codeforces 200 div2 C. Rational Resistance 思路题
1 second
256 megabytes
standard input
standard output
Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.
However, all Mike has is lots of identical resistors with unit resistance R0 = 1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:
- one resistor;
- an element and one resistor plugged in sequence;
- an element and one resistor plugged in parallel.
With the consecutive connection the resistance of the new element equals R = Re + R0. With the parallel connection the resistance of the new element equals . In this case Re equals the resistance of the element being connected.
Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element.
The single input line contains two space-separated integers a and b (1 ≤ a, b ≤ 1018). It is guaranteed that the fraction is irreducible. It is guaranteed that a solution always exists.
Print a single number — the answer to the problem.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the%I64d specifier.
1 1
1
3 2
3
199 200
200
In the first sample, one resistor is enough.
In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance . We cannot make this element using two resistors.
题意:要得到a/b的电阻最小需要多少个电阻;(注意:每次只能串联一个或者并联一个);
思路:每次串联一个得到(a+b)/a,并联一个得到a/(a+b);
#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 100000000000005
#define MAXN 10000010
//#pragma comment(linker, "/STACK:102400000,102400000")
int main()
{
ll x,y,z,i,t;
scanf("%I64d%I64d",&x,&y);
ll ans=;
while()
{
if(x>y)
{
ans+=x/y;
x%=y;
if(x==)
break;
}
else if(x<y)
{
z=x;
x=y-x;
y=z;
ans++;
}
else
{
ans+=;
break;
}
}
cout<<ans<<endl;
return ;
}
codeforces 200 div2 C. Rational Resistance 思路题的更多相关文章
- Bakery CodeForces - 707B (最短路的思路题)
Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. The ...
- Codeforces Round #200 (Div. 1)A. Rational Resistance 数学
A. Rational Resistance Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343 ...
- Codeforces Round #200 (Div. 2) C. Rational Resistance
C. Rational Resistance time limit per test 1 second memory limit per test 256 megabytes input standa ...
- CodeForces Round 200 Div2
这次比赛出的题真是前所未有的水!只用了一小时零十分钟就过了前4道题,不过E题还是没有在比赛时做出来,今天上午我又把E题做了一遍,发现其实也很水.昨天晚上人品爆发,居然排到Rank 55,运气好的话没准 ...
- codeforces #262 DIV2 B题 Little Dima and Equation
题目地址:http://codeforces.com/contest/460/problem/B 这题乍一看没思路.可是细致分析下会发现,s(x)是一个从1到81的数,不管x是多少.所以能够枚举1到8 ...
- codeforces343A A. Rational Resistance
http://http://codeforces.com/problemset/problem/343/A A. Rational Resistance time limit per test 1 s ...
- Codeforces #541 (Div2) - E. String Multiplication(动态规划)
Problem Codeforces #541 (Div2) - E. String Multiplication Time Limit: 2000 mSec Problem Descriptio ...
- Codeforces #180 div2 C Parity Game
// Codeforces #180 div2 C Parity Game // // 这个问题的意思被摄物体没有解释 // // 这个主题是如此的狠一点(对我来说,),不多说了这 // // 解决问 ...
- Codeforces #541 (Div2) - F. Asya And Kittens(并查集+链表)
Problem Codeforces #541 (Div2) - F. Asya And Kittens Time Limit: 2000 mSec Problem Description Inp ...
随机推荐
- 从一个多项目Web工程看Eclipse如何导入Gradle项目
这里再次说一下为什么我们需要熟悉Gradle构建工具,主要原因就是很多开源项目现在都在改用Gradle作为构建工具.一部分的github上的示例代码也在用Gradle构建,如果还是只能用maven,那 ...
- UUID的定义以及作用
UUID含义是通用唯一识别码 (Universally Unique Identifier),这 是一个软件建构的标准,也是被开源软件基金会 (Open Software Foundation, OS ...
- linix防火墙设置之顺序设置问题 -- 解决防火墙规则顺序和插入规则到指定序号的问题
转载于百度经验:https://jingyan.baidu.com/article/ae97a646ce58c2bbfd461d90.html 无论是硬件防火墙还是软件防火墙都会有一个规则序列的问题, ...
- Java开发环境的搭建(jdk,eclipse)
一.java 开发环境的搭建 这里主要说的是在windows 环境下怎么配置环境. 1.首先安装JDK java的sdk简称JDK ,去其官方网站下载最近的JDK即可. http://www.orac ...
- 介绍一种android的裸刷机方法(fastboot刷机实质)
fastboot刷机的前提是你的开发板uboot良好并能正常启动进入fastboot模式,你的开发版的nand分区已存在.对于Android的uboot而言, 已经实现了fastboot命令,当你 ...
- Java Naming and Directory Interface (JNDI) Java 命名和目录接口
https://www.oracle.com/technetwork/java/jndi/index.html Lesson: Overview of JNDI (The Java™ Tutorial ...
- CentOS 6.4下Squid代理服务器的安装与配置(转)
add by zhj: 其实我们主要还是关注它在服务器端使用时,充当反向代理和静态数据缓存.至于普通代理和透明代理,其实相当于客户端做的事,和服务端没有什么关系.另外,Squid的缓存主要是缓存在硬盘 ...
- Git学习-->如何通过Shell脚本自动定时将Gitlab备份文件复制到远程服务器?
一.背景 在我之前的博客 git学习--> Gitlab如何进行备份恢复与迁移? (地址:http://blog.csdn.net/ouyang_peng/article/details/770 ...
- EOS 的网站及资料doc
https://github.com/EOSIO/Documentation/blob/master/zh-CN/Roadmap.md https://bytemaster.github.io/bit ...
- spring登录验证拦截器和根据用户角色登录
大家都知道spring的用户登录拦截器,确实省去了程序员不少的精力,下面说说我在项目中使用的感受. 德安微信管理后台是管理多个微信帐号的平台,登录到平台的用户有三个角色,游客和微信帐号管理员.超级管理 ...