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
- SpringMVC注解@RequestParam解析
1.可以对传入参数指定参数名 1 @RequestParam String inputStr 2 // 下面的对传入参数指定为param,如果前端不传param参数名,会报错 3 @RequestPa ...
- Session的生命同期
一.什么是Session,怎么用 Session是存放用户与web服务器之间的会话,即服务器为浏览器开辟的存储空间. 由于浏览器与服务器之间的会话是无状态(无状态的意思是会话之间无关联性,无法识别该用 ...
- GNU汇编 存储器访问指令
.text .global _start _start: mov r0,#0xff str r0,[r1] ldr r2,[r1]
- 想学习一下node.js,重新安装配置了node
根据这个网站上的教程安装配置的,还不错一次就成功了.觉得安装没什么,就是配置路径的时候容易错. http://www.runoob.com/nodejs/nodejs-install-setup.ht ...
- scrapy--json(喜马拉雅Fm)(二)
学习了对数据的储存,感觉还不够深入,昨天开始对储存数据进行提取.整合和图像化显示.实例还是喜马拉雅Fm,算是对之前数据爬取之后的补充. 明确需要解决的问题 1,蕊希电台全部作品的进行储存 --scra ...
- pageScope、requestScope、sessionScope、applicationScope的区别
https://www.cnblogs.com/qianbaidu/p/6006459.html 1.区别: 1.page指当前页面有效.在一个jsp页面里有效 2.request 指在一次请求的全过 ...
- Huffman Tree -- Huffman编码
#include <stdlib.h> #include <stdio.h> #include <string.h> typedef struct HuffmanT ...
- McNay Art Museum【McNay艺术博物馆】
McNay Art Museum When I was 17, I read a magazine artice about a museum called the McNay, once the h ...
- 012---Django的用户认证组件
知识预览 用户认证 回到顶部 用户认证 auth模块 ? 1 from django.contrib import auth django.contrib.auth中提供了许多方法,这里主要介绍其中的 ...
- issubclasss/type/isinstance/callable/super
issubclass() : 方法用于判断第一个参数是否是第二个参数的子子孙孙类. 语法:issubclass(sub, super) 检查sub类是否是 super 类的派生类 class A: p ...