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Codeforces Round #200 (Div. 二) C. Rational Resistance(脑洞思维)

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Codeforces Round #200 (Div. 2) C. Rational Resistance(脑洞思维)

C. Rational Resistance
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Sample test(s)
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.


题目链接:http://codeforces.com/problemset/problem/344/C

题目大意:最少用多少个阻值为1的电阻组成阻值为a/b的电阻。

解题思路:思维题。抓住串并联的电阻和的特点。并联考虑倒数,真分数小于1,一定是并联。串联考虑整数,带分数的整数部分有串联组成时使用电阻最少。所以分数大于1时分离出整数部分即串联的电阻的个数。分数小于1时取倒,循环这个过程直到没有分数为止。比如3/2,分离整数后:1+1/2,1/2取倒为2,所以答案是1+2=3。

代码如下:

#include <cstdio>
#include <cstring>
#include <iostream>
#define ll long long
using namespace std;
int main(void)
{
	ll a,b,ans=0;
	cin>>a>>b;
	while(a>0&&b>0)
	{
		if(a>b)
		{
			ans+=a/b;
			a%=b;
		}
		else
		{
			ans+=b/a;
			b%=a;
		}
	}
	cout<<ans<<endl;
}


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