Probability One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 379    Accepted Submission(s): 293

Problem 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.
Note: There is a blank space before B.
 
Sample Input
37
38
0
 
Sample Output
1. odd 18
2. even 19
 
Source
 
Recommend
lcy   |   We have carefully selected several similar problems for you:  3351 3352 3353 3355 3356 
 
 //0MS    228K    366 B    G++
/* 题意:
题目看似很难.其实是水题.直接按它的过程计算一遍即可得出解 */
#include<stdio.h>
int main(void)
{
int n;
int k=;
while(scanf("%d",&n),n)
{
int odd=;
n*=;
if(n%){
odd=;
n=(n+)/;
}else n/=;
n*=;
n/=;
if(odd) printf("%d. odd %d\n",k++,n);
else printf("%d. even %d\n",k++,n);
}
return ;
}

hdu 3354 Probability One的更多相关文章

  1. HDU 2131 Probability

    http://acm.hdu.edu.cn/showproblem.php?pid=2131 Problem Description Mickey is interested in probabili ...

  2. HDU 4978 A simple probability problem

    A simple probability problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K ( ...

  3. HDU 6595 Everything Is Generated In Equal Probability (期望dp,线性推导)

    Everything Is Generated In Equal Probability \[ Time Limit: 1000 ms\quad Memory Limit: 131072 kB \] ...

  4. hdu多校第二场 1005 (hdu6595) Everything Is Generated In Equal Probability

    题意: 给定一个N,随机从[1,N]里产生一个n,然后随机产生一个n个数的全排列,求出n的逆序数对的数量,加到cnt里,然后随机地取出这个全排列中的一个非连续子序列(注意这个子序列可以是原序列),再求 ...

  5. hdu 2955 01背包

    http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能 ...

  6. HDU 2955 Robberies 背包概率DP

    A - Robberies Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submi ...

  7. HDU 4816 Bathysphere(数学)(2013 Asia Regional Changchun)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4816 Problem Description The Bathysphere is a spheric ...

  8. HDU 3076:ssworld VS DDD(概率DP)

    http://acm.split.hdu.edu.cn/showproblem.php?pid=3076 ssworld VS DDD Problem Description   One day, s ...

  9. HDU 3853:LOOPS(概率DP)

    http://acm.split.hdu.edu.cn/showproblem.php?pid=3853 LOOPS Problem Description   Akemi Homura is a M ...

随机推荐

  1. Wordpress网站中添加百度统计代码

    百度统计是流量分析平台,帮助收集网站访问数据,提供流量趋势.来源分析.转化跟踪.页面热力图.访问流等多种统计分析服务,同时与百度搜索.百度推广.云服务无缝结合,为网站的精细化运营决策提供数据支持,进而 ...

  2. 前端vue项目部署到tomcat,一刷新报错404解决方法

    公司前端写的后台部署到tomcat webapps目录下后,无法进行刷新,一刷新就会报错404,自动跳的404页面.在网上查了下,官方说是HTML5 History 模式引发的问题,但是解决方案中,并 ...

  3. 《Python语言及其应用》学习笔记

    第二章 ========== 对象的类型决定了可以对它进行的操作.对象的类型还决定了它装着的数据是允许被修改的变量(可变的),还是不可被修改的常量(不可变的). Python是强类型的,你永远无法修改 ...

  4. 数据分析处理库Pandas——数据透视表

    数据 按指定的行列值显示 求和 按行求和 按列求和 数据 求平均 备注:按性别计算每个等级船票的平均价格. 备注:每个等级船舱中每种性别获救的平均值,也就是获救的比例. 备注:每种性别未成年人获救的平 ...

  5. POJ 2079 最大三角形面积(凸包)

    Triangle Description Given n distinct points on a plane, your task is to find the triangle that have ...

  6. 调整图像的亮度和对比度—opencv

    1.理论基础 两个参数  和  一般称作 增益 和 偏置 参数.我们往往用这两个参数来分别控制 对比度 和 亮度 . 你可以把  看成源图像像素,把  看成输出图像像素.这样一来,上面的式子就能写得更 ...

  7. centos安装Linux

    CentOS下安装Redis Redis是一种高级key-value数据库.它跟memcached类似,不过数据可以持久化,而且支持的数据类型很丰富.有字符串,链表,集 合和有序集合.支持在服务器端计 ...

  8. [CodeForces948C]Producing Snow(优先队列)

    Description 题目链接 Solution 将温度做一个前缀和,用一个优先队列依次处理一遍 思路还是很简单的 Code #include <cstdio> #include < ...

  9. 110Balanced Binary Tree

    问题:判断二叉树是否为平衡二叉树分析:树上的任意结点的左右子树高度差不超过1,则为平衡二叉树.         搜索递归,记录i结点的左子树高度h1和右子树高度h2,则i结点的高度为max(h1,h2 ...

  10. P2985 [USACO10FEB]吃巧克力Chocolate Eating

    P2985 [USACO10FEB]吃巧克力Chocolate Eating 题目描述 Bessie has received N (1 <= N <= 50,000) chocolate ...