Codeforces 343A. Rational Resistance

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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: 

  1. one resistor; 
  2. an element and one resistor plugged in sequence; 
  3. an element and one resistor plugged in parallel. 
技术分享

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 技术分享. 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 技术分享. Determine the smallest possible number of resistors he needs to make such an element.

Input

The single input line contains two space-separated integers a and b (1?≤?a,?b?≤?1018). It is guaranteed that the fraction 技术分享 is irreducible. It is guaranteed that a solution always exists.

Output

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.

 

如果当前使用k个能得到a/b,则使用k+1个就能得到(a+b)/b 或 a/(b+a) ,观察形式很像求gcd的形式。
试几组后发现就是gcd的次数
技术分享
#include <bits/stdc++.h>

typedef long long LL;
using namespace std;
#define SIZE 105


LL res = 0;
LL gcd(LL a,LL b){
  if (b == 0) return a;
  res += a/b;
  return gcd(b,a % b);
}

int main(){
  // freopen("test.in","r",stdin);
  ios::sync_with_stdio(false);
  LL a,b;
  cin >> a >> b;
  gcd(a,b);
  cout << res;
  return 0;
}
View Code

 

 

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