Codeforces 343A. Rational Resistance

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. 

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.

Input

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.

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

#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;
}