Rational Resistance CodeForces - 344C(思维)

Rational Resistance

CodeForces - 344C

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 = R_e + R₀. With the parallel connection the resistance of the new element equals . In this case R_e 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.

Examples

Input

1 1

Output

1

Input

3 2

Output

3

Input

199 200

Output

200

Note

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.

这是一道很骚的高中的物理题，要求给你a，b两个数，用最少个数的1欧姆电阻通过串联或者并联的方式得到电阻大小为a/b

根据并联电阻的公式，一、如果a/b大于1，肯定要先串联（int）（a/b）个，

二、剩下的真分数部分a'/b'肯定是需要我们并联得到，那么这部分电阻a'/b' = 1/（b'/a'），

此时b'/a'又是一个假分数，和一开始情形一样了那么继续重复执行步骤1，直到分子为1，最后加上分子为1时那个分数的分母，就是总共需要最小的电阻数

就拿样例3来说199/200的电阻，我们先取倒数得到200/199 = 1 + 1/199，所以我们需要199个1欧姆电阻并联，得到1/199欧

然后1欧的再和1/199欧的并联得到答案也就是 199/200 = 1/（1+1/199）

code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int main(){
    ll a,b;
    ll ans = 0;
    scanf("%lld%lld",&a,&b);
    ans += (a / b);
    a = a % b;
    if(a == 0){//如果整除，那么说明都是串联的
        printf("%lld\n",ans);
        return 0;
    }
    while(a != 1){
        swap(a,b);
        ans += (a / b);
        a = a % b;
        if(a == 0) break;
    }
    if(a == 1) ans += b;
    printf("%lld\n",ans);
    return 0;
}