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.
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.)
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.
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
Recommend
1 //0MS 228K 366 B G++ 2 /* 3 4 题意： 5 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 6 7 */ 8 #include<stdio.h> 9 int main(void) 10 { 11 int n; 12 int k=1; 13 while(scanf("%d",&n),n) 14 { 15 int odd=0; 16 n*=3; 17 if(n%2){ 18 odd=1; 19 n=(n+1)/2; 20 }else n/=2; 21 n*=3; 22 n/=9; 23 if(odd) printf("%d. odd %d\n",k++,n); 24 else printf("%d. even %d\n",k++,n); 25 } 26 return 0; 27 }
