A. Rational Resistance----贪心
来源:互联网 发布:单片机方案公司 编辑:程序博客网 时间:2024/06/06 01:33
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.
题目链接:http://codeforces.com/contest/343/problem/A
从退了队就没有写过一篇博客,两个月没有写。。。感谢QAQ大大
这个题的意思是给你一些电阻为1的小电阻,让你通过串联和并联的方式来组成一个阻值为a/b的大电阻,现在我们考虑第三组数据,199/200,我们考虑并联,1/(1/199+1)不就是199/200,所以我们用199个电阻串联,然后并上一个1欧的电阻,大电阻的值就为199/200,所以我们逆向贪心回去即可,详见代码。
代码:
#include <cstdio>#include <cstring>#include <iostream>#define LL long longusing namespace std;int main(){ LL a,b; scanf("%I64d%I64d",&a,&b); LL sum=0; while(1){ if(a<b){ swap(a,b); } LL k=a/b; a-=k*b; sum+=k; if(a%b==0) break; } printf("%I64d\n",sum); return 0;}
- A. Rational Resistance----贪心
- codeforces343A A. Rational Resistance
- CodeForces 343A Rational Resistance
- CodeForces 343A Rational Resistance
- codeforces 343A Rational Resistance
- Codeforces 344C Rational Resistance【思维+贪心】
- Rational Resistance
- C. Rational Resistance
- Codeforces 344C - Rational Resistance
- codeforces #200 div2.C Rational Resistance [思维]
- Resistance
- Codeforces Round #200 (Div. 2) C. Rational Resistance
- CodeForces - 344C Rational Resistance (模拟题 +递归求解)
- Codeforces Round #200 (Div. 2)344C Rational Resistance(模拟)
- CF#200 div2 C Rational Resistance(math gcd)
- Codeforces 501C Rational Resistance 迭代+思维
- Codeforces Round #200 (Div. 2) C. Rational Resistance(脑洞思维)
- 贪心A
- Unity3D 利用Highlighing System插件使物体轮廓高亮显示
- java用户metadata-extractor读取照片信息错误Exception in thread “main” java.lang.NoClassDefFoundError: com/adobe
- v_layout
- Dynamic Web Module 3.0 requires Java 1.6 or newer
- MVC过滤器的使用
- A. Rational Resistance----贪心
- 命令行更新scikit-learn
- 实时加载页面的内容loading(主要针对图片)
- 软件自动更新的实现
- 动态规划中级教程300.Longest Increasing Subsequence
- ScrollView中ViewPager无法正常滑动问题
- Ubuntu中的查找命令
- 杭电OJ 1001
- JAVA正则表达式