&&n)
{
for(i=;i<=n;i++)
scanf("%d",&num[i]);
sum[]=num[];
ans=sum[];
for(i=;i<=n;i++)
{
sum[i]=num[i];
for(j=;j<i;j++)
{
if(num[j]<num[i]&&sum[j]+num[i]>sum[i])
sum[i]=sum[j]+num[i];
}
if(ans<sum[i])
ans=sum[i];
}
printf("%d\n",ans); }
return ;
}

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

  1. HDU - 1087 Super Jumping!Jumping!Jumping!(dp求最长上升子序列的和)

    传送门:HDU_1087 题意:现在要玩一个跳棋类游戏,有棋盘和棋子.从棋子st开始,跳到棋子en结束.跳动棋子的规则是下一个落脚的棋子的号码必须要大于当前棋子的号码.st的号是所有棋子中最小的,en ...

  2. E - Super Jumping! Jumping! Jumping!

    /* Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is very popula ...

  3. DP专题训练之HDU 1087 Super Jumping!

    Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is ve ...

  4. hdu 1087 Super Jumping! Jumping! Jumping! 简单的dp

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

  5. Super Jumping! Jumping! Jumping!——E

    E. Super Jumping! Jumping! Jumping! Time Limit: 1000ms Memory Limit: 32768KB 64-bit integer IO forma ...

  6. HDU 1087 Super Jumping! Jumping! Jumping! 最大递增子序列

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

  7. 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 ...

  8. HDU 1087 Super Jumping! Jumping! Jumping! (DP)

    C - Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format: ...

  9. HDU 1087 Super Jumping! Jumping! Jumping!(动态规划)

    Super Jumping! Jumping! Jumping! Problem Description Nowadays, a kind of chess game called “Super Ju ...

随机推荐

  1. 几种经典排序算法的R语言描述

    1.数据准备 # 测试数组 vector = c(,,,,,,,,,,,,,,) vector ## [] 2.R语言内置排序函数 在R中和排序相关的函数主要有三个:sort(),rank(),ord ...

  2. CodeForces 518B. Tanya and Postcard

    B. Tanya and Postcard time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  3. 使用Git进行项目管理

    首先在https://git.oschina.net进行注册以及登陆 登陆进去之后,如果想要创建项目,可以在 点击加号按钮,进行项目创建 3.这里以创建私有项目为例: 输入完成后,点击“创建”,进入下 ...

  4. jquery 抽奖示例

    jquery 抽奖示例: <%@ page language="java" import="java.util.*" pageEncoding=" ...

  5. vmstat命令学习

    vmstat 是Linux/Unix系统用来进行系统监控的工具 监控的目标主要有目标服务器的cpu使用率.内存的使用情况.虚拟内存交换情况,IO读写情况. 可以通过vmstat --help来获得该命 ...

  6. java list去重

    1.不带类型写法: 1 List listWithoutDup = new ArrayList(new HashSet(listWithDup)); 2.带类型写法(以String类型为例):1)Ja ...

  7. [uva12170]Easy Climb

    还是挺难的一个题,看了书上的解析以后还是不会写,后来翻了代码仓库,发现lrj又用了一些玄学的优化技巧. #include <algorithm> #include <iostream ...

  8. Java基础语法

    java基础学习总结——基础语法1 一.标识符

  9. Orcal函数

    where b.rn between 4 and 6--日期函数select sysdate from dual--返回两个日期select months_between(to_date('2017- ...

  10. SET QUOTED_IDENTIFIER ON和SET ANSI_NULLS ON

    distinct是sqlserver的标识符,如果想以distinct为表时,在QUOTED_IDENTIFIER为off的情况下,是不能创建表名为distinct的表的,因为在QUOTED_IDEN ...