Probability One
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 1674   Accepted: 1151

Description

Number guessing is a popular game between elementary-school kids. Teachers encourage pupils to play the game as it enhances their arithmetic skills, logical thinking, and following-up simple procedures. We think that, most probably, you too will master in few
minutes. Here’s one example of how you too can play this game: Ask a friend to think of a number, let’s call it n0. Then:

  1. Ask your friend to compute n1 = 3 * n0 and to tell you if n1 is even or odd.
  2. If n1 is even, ask your friend to compute n2 = n1/2. If, otherwise, n1 was odd then let your friend compute n2 = (n1 + 1)/2.
  3. Now ask your friend to calculate n3 = 3 * n2.
  4. Ask your friend to tell tell you the result of n4 = n3/9. (n4 is the quotient of the division operation. In computer lingo, ’/’ is the integer-division operator.)
  5. Now you can simply reveal the original number by calculating n0 = 2 * n4 if n1 was even, or n0 = 2 * n4 + 1 otherwise.

Here’s an example that you can follow: If n0 = 37, then n1 = 111 which is odd. Now we can calculate n2 = 56, n3 = 168, and n4 = 18, which is what your friend will tell you. Doing the calculation 2 * n4 +
1 = 37 reveals n0.

Input

Your program will be tested on one or more test cases. Each test case is made of a single positive number (0 < n0 < 1,000,000). 

The last line of the input file has a single zero (which is not part of the test cases.)

Output

For each test case, print the following line: 

k. B Q 

Where k is the test case number (starting at one,) B is either ’even’ or ’odd’ (without the quotes) depending on your friend’s answer in step 1. Q is your friend’s answer to step 4.

Sample Input

37
38
0

Sample Output

1. odd 18
2. even 19

把整个过程换算完了就是把原数除以2。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std; int main()
{
int num,i=1;
while(cin>>num)
{
if(num==0)
break;
cout<<i<<". ";
i++;
if(num%2)
cout<<"odd ";
else
cout<<"even ";
cout<<num/2<<endl;
}
return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

POJ 3994:Probability One的更多相关文章

  1. POJ 3321:Apple Tree + HDU 3887:Counting Offspring(DFS序+树状数组)

    http://poj.org/problem?id=3321 http://acm.hdu.edu.cn/showproblem.php?pid=3887 POJ 3321: 题意:给出一棵根节点为1 ...

  2. POJ 3252:Round Numbers

    POJ 3252:Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10099 Accepted: 36 ...

  3. POJ 1459:Power Network(最大流)

    http://poj.org/problem?id=1459 题意:有np个发电站,nc个消费者,m条边,边有容量限制,发电站有产能上限,消费者有需求上限问最大流量. 思路:S和发电站相连,边权是产能 ...

  4. POJ 3436:ACM Computer Factory(最大流记录路径)

    http://poj.org/problem?id=3436 题意:题意很难懂.给出P N.接下来N行代表N个机器,每一行有2*P+1个数字 第一个数代表容量,第2~P+1个数代表输入,第P+2到2* ...

  5. POJ 2195:Going Home(最小费用最大流)

    http://poj.org/problem?id=2195 题意:有一个地图里面有N个人和N个家,每走一格的花费是1,问让这N个人分别到这N个家的最小花费是多少. 思路:通过这个题目学了最小费用最大 ...

  6. POJ 3281:Dining(最大流)

    http://poj.org/problem?id=3281 题意:有n头牛,f种食物,d种饮料,每头牛有fnum种喜欢的食物,dnum种喜欢的饮料,每种食物如果给一头牛吃了,那么另一个牛就不能吃这种 ...

  7. POJ 3580:SuperMemo(Splay)

    http://poj.org/problem?id=3580 题意:有6种操作,其中有两种之前没做过,就是Revolve操作和Min操作.Revolve一开始想着一个一个删一个一个插,觉得太暴力了,后 ...

  8. POJ 3237:Tree(树链剖分)

    http://poj.org/problem?id=3237 题意:树链剖分.操作有三种:改变一条边的边权,将 a 到 b 的每条边的边权都翻转(即 w[i] = -w[i]),询问 a 到 b 的最 ...

  9. POJ 2763:Housewife Wind(树链剖分)

    http://poj.org/problem?id=2763 题意:给出 n 个点, n-1 条带权边, 询问是询问 s 到 v 的权值, 修改是修改存储时候的第 i 条边的权值. 思路:树链剖分之修 ...

随机推荐

  1. VueCli3 项目结构和具体作用

  2. Vue二次精度随笔(2)

    1.vue中数组更新是否会引起视图刷新的研究 (1)vue中修改数组可以引起视图刷新的方法 (2)不会引起数组刷新的方法,需要手动进行赋值 (3)有些数组的变化是不能够引起视图的刷新的,一个是修改数组 ...

  3. golang的传值调用和传引用调用

    传值还是传引用 调用函数时, 传入的参数的 传值 还是 传引用, 几乎是每种编程语言都会关注的问题. 最近在使用 golang 的时候, 由于 传值 和 传引用 的方式没有弄清楚, 导致了 BUG. ...

  4. Git闪退问题

    打开Git 会一闪而过.并出现一个错误的日志文件.自己尝试安装了几个不同的版本Git还是解决不了问题.后来自己在网上找了一些办法,并进行总结 1. 进入git目录下的bin目录执行rebase -b ...

  5. HDU3306-Another kind of Fibonacci(矩阵构造)

    Another kind of Fibonacci Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Jav ...

  6. PHP使用ElasticSearch做搜索

    PHP 使用 ElasticSearch 做搜索 https://blog.csdn.net/zhanghao143lina/article/details/80280321 https://www. ...

  7. 提升Essay写作说服力,需要注意这几个细节

    很多留学生对于essay写作都不精通,能够勉强通过就不错了.那么Essay写作到底该怎么提分呢?可以从哪些方面入手?小编给同学们指几条路,相信可以帮到大家. 在有说服力的Essay中总结您的论点.尽管 ...

  8. thinkphp配置到二级目录,不配置到根目录,访问除首页的其他路径都是404报错

    1.在nginx的配置里面,进行重定向 vi /etc/nginx/conf.d/default.conf 2.进入编辑 location /thinkphp/public { if (!-e $re ...

  9. Codeforces 448C:Painting Fence 刷栅栏 超级好玩的一道题目

    C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard in ...

  10. Si7006主要面向传统上使用的分立RH / T传感器的低精度的应用

    Silicon Labs的Si7006 / 13/20/21个I 2 C相对湿度及温度传感器结合充分工厂校准湿度和温度传感器元件与模拟-数字转换器,信号处理和一个I 2 C主机接口.专利使用业界标准低 ...