codeforces 343A Rational Resistance

A. Rational Resistance

Mad scientist Mikeis building a time machine in his spare time. To finish the work, he needs aresistor with a certain resistance value.

However, allMike has is lots of identical resistors with unit resistance R₀ = 1. Elements withother resistance can be constructed from these resistors. In this problem, wewill consider the following as elements:

1.   one resistor;

2.   an element and one resistor plugged insequence;

3.   an element and one resistor plugged inparallel.

With theconsecutive connection the resistance of the new element equals R = R_e + R₀. With theparallel connection the resistance of the new element equals . In this case R_e equals theresistance of the element being connected.

Mike needs toassemble an element with a resistance equal to the fraction . Determine the smallest possible numberof resistors he needs to make such an element.

Input

The single inputline contains two space-separated integers a and b (1 ≤ a, b ≤ 10¹⁸). It isguaranteed that the fraction  is irreducible. It is guaranteedthat a solution always exists.

Output

Print a singlenumber — the answer to the problem.

Please do notuse the %lld specifierto read or write 64-bit integers in С++. It is recommended to use the cin, cout streams orthe%I64d specifier.

Sample test(s)

input

1 1

output

1

input

3 2

output

3

input

199 200

output

200

Note

In the firstsample, one resistor is enough.

In the secondsample one can connect the resistors in parallel, take the resulting elementand connect it to a third resistor consecutively. Then, we get an element withresistance . We cannot make this element using tworesistors.

 

 

题意:

         给出 两个数A，B；要求用最少的阻值为1的电阻串联、并联在一起组成阻值为A/B的电阻；输出最少的电阻数量；

 

思路：

         对于电阻来说将k个电阻串联和并联关系完全反转的话，阻值的变化特性是变为原阻值的倒数，因而我们可以利用这一特性不停的将并联关系转化为串联关系，而A/B可化为假分数形式n+a/B，期中整数部分由n个电阻串联，分数部分由一个并联电阻组组成，将并联化为串联关系，即将分数部分倒过来，不断反复，累加得到答案；

 

代码

 

#include <bits/stdc++.h>using namespace std;typedef long long ll;int main(){   ll a,b;   cin >>a>>b;   ll tmp, ans = 0;   while(b){       ans += a/b;       a %= b;       swap(a,b);    }   cout << ans;}