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  • Codeforces Round #200 (Div. 1)A. Rational Resistance 数学

    A. Rational Resistance

    Time Limit: 1 Sec  

    Memory Limit: 256 MB

    题目连接

    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:

    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 Input

    199 200

    Sample Output

    200

    HINT

    题意

    你有无数个1欧的电阻,要求你用并联和串联构成a/b欧的电阻

    问你最少需要多少个电阻

    题解:

    首先结构肯定是串联加并联啦

    a/b,整数部分由串联构成,分数部分由并联构成就好了

    这样就可以不停的递归了其实

    代码:

    #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
    ",ans);
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4870943.html
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