POJ3994 HDU3354 UVALive4736 Probability One水题

Posted 2021-02-02 tigerisland45

Probability One
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 1944 Accepted: 1335

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:

Ask your friend to compute n1 = 3 * n0 and to tell you if n1 is even or odd.

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.

Now ask your friend to calculate n3 = 3 * n2.

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.)

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.

Sample Input

37
38
0

Sample Output

1. odd 18
2. even 19

Source

anarc 2009

问题链接：POJ3994 HDU3354 UVALive4736 Probability One
问题简述：（略）
问题分析：
????这个题是一个水题。然而，计算方法有两种，一是按照题意计算；二是进行计算公式推导后计算，n4=n3/9=3n2/9=3n1/2/9=33n0/2/9=n0/2，另外输出的奇偶是n1的奇偶，而n1=3*n0，所以n1的奇偶同n0的奇偶。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C语言程序（推导计算公式计算）如下：

/* POJ3994 HDU3354 UVALive4736 Probability One */

#include <stdio.h>

int main(void)
{
    int caseno = 0, n;
    while(scanf("%d", &n)  != EOF && n)
        printf("%d. %s %d
", ++caseno, n & 1 ? "odd" : "even", n / 2);

    return 0;
}

AC的C语言程序（按照题意计算）如下：

/* POJ3994 HDU3354 UVALive4736 Probability One */

#include <stdio.h>

int main(void)
{
    int caseno = 0, n, odd;
    while(scanf("%d", &n)  != EOF && n) {
        n *= 3;
        odd = n & 1;
        n >>= 1;
        n *= 3;
        n /= 9;
        printf("%d. %s %d
", ++caseno, odd ? "odd" : "even", n);
    }
    return 0;
}