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

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.

水题,可 以转化成子问题!如果,是大于1,取整部分当成k个串联电阻,如果小于1部分,倒过来,重复上过程 ,一定可以得到结果!

#include <iostream>
#pragma comment(linker,"/STACK:10240000000,10240000000")
#include <stdio.h>
#include <string.h>
using namespace std;

__int64 gcd(__int64 a,__int64 b)
{
    if(a==0)return b;
    return gcd(b%a,a);
}
__int64 ff(__int64 a,__int64 b)
{
    __int64 answer=0,temp;
   while(a&&b)
   {
    __int64 c=gcd(a,b);
    if(c)
    a=a/c,b=b/c;__int64 k=a/b;answer+=k;temp=b;b=a-b*k;a=temp;
   }
    return answer;
}
int main()
{
    __int64 a,b;

    while(scanf("%I64d%I64d",&a,&b)!=EOF){
      __int64 c=ff(a,b);
      printf("%I64d\n",c);
    }
    return 0;
}