C. Rational Resistance

最新推荐文章于 2022-01-03 21:20:38 发布

UMR小豪

最新推荐文章于 2022-01-03 21:20:38 发布

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分类专栏： CF 思维文章标签： codeforces 脑洞思维

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科学家Mike需要通过串联或并联的方式组合单位电阻来获得特定阻值。本篇介绍了一种算法，通过递归减少分数的方式来确定达到所需阻值所需的最少电阻数量。

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:

one resistor;
an element and one resistor plugged in sequence;
an element and one resistor plugged in parallel.

$\text{[math]}$

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 $\text{[math]}$ . 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 $\text{[math]}$ . 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 $\text{[math]}$ 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 $\text{[math]}$ . We cannot make this element using two resistors.

感觉很像gcd的样子，要使得电阻最小那么，分子大于分母的时候，把整数部分剥离出来。

然后分子小于分母的时候必定是并联，那么取倒数，重复上一个过程就可以了。

脑洞加思维。

#include <cstdio>
#include <queue>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const long long inf = 1e18;
const int MAXN = 20+5;

int main()
{
    long long a,b,ans=0;
    scanf("%I64d%I64d",&a,&b);
    while(a > 0 && b > 0)
    {
        if(a > b)
        {
            ans += a/b;
            a%=b;
        }
        else
        {
            ans += b/a;
            b%=a;
        }
    }
    cout << ans <<endl;
    return 0;
}