zoukankan      html  css  js  c++  java
  • Light Oj 1116

    Ekka Dokka
    Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
    Submit

    Status
    Description
    Ekka and his friend Dokka decided to buy a cake. They both love cakes and that's why they want to share the cake after buying it. As the name suggested that Ekka is very fond of odd numbers and Dokka is very fond of even numbers, they want to divide the cake such that Ekka gets a share of N square centimeters and Dokka gets a share of M square centimeters where N is odd and M is even. Both N and M are positive integers.

    They want to divide the cake such that N * M = W, where W is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.

    Input
    Input starts with an integer T (≤ 10000), denoting the number of test cases.

    Each case contains an integer W (2 ≤ W < 263). And W will not be a power of 2.

    Output
    For each case, print the case number first. After that print "Impossible" if they can't buy their desired cake. If they can buy such a cake, you have to print N and M. If there are multiple solutions, then print the result where M is as small as possible.

    Sample Input
    3
    10
    5
    12
    Sample Output
    Case 1: 5 2
    Case 2: Impossible
    Case 3: 3 4

    <span style="color:#6600cc;">/****************************************
    
         author   :   Grant Yuan
         time     :   2014.8.6
         algorithm:   数论
         source   :   Light Oj 1116
         explain  :   如果这个数为奇数则一定不可能;
                      如果为偶数,要保证所得偶数最小,则奇数要最大
                      用原来的数不断除以2,直到所得的数为奇数,再用
                      原数除以奇数便可得到偶数;
                      
    *****************************************/
    
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    
    using namespace std;
    long long n,a,b,c;
    
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
    
            scanf("%lld",&a);
            if(a%2)
                printf("Case %d: Impossible
    ",i);
            else{
                c=a;
                while(c%2==0)
                {
                 c=c/2;}
                b=a/c;
                printf("Case %d: %lld %lld
    ",i,c,b);
            }
        }
        return 0;
    }
    </span>


     

  • 相关阅读:
    年近30,朋友聚会都聊什么?
    2016世界最热门的编程语言与薪资揭秘
    程序员的春天来了,最美赏花旅游地十大攻略
    雄联盟工程师独家分享:如何使开发更有效率
    小偷被抓叫嚣:我不偷警察没饭吃
    3.7女生节:被程序员男友送的奇葩礼物宠哭了
    最适合程序员加班吃的6大营养美食
    谷歌汽车出误判曝光 6大奇葩科技更牛
    【程序员的爱情】彼岸花开谁又种下了执念
    分享10个免费或便宜的Photoshop替代工具
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254459.html
Copyright © 2011-2022 走看看