北大ACM poj3994 Probability One

Probability One

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. 
   
2. Ask your friend to compute n₁ = 3 * n₀ and to tell you if n₁ is even or odd.
3. 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.
4. Now ask your friend to calculate n₃ = 3 * n₂.
5. 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.)
6. 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 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.

Sample Input

37 38 0

Sample Output

1. odd 18 2. even 19

一道水得没法再水的题。

#include<stdio.h>
main()
{
	int i=0,n;
	while(scanf("%d",&n),n)
	{
		i++;
		printf("%d. %s %d\n",i,n%2?"odd":"even",n/2);
	}
}