hdu 3354 Probability One

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

2009 ANARC

 

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//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=1;
    while(scanf("%d",&n),n)
    {
        int odd=0;
        n*=3;
        if(n%2){
            odd=1;
            n=(n+1)/2;
        }else n/=2;
        n*=3;
        n/=9;
        if(odd) printf("%d. odd %d
",k++,n);
        else printf("%d. even %d
",k++,n);
    }
    return 0;
}