Codeforces Round #200 (Div. 1)A. Rational Resistance 数学

A. Rational Resistance

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/343/problem/A

Description

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.

Sample Input

199 200

Sample Output

200

 

HINT

 

题意

你有无数个1欧的电阻，要求你用并联和串联构成a/b欧的电阻

问你最少需要多少个电阻

题解：

首先结构肯定是串联加并联啦

a/b，整数部分由串联构成，分数部分由并联构成就好了

这样就可以不停的递归了其实

代码：

#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;

int main()
{
    long long a,b;
    scanf("%lld%lld",&a,&b);
    if(a<b)swap(a,b);
    long long ans = 0;
    while(a&&b)
    {
        ans += a/b;
        a%=b;
        swap(a,b);
    }
    printf("%lld\n",ans);
}