Codeforces 344C Rational Resistance【思维+贪心】

最新推荐文章于 2020-07-18 11:01:20 发布

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最新推荐文章于 2020-07-18 11:01:20 发布

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分类专栏：贪心思维文章标签： Codeforces 344C

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本文介绍了一个有趣的问题：如何使用单位电阻构建一个等效电阻值为a/b的电阻网络，并尽可能减少使用的电阻数量。讨论了通过串联和并联操作实现目标电阻值的方法。

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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:

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 = R_e + R₀. With the parallel connection the resistance of the new element equals $\text{[math]}$ . 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 $\text{[math]}$ . 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

Input

3 2

Output

Input

199 200

Output

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.

题目大意：

用最少的阻值为1的电阻来构成一个网络，使得其总阻值为a/b。我们可以并联，也可以串联。

思路：

1、通过枚举几种情况我们贪心去处理，发现：

①如果a/b>1，那么我们用一个阻值为1的串联起来，此时剩余要处理的阻值为（a-b）/b.那么如果此时（a-b）/b仍然大于1.那么我们继续用一个阻值为1的串联起来，直到我们a/b<1为止。

②那么如果a/b<1（a<b），那么我们考虑并联一堆电阻的同时再串联起来一个阻值为1的电阻，和上述串联过程相似，直到我们再有a/b>1的时候停止。

③那么考虑整个过程，其实就是搞，用a，b中较大值不断的减去a，b中的较小值，直到不能减为止（一方变成了1，那么此时考虑并联较大值个电阻即可）。

2、那么我们接下来模拟这个过程即可：

①如果a==b输出1即可。

②设定output=0；

③如果a>b，swap（a，b）【我们保证a是较小值好处理】；

④如果a==1停止过程，output+=b；如果a！=1，那么对应output+=b/a；b%=a；

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll __int64
int main()
{
    ll a,b;
    while(~scanf("%I64d%I64d",&a,&b))
    {
        if(a==b)
        {
            printf("1\n");
            continue;
        }
        ll output=0;
        while(1)
        {
            if(a>b)swap(a,b);
            if(a==1)break;
            else
            {
                output+=b/a;
                b%=a;
            }
        }
        output+=b;
        printf("%I64d\n",output);
    }
}