A. Rational Resistance

Time Limit: 1 Sec ?

Memory Limit: 256 MB

題目連接

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:

one resistor;

an element and one resistor plugged in sequence;

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

?

題意

你有無數(shù)個1歐的電阻，要求你用并聯(lián)和串聯(lián)構(gòu)成a/b歐的電阻

問你最少需要多少個電阻

題解：

首先結(jié)構(gòu)肯定是串聯(lián)加并聯(lián)啦

a/b，整數(shù)部分由串聯(lián)構(gòu)成，分?jǐn)?shù)部分由并聯(lián)構(gòu)成就好了

這樣就可以不停的遞歸了其實

代碼：

#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); }

?

轉(zhuǎn)載于:https://www.cnblogs.com/qscqesze/p/4870943.html

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