• Codeforces Beta Round #16 div 2 C.Monitor最大公约数


    C. Monitor
    time limit per test
    0.5 second
    memory limit per test
    64 megabytes
    input
    standard input
    output
    standard output

    Reca company makes monitors, the most popular of their models is AB999 with the screen size a × b centimeters. Because of some production peculiarities a screen parameters are integer numbers. Recently the screen sides ratio x: y became popular with users. That's why the company wants to reduce monitor AB999 size so that its screen sides ratio becomes x: y, at the same time they want its total area to be maximal of all possible variants. Your task is to find the screen parameters of the reduced size model, or find out that such a reduction can't be performed.

    Input

    The first line of the input contains 4 integers — abx and y (1 ≤ a, b, x, y ≤ 2·109).

    Output

    If the answer exists, output 2 positive integers — screen parameters of the reduced size model. Output 0 0 otherwise.

    Examples
    input
    800 600 4 3
    output
    800 600
    input
    1920 1200 16 9
    output
    1920 1080
    input
    1 1 1 2
    output
    0 0
    思路:先将x,y比例化成最简,然后求最大倍数;
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    #define true ture
    #define false flase
    using namespace std;
    #define ll __int64
    #define inf 0xfffffff
    int scan()
    {
        int res = 0 , ch ;
        while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
        {
            if( ch == EOF )  return 1 << 30 ;
        }
        res = ch - '0' ;
        while( ( ch = getchar() ) >= '0' && ch <= '9' )
            res = res * 10 + ( ch - '0' ) ;
        return res ;
    }
    ll gcd(ll x,ll y)
    {
        return y==0?x:gcd(y,x%y);
    }
    int main()
    {
        ll a,b,c,x,y;
        cin>>a>>b>>x>>y;
        c=gcd(x,y);
        x/=c;
        y/=c;
        c=min(a/x,b/y);
        cout<<x*c<<" "<<y*c<<endl;
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/jhz033/p/5459178.html
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