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 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

37380

 

 

Sample Output

1. odd 182. 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;}