ZeptoLab Code Rush 2015 B. Om Nom and Dark Park
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the
other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
The park consists of 2n + 1 - 1 squares
connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1.
The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and
these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1)
there is a road to the square
.
Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.

To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square
has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number
of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add
an arbitrary number of street lights to each of the roads.
The first line contains integer n (1 ≤ n ≤ 10) —
the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 —
the initial numbers of street lights on each road of the park. Here ai is
the number of street lights on the road between squares i and
.
All numbers ai are
positive integers, not exceeding 100.
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
2
1 2 3 4 5 6
5
Picture for the sample test. Green color denotes the additional street lights.
这题可以把父节点和子节点之间的距离全部算作是子节点上的值,这样除了根节点外,子节点恰好每个有一个值。因为每一个父节点的两个子节点最后的值不可能同时增加(因为如果有同时增加的情况,可以直接加在父节点上),所以每次都是一个子节点的增加相应的值和另一个子节点相同,所以可以从倒数第二层开始向前递归。
#include<stdio.h>
#include<math.h>
#include<string.h>
int max(int a,int b)
{
return a>b?a:b;
}
int a[3000];
int main()
{
int n,m,i,j,b,ans;
while(scanf("%d",&n)!=EOF)
{
for(i=2;i<=pow(2,n+1)-1;i++)
{
scanf("%d",&b);
a[i]=b;
}
ans=0;
for(i=pow(2,n)-1;i>=1;i--)
{
a[i]+=max(a[i*2],a[i*2+1]);
ans=ans+2*max(a[i*2],a[i*2+1])-a[i*2]-a[i*2+1];
//printf("%d\n",ans);
}
printf("%d\n",ans);
}
return 0;
}
ZeptoLab Code Rush 2015 B. Om Nom and Dark Park的更多相关文章
- ZeptoLab Code Rush 2015 B. Om Nom and Dark Park DFS
B. Om Nom and Dark Park Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
- Codeforces - ZeptoLab Code Rush 2015 - D. Om Nom and Necklace:字符串
D. Om Nom and Necklace time limit per test 1 second memory limit per test 256 megabytes input standa ...
- ZeptoLab Code Rush 2015 C. Om Nom and Candies 暴力
C. Om Nom and Candies Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/526 ...
- ZeptoLab Code Rush 2015 C. Om Nom and Candies [ 数学 ]
传送门 C. Om Nom and Candies time limit per test 1 second memory limit per test 256 megabytes input sta ...
- Codeforces ZeptoLab Code Rush 2015 D.Om Nom and Necklace(kmp)
题目描述: 有一天,欧姆诺姆发现了一串长度为n的宝石串,上面有五颜六色的宝石.他决定摘取前面若干个宝石来做成一个漂亮的项链. 他对漂亮的项链是这样定义的,现在有一条项链S,当S=A+B+A+B+A+. ...
- CodeForces ZeptoLab Code Rush 2015
拖了好久的题解,想想还是补一下吧. A. King of Thieves 直接枚举起点和5个点之间的间距,进行判断即可. #include <bits/stdc++.h> using na ...
- B. Om Nom and Dark Park
B. Om Nom and Dark Park 在满二叉树上的某些边上添加一些值.使得根节点到叶子节点的路径上的权值和都相等.求最少需要添加多少. 我们利用性质解题. 考察兄弟节点.由于他们从跟节 ...
- Zepto Code Rush 2014 B - Om Nom and Spiders
注意题目给的是一个nxm的park,设元素为aij,元素aij 有4种可能U(上移),D(下移),L(左移),R(右移) 假设第i行第j列元素aij(注意元素的索引是从0开始的) 当aij为D时,此时 ...
- CF Zepto Code Rush 2014 B. Om Nom and Spiders
Om Nom and Spiders time limit per test 3 seconds memory limit per test 256 megabytes input standard ...
随机推荐
- 【Java基础】IO 流
IO 流 File 类 java.io.File 类是文件和文件目录路径的抽象表示形式,与平台无关. File 能新建.删除.重命名文件和目录,但 File 不能访问文件内容本身. 如果需要访问文件内 ...
- python基础学习总结
python管理cisco设备:http://www.linuxeye.com/program/1680.html 学习:https://www.liaoxuefeng.com/wiki/001431 ...
- 【Spring】Spring的数据库开发 - 2、Spring JdbcTemplate的常用方法(execute、update、query)
Spring JdbcTemplate的常用方法 文章目录 Spring JdbcTemplate的常用方法 execute() update() query() 简单记录-Java EE企业级应用开 ...
- 目录遍历 - Pikachu
概述: 在web功能设计中,很多时候我们会要将需要访问的文件定义成变量,从而让前端的功能便的更加灵活. 当用户发起一个前端的请求时,便会将请求的这个文件的值(比如文件名称)传递到后台,后台再执行其对应 ...
- 汇编学习笔记——DOS及DEBUG介绍
转自:https://www.shiyanlou.com/courses/running/332 一.课程简介 声明:该课程基于<汇编语言(第2版)>郑晓薇 编著,机械工业出版社.本节实验 ...
- 微服务网关2-搭建Gateway服务
一.创建父模块infrastructure 1.创建模块 在guli_parent下创建普通maven模块 Artifact:infrastructure 2.删除src目录 二.创建模块api_ga ...
- 量子化学Gaussian技术实战课 2021年4月9号--12号 远程在线教学
材料模拟分子动力学课程 3月19号--22号 远程在线课 lammps分子动力学课程 3月12号--15号 远程在线课 第一性原理VASP实战课 3月25号-28号 远程在线课 量子化学Gaussia ...
- cfsetispeed、cfsetospeed和cfsetspeed探究
在我https://www.cnblogs.com/Suzkfly/p/11055532.html这篇博客中有一个疑问,就是在串口设置波特率的域中,没有将输入输出波特率分开,那为什么会有几个不同的设置 ...
- Scheduling Multithreaded Computations by Work Stealing
steal.pdf http://supertech.csail.mit.edu/papers/steal.pdf This paper studies the problem of eciently ...
- RPM 和YUM总结
RPM RPM命名: 安装 rpm -ihv 其他常用的选项: 1. 重新安装 --replacepkgs (或者 --force ) 2. 不考虑依赖 --nodeps (不推荐) 升级: 查询: ...