B. Pyramid of Glasses
原题链接
B. Pyramid of Glasses
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
大致题意
有n层高的杯子,每秒钟从顶部倒一杯水下来,问经过t秒,有多少个杯子的水满了。
思路
很有趣的模拟题。首先考虑把每个杯子看成‘1’,然后将每个杯子写进double类型的二维数组中,如果杯子满了,则将溢出的分成相等的两份然后分别流进下层对应的两个杯子中,同时对满了的杯子计数。
AC代码
#include<iostream>
using namespace std;
double v[15][15];
void pour(int n)
{
v[1][1]+=1;
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
{
if(v[i][j]<=1) continue;
double p=v[i][j]-1;
v[i][j]=1;
v[i+1][j]+=p/2;
v[i+1][j+1]+=p/2;
}
}
int main(void)
{
int n,t;
cin>>n>>t;
for(int i=1;i<=t;i++)
pour(n);
int full=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
{
if(v[i][j]>0.9999)
full++;
}
cout<<full<<endl;
return 0;
}
如有疑问欢迎私信!!
B. Pyramid of Glasses的更多相关文章
- CF 676B Pyramid of Glasses[模拟]
B. Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standar ...
- codeforces 676B B. Pyramid of Glasses(模拟)
题目链接: B. Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input s ...
- Pyramid of Glasses(递推)
Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #354 (Div. 2) B. Pyramid of Glasses 模拟
B. Pyramid of Glasses 题目连接: http://www.codeforces.com/contest/676/problem/B Description Mary has jus ...
- [Codeforces676B]Pyramid of Glasses(递推,DP)
题目链接:http://codeforces.com/problemset/problem/676/B 递推,dp(i, j)表示第i层第j个杯子,从第一层开始向下倒,和数塔一样的题.每个杯子1个时间 ...
- 【CF 676B Pyramid of Glasses】模拟,递归
题目链接:http://codeforces.com/problemset/problem/676/B 题意:一个n层的平面酒杯金字塔,如图,每个杯子的容量相同.现在往最顶部的一个杯子倒 t 杯酒,求 ...
- Codeforces Round #354 (Div. 2) ABCD
Codeforces Round #354 (Div. 2) Problems # Name A Nicholas and Permutation standard input/out ...
- Codeforces Round #354 (Div. 2)
贪心 A Nicholas and Permutation #include <bits/stdc++.h> typedef long long ll; const int N = 1e5 ...
- Codeforces Round #354 (Div. 2)-B
B. Pyramid of Glasses 题目链接:http://codeforces.com/contest/676/problem/B Mary has just graduated from ...
随机推荐
- java日志概述和原理
OK,现在我们来研究下Java相关的日志. 日志记录是应用程序运行中必不可少的一部分.具有良好格式和完备信息的日志记录可以在程序出现问题时帮助开发人员迅速地定位错误的根源.对于开发人员来说,在程序中使 ...
- CentOS如何把deb转为rpm
说明:可以转换,但不一定可用,可以根据报错提示,安装需要的依赖. 1 安装alien工具,下载地址http://ftp.de.debian.org/debian/pool/main/a/alien/ ...
- SSMS 2005 连接 SQL SERVER 2008问题
用本机的 Microsoft SQL Server Management Studio 2005 客户端连接数据库服务器时报错:"This version of Microsoft SQL ...
- java 表现层:jsp、freemarker、velocity
在java领域,表现层技术主要有三种:jsp.freemarker.velocity. jsp是大家最熟悉的技术 优点: 1.功能强大,可以写java代码 2.支持jsp标签(jsp tag) 3.支 ...
- 获取android手机屏幕的宽高、density
public static String getDisplayMetrics(Context cx) { String str = ""; DisplayMetrics dm = ...
- Shader 入门笔记(一)
本笔记,是根据自己学习shader的笔记,主要是参照冯乐乐的<Shader 入门精要> 和游戏蛮牛shaderLad视频 和网上一些博客. 为啥要学习这个呐? 自己其实之前学过一段时间的s ...
- [DeeplearningAI笔记]改善深层神经网络_优化算法2.6_2.9Momentum/RMSprop/Adam优化算法
Optimization Algorithms优化算法 觉得有用的话,欢迎一起讨论相互学习~Follow Me 2.6 动量梯度下降法(Momentum) 另一种成本函数优化算法,优化速度一般快于标准 ...
- 从iconfont下载项目所需的图标资源
前端开发中,经常会用到各种各样的图标(icon).这些icon,如果每个都要自己去做,那真的是耗时又耗力.但是,有了阿里巴巴矢量图标库这样的平台后,一切都变得简单了起来. 本文以此平台为例,演示如何搜 ...
- 单元测试系列:JUnit单元测试规范
更多原创测试技术文章同步更新到微信公众号 :三国测,敬请扫码关注个人的微信号,感谢! 原文链接:http://www.cnblogs.com/zishi/p/6762032.html Junit测试代 ...
- 安装Mercurial进行版本管理
mercurial是又一个去中心化的版本管理软件,类似git 先介绍如何安装mercurial yum -y install mercurial mercurial需要一个用户名来记录commit动作 ...