HDU-4972 A simple dynamic programming problem
http://acm.hdu.edu.cn/showproblem.php?pid=4972
++和+1还是有区别的,不可大意。
A simple dynamic programming problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 307 Accepted Submission(s):
117
Heat.
Here's an introduction of basketball
game:http://en.wikipedia.org/wiki/Basketball. However the game in Dragon's
version is much easier:
"There's two teams fight for the winner. The only
way to gain scores is to throw the basketball into the basket. Each time after
throwing into the basket, the score gained by the team is 1, 2 or 3. However due
to the uncertain factors in the game, it’s hard to predict which team will get
the next goal".
Dragon is a crazy fan of Miami Heat so that after each
throw, he will write down the difference between two team's score regardless of
which team keeping ahead. For example, if Heat's score is 15 and the opposite
team's score is 20, Dragon will write down 5. On the contrary, if Heat has 20
points and the opposite team has 15 points, Dragon will still write down
5.
Several days after the game, Dragon finds out the paper with his
record, but he forgets the result of the game. It's also fun to look though the
differences without knowing who lead the game, for there are so many uncertain!
Dragon loves uncertain, and he wants to know how many results could the
game has gone?
the number of test cases. Following T blocks, each block describe one test
case.
For each test case, the first line contains only one integer
N(N<=100000), which means the number of records on the paper. Then there
comes a line with N integers (a1, a2, a3, ... ,
an). ai means the number of i-th record.
start with "Case #i: ", with i implying the case number. Then for each case just
puts an integer, implying the number of result could the game has
gone.
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int abs(int x)
{
if(x>)
return x;
else
return -x;
}
int main()
{
int i,t,n,a[],ans,k=;
scanf("%d",&t);
while(t--)
{
memset(a,,sizeof(a));
ans=;
scanf("%d",&n);
for(i=;i<n;i++)
scanf("%d",&a[i]);
int cnt=;
int flag=;
for(i=;i<n&&(i+)<n;i++)
{
if(((a[i+]==a[i])&&a[i]!=)||abs(a[i+]-a[i])>)
{
flag=;
break;
}
if((a[i]==&&a[i+]==)||(a[i]==&&a[i+]==))
cnt++;
} if(flag==)
{
printf("Case #%d: %d\n",k++,ans);
continue;
}
if(a[n-]==)
ans=cnt+;
else
ans=*cnt+;
printf("Case #%d: %d\n",k++,ans);
}
return ;
}
HDU-4972 A simple dynamic programming problem的更多相关文章
- hdu 4972 A simple dynamic programming problem(高效)
pid=4972" target="_blank" style="">题目链接:hdu 4972 A simple dynamic progra ...
- 2014多校第十场1002 || HDU 4972 A simple dynamic programming problem
题目链接 题意 : 每次无论哪个队投进一个篮球,就记下现在两队比分的差值,问你最后的结果有多少种情况. 思路 : 该题实在是不好理解,最后的结果有多少种情况就是说不管中间过程怎么来的,只要最后结果不一 ...
- hdu 4972 A simple dynamic programming problem (转化 乱搞 思维题) 2014多校10
题目链接 题意:给定一个数组记录两队之间分差,只记分差,不记谁高谁低,问最终有多少种比分的可能性 分析: 类似cf的题目,比赛的时候都没想出来,简直笨到极点..... 最后的差确定,只需要计算和的种类 ...
- 【HDOJ】4972 A simple dynamic programming problem
水题. #include <cstdio> #include <cstring> #include <cstdlib> int abs(int x) { ? -x: ...
- HDU 4975 A simple Gaussian elimination problem.
A simple Gaussian elimination problem. Time Limit: 1000ms Memory Limit: 65536KB This problem will be ...
- HDU 4971 A simple brute force problem.
A simple brute force problem. Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged o ...
- hdu 4975 A simple Gaussian elimination problem.(网络流,推断矩阵是否存在)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4975 Problem Description Dragon is studying math. One ...
- hdu - 4975 - A simple Gaussian elimination problem.(最大流量)
意甲冠军:要在N好M行和列以及列的数字矩阵和,每个元件的尺寸不超过9,询问是否有这样的矩阵,是独一无二的N(1 ≤ N ≤ 500) , M(1 ≤ M ≤ 500). 主题链接:http://acm ...
- hdu 4975 A simple Gaussian elimination problem 最大流+找环
原题链接 http://acm.hdu.edu.cn/showproblem.php?pid=4975 这是一道很裸的最大流,将每个点(i,j)看作是从Ri向Cj的一条容量为9的边,从源点除法连接每个 ...
随机推荐
- web 自定义监听器中设置加载系统相关的静态变量及属性
直接上代码: 在src下新建一个StartListener 实现接口ServletContextListener,: /** * @Title:StartListener.java * @Packag ...
- WPF 窗体中的 Canvas 限定范围拖动 鼠标滚轴改变大小
xaml代码: <Canvas Name="movBg"> <Canvas.Background> <LinearGradientBrush EndP ...
- ThinkPHP调试模式与日志记录
1.可以在config.php中进行设置,默认为关闭状态. 'APP_DEBUG' => true 打开\ThinkPHP\Common\debug.php文件可以查看debug的默认设置 ...
- 交叉编译tslib1.4
cross-compiler: arm-linux-gcc V4.2.1 source code: tslib-1.4.tar.gz #tar zxvf tslib-1.4.tar.gz #./aut ...
- session cookie 相结合实现
数据库配置文件 config.php <?php// config.php 数据库连接文件define('DB_HOST', 'localhost');define('DB_USER', 'ro ...
- linux安装composer
1,确保php已成功安装,并且php可以被访问php -r "copy('https://getcomposer.org/installer', 'composer-setup.php'); ...
- ecshop 报错
ECShop出现Strict Standards: Only variables should be passed b (2014-06-04 17:00:37) 转载▼ 标签: ecshop 报错 ...
- 帝国cms 列表页分页样式修改美化【1】
[1]自己修改帝国cms默认的分页样式(css),这样做的好处是你不用去改动帝国的核心文件,方便以后升级. [2]自己动手去修改帝国的分页(php+css),帝国的分页在e>class>下 ...
- Administration Commands
Commands useful for administrators of a hadoop cluster. balancer Runs a cluster balancing utility. A ...
- git+Coding.netの小试牛刀
一.将本地项目推送到Coding中 1.在Coding中新建项目,填写项目名称和项目描述,设置属性,勾选初始化仓库