POJ 3994:Probability One
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1674 | Accepted: 1151 |
Description
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:
- Ask your friend to compute n1 = 3 * n0 and to tell you if n1 is even or odd.
- 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.
- Now ask your friend to calculate n3 = 3 * n2.
- 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.)
- 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
The last line of the input file has a single zero (which is not part of the test cases.)
Output
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
- 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 ...
- POJ 3252:Round Numbers
POJ 3252:Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10099 Accepted: 36 ...
- 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 ...
- 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* ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- Android APK反编译就这么简单 详解(附图)--转
转自:http://blog.csdn.net/vipzjyno1/article/details/21039349/ 在学习Android开发的过程你,你往往会去借鉴别人的应用是怎么开发的,那些漂亮 ...
- 五、生产者消费者模型_ThreadLocal
1.生产者消费者模型作用和示例如下:1)通过平衡生产者的生产能力和消费者的消费能力来提升整个系统的运行效率 ,这是生产者消费者模型最重要的作用2)解耦,这是生产者消费者模型附带的作用,解耦意味着生产者 ...
- tomcat安装apr报错解决
参考http://www.cnblogs.com/nuccch/p/7598361.html 1.no c complie 安装gcc解决 2.rm: cannot remove `libtoolT' ...
- L/SQL Developer 和 instantclient客户端安装配置
PL/SQL Developer 和 instantclient客户端安装配置(图文) 一: PL/SQL Developer 安装 下载安装文件安装,我这里的版本号是PLSQL7.1.4.1391, ...
- 关于pgsql 几个操作符的效率测试比较
关于pgsql 几个操作符的效率测试比较1. json::->> 和 ->> 测试方法:单次运行100次,运行10个单次取平均时间.测试结果:->> 效率高 5% ...
- 二 Hibernate 改写学生管理系统的业务功能
public class StudentDaoImpl implements StudentDao { @Override /** * 查询所有学生 * * @throws SQLException ...
- 攻防世界web新手练习区(2)
弱认证:http://111.198.29.45:43769/ 打开是这个页面: 用户名输入1,密码输入2,试试看.会提示你用户名为admin.接下来用burp对密码进行爆破,发现弱口令0123456 ...
- js左右选项移动
<!--网页代码--><div class="modal" id="modal-primary7"> <div class=&qu ...
- POJ 1979 Red and Black 四方向棋盘搜索
Red and Black Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 50913 Accepted: 27001 D ...
- java 寒假作业
寒假作业 现在小学的数学题目也不是那么好玩的. 看看这个寒假作业: □ + □ = □ □ - □ = □ □ × □ = □ □ ÷ □ = □ (如果显示不出来,可以参见[图1.jpg]) 每个方 ...