HDU_3354 Probability One

Probability One

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 n ₀.
Then:
1. Ask your friend to compute n ₁ = 3 * n ₀ and to tell you if n1 is even or odd.
2. If n ₁ is even, ask your friend to compute n ₂ = n ₁/2. If, otherwise, n ₁ was odd then let your friend compute n ₂ = (n ₁ + 1)/2.
3. Now ask your friend to calculate n ₃ = 3 * n ₂.
4. Ask your friend to tell tell you the result of n ₄ = n ₃/9. (n ₄ 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 n ₀ = 2 * n ₄ if n ₁ was even, or n ₀ = 2 * n ₄ + 1 otherwise.
Here’s an example that you can follow: If n ₀ = 37, then n ₁ = 111 which is odd. Now we can calculate n ₂ = 56, n ₃= 168, and n ₄ = 18, which is what your friend will tell you. Doing the calculation 2 × n ₄ + 1 = 37 reveals n ₀.

 

 

Input

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

 

 

//位运算水题 
#include<stdio.h>
int main()
{
    int count=0,n;
    while(scanf("%d",&n),n!=0)
    {
        printf("%d. ",++count);
        if(n&1)  printf("odd ");   //奇数 
        else printf("even ");
        if(n&1) printf("%d\n",(n-1)>>1);  //  除以2，右移一位 
        else  printf("%d\n",n>>1);
    }
    return 0;
}